- Sine is the ratio of the opposite leg to the hypotenuse. - Identify the right angle first so you can locate the hypotenuse.

1. Side \(z\) is the hypotenuse. 2. The side opposite \(\alpha\) is \(y\), so \(\sin(\alpha)=\frac{y}{z}\). 3. The side opposite \(\gamma\) is \(x\), so \(\sin(\gamma)=\frac{x}{z}\).

\(\sin(\alpha)=\frac{y}{z}\) \(\sin(\gamma)=\frac{x}{z}\)

- Find the side opposite the right angle. - For each acute angle, identify the leg that touches it but is not the hypotenuse.

1. Side \(p\) is opposite the right angle, so it is the hypotenuse. 2. The leg adjacent to \(\delta\) is \(r\), so \(\cos(\delta)=\frac{r}{p}\). 3. The leg adjacent to \(\epsilon\) is \(q\), so \(\cos(\epsilon)=\frac{q}{p}\).

\(\cos(\delta)=\frac{r}{p}\) \(\cos(\epsilon)=\frac{q}{p}\)

- Tangent uses the two legs, not the hypotenuse. - Identify the side opposite \(\mu\) first.

1. Tangent is the ratio of the opposite leg to the adjacent leg. 2. Relative to \(\mu\), \(h\) is opposite and \(g\) is adjacent, so \(\tan(\mu)=\frac{h}{g}\). 3. Relative to \(\nu\), \(g\) is opposite and \(h\) is adjacent, so \(\tan(\nu)=\frac{g}{h}\).

\(\tan(\mu)=\frac{h}{g}\) \(\tan(\nu)=\frac{g}{h}\)

- Identify the hypotenuse from the right-angle mark. - Decide which marked acute angle has side \(e\) as its adjacent leg.

1. The right-angle mark at \(A\) shows that \(f\) is the hypotenuse. 2. Cosine is adjacent leg over hypotenuse. 3. Side \(e\) is adjacent to angle \(\tau\), so \(\cos(\tau)=\frac{e}{f}\).

\(\tau\)

- Identify the hypotenuse from the right-angle mark. - Which marked acute angle is opposite side \(k\)?

1. The right-angle mark at \(K\) shows that \(m\) is the hypotenuse. 2. Sine is opposite leg over hypotenuse. 3. Side \(k\) is opposite angle \(\zeta\), so \(\sin(\zeta)=\frac{k}{m}\).

\(\zeta\)

- Find the vertex with the right-angle mark. - The hypotenuse does not touch that vertex.

1. The right angle is at \(B\). 2. The hypotenuse is the side opposite the right angle, \(AC\).

\(AC\)

- An opposite side does not touch the chosen angle. - Change the angle you are looking from, then name the leg again.

1. The leg across from \(A\) is \(BC\). 2. Leg \(BC\) touches angle \(B\), so it is adjacent to \(B\).

\(BC\) is opposite \(A\) and adjacent to \(B\).

- Identify the two legs; the hypotenuse is not used in tangent. - Look at the legs from each acute vertex in turn.

1. For angle \(A\), the opposite leg is \(BC\) and the adjacent leg is \(AC\), so \(\tan A=\frac{6}{8}=\frac{3}{4}\). 2. For angle \(B\), the opposite leg is \(AC\) and the adjacent leg is \(BC\), so \(\tan B=\frac{8}{6}=\frac{4}{3}\). 3. Changing the reference angle switches the opposite and adjacent legs.

\(\tan A=\frac{BC}{AC}=\frac{3}{4}\); \(\tan B=\frac{AC}{BC}=\frac{4}{3}\). The leg roles switch when the angle changes.

- Separate the tower's total height into the part above eye level and the eye height. - Which right-triangle sides are opposite and adjacent to the marked angle of elevation? - After using tangent, remember that the result is only the height above eye level.

1. Let \(h\) be the vertical distance from the observer's eye level to the top of the tower. The horizontal distance is \(32\,\text{m}\), so \(\tan(38^\circ)=\frac{h}{32}\). 2. Solve for \(h\): \(h=32\tan(38^\circ)\approx25.00\,\text{m}\). 3. Add the eye height shown in the diagram: \(25.00\,\text{m}+1.60\,\text{m}\approx26.60\,\text{m}\).

Approximately \(26.60\,\text{m}\)

- For each named angle, identify which leg is opposite and which is adjacent. - Substitute \(b=2a\) into each tangent ratio. - Compare the two tangent values after simplifying.

1. For \(\alpha=\angle A\), \(\tan(\alpha)=\frac{a}{b}\). For \(\beta=\angle B\), \(\tan(\beta)=\frac{b}{a}\). 2. Since \(b=2a\), \(\tan(\alpha)=\frac{a}{2a}=\frac{1}{2}=0.5\). 3. Also, \(\tan(\beta)=\frac{2a}{a}=2\).

\(\tan(\alpha)=0.5\) and \(\tan(\beta)=2\)

- Find the third angle in each triangle. - Think about the side lengths after an equilateral triangle is cut in half. - What happens to a fraction when its numerator and denominator are multiplied by the same number?

1. Each triangle has angles of \(90^\circ\), \(30^\circ\), and \(60^\circ\), so the triangles are similar by AA. 2. Let the equilateral triangle have side length \(c\). Cutting it in half creates a right triangle whose hypotenuse is \(c\) and whose side opposite the \(30^\circ\) angle is \(\frac{c}{2}\). 3. Thus, \(\frac{\text{opposite leg}}{\text{hypotenuse}}=\frac{c/2}{c}=\frac{1}{2}\). 4. Doubling all side lengths multiplies both parts of the ratio by \(2\), so \(\frac{2a}{2c}=\frac{a}{c}\).

a) The triangles are similar by AA because both have angles of \(90^\circ\), \(30^\circ\), and \(60^\circ\). b) The ratio is \(\frac{1}{2}\). c) Scaling multiplies both the numerator and denominator by the same factor, so the ratio does not change.

- How does percent grade compare the rise with horizontal run? - Which trigonometric ratio relates rise and horizontal run? - To work from a percent back to an angle, what operation reverses tangent?

1. A \(15\%\) grade means \(\frac{\text{rise}}{\text{run}}=0.15\). Since this ratio is \(\tan(\alpha)\), \(\alpha=\tan^{-1}(0.15)\approx8.53^\circ\). 2. For a \(10^\circ\) incline, the decimal grade is \(\tan(10^\circ)\approx0.17632698\). 3. Convert to percent: \(100\tan(10^\circ)\approx17.63\%\).

a) \(\alpha\approx8.53^\circ\) b) \(\approx17.63\%\)

- Which two sides are related by sine? - Rearrange the sine ratio to isolate the unknown side. - Interpret \(0.6\) as a fraction or percent.

1. By definition, \(\sin(\alpha)=\frac{a}{c}\), so \(a=c\sin(\alpha)\). 2. Substitute the given values: \(a=15\cdot 0.6=9\,\text{cm}\). 3. The value \(0.6\) means the opposite leg is \(60\%\) of the hypotenuse, or that their ratio is \(3:5\).

1. The opposite leg is \(9\,\text{cm}\). 2. The opposite leg is \(0.6\) times as long as the hypotenuse, or \(\frac{3}{5}\) of its length.

- How are the two acute angles in a right triangle related? - Which trigonometric ratio relates the side opposite \(\alpha\) to the hypotenuse? - Which ratio relates the two legs? - Make sure your calculator is in degree mode.

1. The acute angles are complementary, so \(\beta=90^\circ-35^\circ=55^\circ\). 2. Since \(\sin(\alpha)=\frac{a}{c}\), \(c=\frac{a}{\sin(\alpha)}=\frac{8.4}{\sin(35^\circ)}\approx 14.64\,\text{cm}\). 3. Since \(\tan(\alpha)=\frac{a}{b}\), \(b=\frac{a}{\tan(\alpha)}=\frac{8.4}{\tan(35^\circ)}\approx 12.00\,\text{cm}\).

\(\beta=55^\circ\), \(b\approx 12.00\,\text{cm}\), and \(c\approx 14.64\,\text{cm}\)

- Express the leg ratio as a fraction. - Which angle is opposite the shorter leg? - How are the two acute angles of a right triangle related?

1. The ratio \(a:b=2:3\) means \(\frac{a}{b}=\frac{2}{3}\). 2. Since \(\tan(\alpha)=\frac{a}{b}\), \(\alpha=\tan^{-1}\left(\frac{2}{3}\right)\approx33.69^\circ\). 3. The acute angles are complementary, so \(\beta=90^\circ-\alpha\approx56.31^\circ\).

\(\alpha\approx33.69^\circ\) and \(\beta\approx56.31^\circ\)

- Use sine to relate the opposite leg to the fixed hypotenuse. - Calculate the leg length for both angles. - Compare the second length with twice the first.

1. For \(\alpha=30^\circ\), the opposite leg is \(a_1=10\sin(30^\circ)=5\,\text{cm}\). 2. Doubling the angle gives \(60^\circ\). The new opposite leg is \(a_2=10\sin(60^\circ)=5\sqrt{3}\,\text{cm}\approx 8.66\,\text{cm}\). 3. Twice the original leg would be \(10\,\text{cm}\), not approximately \(8.66\,\text{cm}\). 4. Therefore, the claim is false. Sine is not proportional to the angle measure.

The claim is false. The opposite leg changes from \(5\,\text{cm}\) at \(30^\circ\) to \(5\sqrt{3}\,\text{cm}\approx 8.66\,\text{cm}\) at \(60^\circ\), not to \(10\,\text{cm}\).

- Substitute \(2a\) and \(2b\) into the tangent ratio. - Simplify the new ratio. - What happens to angles when a figure is scaled uniformly?

1. After doubling both legs, \(\tan(\alpha_{\text{new}})=\frac{2a}{2b}=\frac{a}{b}\). 2. The tangent ratio is unchanged, so the acute angle is unchanged: \(\alpha_{\text{new}}=\alpha\). 3. Geometrically, multiplying both legs by the same scale factor produces a triangle similar to the original. Corresponding angles in similar triangles are congruent.

Angle \(\alpha\) does not change. The ratio \(\frac{2a}{2b}\) equals \(\frac{a}{b}\), and the new triangle is a scaled copy of the original.

- Divide the isosceles triangle into two right triangles. - Which theorem finds a missing side when both legs are known? - Which trigonometric ratio compares the opposite and adjacent legs?

1. The altitude bisects the \(8.00\,\text{m}\) base, forming a right triangle with legs \(4.00\,\text{m}\) and \(3.00\,\text{m}\). 2. By the Pythagorean theorem, the rafter length is \(s=\sqrt{4.00^2+3.00^2}=5.00\,\text{m}\). 3. For the base angle, \(\tan(\alpha)=\frac{3.00}{4.00}\), so \(\alpha=\tan^{-1}(0.75)\approx 36.87^\circ\). 4. The gable area is \(A=\frac{1}{2}(8.00)(3.00)=12.00\,\text{m}^2\).

a) Each rafter is \(5.00\,\text{m}\) long. b) \(\alpha\approx 36.87^\circ\) c) The area is \(12.00\,\text{m}^2\).

- Identify the hypotenuse from the right angle. - For each acute angle, name its opposite and adjacent legs. - How are the two acute angles of a right triangle related?

1. The right angle is at \(C\), so \(w\) is the hypotenuse. 2. Relative to \(\epsilon\) at \(B\), \(v\) is opposite and \(u\) is adjacent. Thus, \(\sin(\epsilon)=\frac{v}{w}\). 3. Relative to \(\phi\) at \(A\), \(u\) is opposite and \(v\) is adjacent. Thus, \(\sin(\phi)=\frac{u}{w}\), \(\tan(\phi)=\frac{u}{v}\), and \(\cos(\phi)=\frac{v}{w}\). 4. The two acute angles in a right triangle add to \(90^\circ\).

a) \(\sin(\epsilon)=\frac{v}{w}\) b) \(\sin(\phi)=\frac{u}{w}\) c) \(\tan(\phi)=\frac{u}{v}\) d) \(\cos(\phi)=\frac{v}{w}\) e) \(\epsilon+\phi=90^\circ\)

- Decide whether each fraction is opposite over hypotenuse, adjacent over hypotenuse, or opposite over adjacent. - Match each ratio to the marked angle \(\delta\) or \(\eta\). - Remember that the same two side lengths can represent sine of one acute angle and cosine of the other.

1. The right angle is at \(A\), so \(f\) is the hypotenuse. 2. Relative to \(\delta\) at \(C\), \(d\) is opposite and \(e\) is adjacent. 3. Relative to \(\eta\) at \(B\), \(e\) is opposite. 4. Therefore, \(\cos(\delta)=\frac{e}{f}\), \(\sin(\eta)=\frac{e}{f}\), \(\tan(\delta)=\frac{d}{e}\), and \(\sin(\delta)=\frac{d}{f}\).

a) \(\cos(\delta)=\frac{e}{f}\) b) \(\sin(\eta)=\frac{e}{f}\) c) \(\tan(\delta)=\frac{d}{e}\) d) \(\sin(\delta)=\frac{d}{f}\)

- Read the right angle, side lengths, and acute-angle labels from the diagram. - Identify opposite, adjacent, and hypotenuse relative to \(\alpha\). - When you switch to \(\beta\), which two leg roles switch?

1. Relative to \(\alpha\), the diagram shows opposite leg \(12\), adjacent leg \(35\), and hypotenuse \(37\). 2. Thus, \(\sin(\alpha)=\frac{12}{37}\approx0.324\), \(\cos(\alpha)=\frac{35}{37}\approx0.946\), and \(\tan(\alpha)=\frac{12}{35}\approx0.343\). 3. Relative to \(\beta\), the opposite and adjacent legs switch roles. 4. Thus, \(\sin(\beta)=\frac{35}{37}\approx0.946\), \(\cos(\beta)=\frac{12}{37}\approx0.324\), and \(\tan(\beta)=\frac{35}{12}\approx2.917\).

a) \(\sin(\alpha)=\frac{12}{37}\approx0.324\), \(\cos(\alpha)=\frac{35}{37}\approx0.946\), and \(\tan(\alpha)=\frac{12}{35}\approx0.343\) b) \(\sin(\beta)=\frac{35}{37}\approx0.946\), \(\cos(\beta)=\frac{12}{37}\approx0.324\), and \(\tan(\beta)=\frac{35}{12}\approx2.917\)

- Read the marked angle and hypotenuse from the diagram. - Which unknown leg is opposite the marked angle, and which is adjacent? - Choose sine for one leg and cosine for the other.

1. The diagram shows a hypotenuse of \(14\,\text{cm}\) and an acute angle of \(32^\circ\). 2. Side \(x\) is opposite the angle, so \(x=14\sin(32^\circ)\approx7.42\,\text{cm}\). 3. Side \(y\) is adjacent to the angle, so \(y=14\cos(32^\circ)\approx11.87\,\text{cm}\).

\(x\approx7.42\,\text{cm}\) and \(y\approx11.87\,\text{cm}\)

- Read the string length and ground angle from the diagram. - Which side is the hypotenuse, and which side represents height? - Choose the trig ratio that relates the opposite side to the hypotenuse.

1. The diagram shows that the string is the hypotenuse and the kite's height \(h\) is opposite the marked angle. 2. Use sine: \(\sin(55^\circ)=\frac{h}{100}\). 3. Therefore, \(h=100\sin(55^\circ)\approx81.92\,\text{ft}\).

The kite is approximately \(81.92\,\text{ft}\) above the ground.

- Read the known leg and angle from the diagram. - Relative to the marked angle, identify the opposite leg, adjacent leg, and hypotenuse. - Use sine for \(u\) and tangent or the Pythagorean theorem for \(v\).

1. The diagram shows that the \(9\,\text{cm}\) leg is opposite the \(40^\circ\) angle. 2. Since \(\sin(40^\circ)=\frac{9}{u}\), \(u=\frac{9}{\sin(40^\circ)}\approx14.00\,\text{cm}\). 3. Since \(\tan(40^\circ)=\frac{9}{v}\), \(v=\frac{9}{\tan(40^\circ)}\approx10.73\,\text{cm}\).

\(u\approx14.00\,\text{cm}\) and \(v\approx10.73\,\text{cm}\)

- Read the rise and incline angle from the diagram. - Which side represents the ramp surface? - Which trigonometric ratio links the rise to the sloped side?

1. The ramp surface is the hypotenuse, and the vertical rise is opposite the marked angle. 2. Use sine: \(\sin(4^\circ)=\frac{4}{l}\). 3. Solve for \(l\): \(l=\frac{4}{\sin(4^\circ)}\approx57.34\,\text{ft}\).

\(l\approx57.34\,\text{ft}\)

- Identify the hypotenuse and the marked acute angle. - Which leg lies opposite that angle, and which lies next to it? - Choose between sine and cosine for each leg.

1. The diagram shows a hypotenuse of \(10\,\text{cm}\) and a marked angle of \(50^\circ\). 2. Side \(m\) is adjacent to the angle, so \(m=10\cos(50^\circ)\approx6.43\,\text{cm}\). 3. Side \(k\) is opposite the angle, so \(k=10\sin(50^\circ)\approx7.66\,\text{cm}\).

\(k\approx7.66\,\text{cm}\) and \(m\approx6.43\,\text{cm}\)

- Read the shadow length and sunlight angle from the diagram. - Identify the opposite and adjacent legs relative to that angle. - Choose the trigonometric ratio that uses those two legs.

1. The diagram shows that the shadow is adjacent to the marked ground angle and the pole's height is opposite. 2. Use tangent: \(\tan(35^\circ)=\frac{h}{25}\). 3. Therefore, \(h=25\tan(35^\circ)\approx17.51\,\text{ft}\).

The pole is approximately \(17.51\,\text{ft}\) tall.

- Read the two leg lengths from the diagram and identify which is opposite \(\alpha\). - Use the tangent ratio to determine \(\alpha\). - Use the relationship between the two acute angles of a right triangle for \(\beta\).

1. In the right triangle formed by the diagonal, the side opposite \(\alpha\) is \(8\,\text{cm}\) and the adjacent side is \(15\,\text{cm}\). 2. Thus, \(\tan(\alpha)=\frac{8}{15}\approx 0.533\). 3. Then \(\alpha=\tan^{-1}\left(\frac{8}{15}\right)\approx 28.1^\circ\). 4. The acute angles are complementary, so \(\beta=90^\circ-\alpha\approx 61.9^\circ\).

a) \(\tan(\alpha)=\frac{8}{15}\approx 0.533\) b) \(\alpha\approx 28.1^\circ\) and \(\beta\approx 61.9^\circ\)

- Identify which right triangle is used in each part. - Find the hypotenuse in that triangle. - Identify the opposite and adjacent legs relative to the given angle.

1. In right triangle \(ABC\), \(r\) is the hypotenuse. Relative to \(\alpha\), \(p\) is adjacent and \(q\) is opposite. Thus, \(\cos(\alpha)=\frac{p}{r}\) and \(\tan(\alpha)=\frac{q}{p}\). 2. In right triangle \(ACD\), \(t\) is the hypotenuse. Relative to \(\beta\), \(s\) is opposite and \(r\) is adjacent. Thus, \(\sin(\beta)=\frac{s}{t}\) and \(\cos(\beta)=\frac{r}{t}\).

a) \(\cos(\alpha)=\frac{p}{r}\) b) \(\sin(\beta)=\frac{s}{t}\) c) \(\tan(\alpha)=\frac{q}{p}\) d) \(\cos(\beta)=\frac{r}{t}\)

- Determine how sides \(g\) and \(i\) relate to each acute angle. - Check whether each question mark stands for an angle or a trigonometric function. - Use the same side ratio from the viewpoint of both acute angles when useful.

1. Side \(h\) is the hypotenuse. 2. Relative to \(\epsilon\), \(g\) is adjacent and \(i\) is opposite. Thus, \(\cos(\epsilon)=\frac{g}{h}\), \(\tan(\epsilon)=\frac{i}{g}\), and \(\sin(\epsilon)=\frac{i}{h}\). 3. Relative to \(\delta\), \(g\) is opposite and \(i\) is adjacent. Thus, \(\sin(\delta)=\frac{g}{h}\) and \(\cos(\delta)=\frac{i}{h}\).

a) \(\cos(\epsilon)=\frac{g}{h}\) b) \(\sin(\delta)=\frac{g}{h}\) c) \(\tan(\epsilon)=\frac{i}{g}\) d) \(\sin(\epsilon)=\frac{i}{h}\) e) \(\cos(\delta)=\frac{i}{h}\)

- Identify the hypotenuse and the given leg. - Which trigonometric ratio connects them to \(\beta\)? - Once one acute angle is known, find its complement.

1. The hypotenuse is \(11.2\,\text{cm}\), and the leg opposite \(\beta\) is \(6.5\,\text{cm}\). 2. Thus, \(\sin(\beta)=\frac{6.5}{11.2}\), so \(\beta=\sin^{-1}\left(\frac{6.5}{11.2}\right)\approx 35.5^\circ\). 3. The acute angles are complementary, so \(\alpha=90^\circ-35.5^\circ\approx 54.5^\circ\).

\(\alpha\approx 54.5^\circ\) and \(\beta\approx 35.5^\circ\)

- Use the angle sum to find the other acute angle. - Identify the opposite and adjacent legs relative to \(28^\circ\). - Use sine and cosine with the hypotenuse.

1. The acute angles are complementary, so \(\beta=90^\circ-28^\circ=62^\circ\). 2. Side \(a\) is opposite the \(28^\circ\) angle, so \(a=12\sin(28^\circ)\approx 5.63\,\text{cm}\). 3. Side \(b\) is adjacent to the \(28^\circ\) angle, so \(b=12\cos(28^\circ)\approx 10.60\,\text{cm}\).

\(a\approx 5.63\,\text{cm}\) \(b\approx 10.60\,\text{cm}\) \(\beta=62^\circ\)

- Read the side length and diagonal length from the figure. - Relative to \(\alpha\), decide which given length is adjacent and which is the hypotenuse. - Use the inverse of the corresponding trigonometric ratio.

1. The diagonal forms a right triangle in which the \(9\,\text{cm}\) side is adjacent to \(\alpha\) and the \(11\,\text{cm}\) diagonal is the hypotenuse. 2. Thus, \(\cos(\alpha)=\frac{9}{11}\). 3. Therefore, \(\alpha=\cos^{-1}\left(\frac{9}{11}\right)\approx 35.1^\circ\).

\(\alpha\approx 35.1^\circ\)

- Read the three side lengths from the diagram and identify the hypotenuse. - The opposite and adjacent legs change when you switch from \(\alpha\) to \(\beta\). - Compare \(\sin(\beta)\) with \(\cos(\alpha)\) after writing both ratios.

1. Side \(c\) is the hypotenuse. Relative to \(\alpha\), \(a\) is opposite and \(b\) is adjacent. 2. \(\sin(\alpha)=\frac{2}{5.2}\approx 0.385\). 3. Relative to \(\beta\), \(b\) is opposite, so \(\sin(\beta)=\frac{4.8}{5.2}\approx 0.923\). 4. Also, \(\cos(\alpha)=\frac{4.8}{5.2}\approx 0.923\).

\(\sin(\alpha)\approx 0.385\), \(\sin(\beta)\approx 0.923\), and \(\cos(\alpha)\approx 0.923\)

- Read the side lengths from the diagram and identify the hypotenuse from the right angle. - Determine opposite and adjacent legs separately for \(\sigma\) and \(\upsilon\). - Compare the first two ratios after writing them from the two angle viewpoints.

1. Side \(t\) is the hypotenuse. Relative to \(\sigma\), \(s\) is opposite and \(u\) is adjacent. 2. Thus, \(\sin(\sigma)=\frac{4}{4.1}\approx 0.976\). 3. Relative to \(\upsilon\), \(s\) is adjacent, so \(\cos(\upsilon)=\frac{4}{4.1}\approx 0.976\). 4. Finally, \(\tan(\sigma)=\frac{4}{0.9}\approx 4.444\).

\(\sin(\sigma)\approx 0.976\), \(\cos(\upsilon)\approx 0.976\), and \(\tan(\sigma)\approx 4.444\)

- Identify the hypotenuse first. - Determine the opposite and adjacent legs separately for each acute angle. - Apply the right-triangle definitions of sine, cosine, and tangent.

1. Side \(z\) is opposite the right angle, so it is the hypotenuse. 2. Relative to \(\alpha\), \(x\) is opposite and \(y\) is adjacent. Therefore, \(\sin(\alpha)=\frac{x}{z}\), \(\cos(\alpha)=\frac{y}{z}\), and \(\tan(\alpha)=\frac{x}{y}\). 3. Relative to \(\beta\), the legs switch roles. Therefore, \(\sin(\beta)=\frac{y}{z}\), \(\cos(\beta)=\frac{x}{z}\), and \(\tan(\beta)=\frac{y}{x}\).

\(\sin(\alpha)=\frac{x}{z}\), \(\cos(\alpha)=\frac{y}{z}\), and \(\tan(\alpha)=\frac{x}{y}\) \(\sin(\beta)=\frac{y}{z}\), \(\cos(\beta)=\frac{x}{z}\), and \(\tan(\beta)=\frac{y}{x}\)

- Identify the side adjacent to the marked angle and the hypotenuse. - Choose the right-triangle ratio that relates exactly those two sides. - Keep the trigonometric expression intact until the final decimal evaluation.

1. Relative to the marked \(41^\circ\) angle, \(AB\) is the adjacent leg and \(AC=x\) is the hypotenuse. 2. Therefore, \(\cos(41^\circ)=\frac{5}{x}\). 3. Solving gives \(x=\frac{5}{\cos(41^\circ)}\,\text{cm}\approx6.63\,\text{cm}\).

\(x=\frac{5}{\cos(41^\circ)}\,\text{cm}\approx6.63\,\text{cm}\)

- Focus on the smaller right triangle containing the displayed side length and angle. - Relative to the marked angle, decide which unknown is opposite and which is adjacent. - Use the hypotenuse shown in that smaller triangle with two right-triangle ratios.

1. In right triangle \(BCD\), \(BC=10\,\text{cm}\) is the hypotenuse. 2. The altitude \(x=CD\) is opposite \(35^\circ\), so \(x=10\sin(35^\circ)\,\text{cm}\approx 5.74\,\text{cm}\). 3. Segment \(y=BD\) is adjacent to \(35^\circ\), so \(y=10\cos(35^\circ)\,\text{cm}\approx 8.19\,\text{cm}\).

\(x\approx 5.74\,\text{cm}\) and \(y\approx 8.19\,\text{cm}\)

- Focus on the smaller right triangle that contains \(y\). - Read its hypotenuse length and acute angle from the diagram. - Choose the ratio that compares the adjacent leg with the hypotenuse.

1. Because \(CD\) is an altitude, \(\triangle ADC\) is a right triangle with hypotenuse \(AC\). 2. Relative to the marked \(40^\circ\) angle, \(AD\) is the adjacent leg. Therefore, \(\cos(40^\circ)=\frac{AD}{AC}=\frac{y}{9}\). 3. Solving for \(y\), \(y=9\cos(40^\circ)\approx 6.89\,\text{cm}\).

\(y\approx 6.89\,\text{cm}\)

- What is the sum of the two acute angles in a right triangle? - Pick one leg and describe its role relative to each acute angle. - Compare the definitions of sine and cosine after identifying that leg.

1. The acute angles of a right triangle are complementary, so \(\theta+\phi=90^\circ\). 2. Choose the leg opposite \(\theta\). That same leg is adjacent to \(\phi\). 3. Both ratios use the same hypotenuse, so \(\sin(\theta)=\frac{\text{opposite to }\theta}{\text{hypotenuse}}\) and \(\cos(\phi)=\frac{\text{adjacent to }\phi}{\text{hypotenuse}}\) are the same ratio. 4. Therefore, \(\sin(\theta)=\cos(\phi)=\cos(90^\circ-\theta)\).

The leg opposite \(\theta\) is the same leg that is adjacent to the complementary angle \(\phi\), and both ratios use the same hypotenuse. Therefore, \(\sin(\theta)=\cos(\phi)\).

- Find the side opposite the right angle first. - Now look from vertex \(A\). Which leg lies across from this angle, and which touches it?

1. The hypotenuse is \(AB\). For angle \(A\), \(BC\) is opposite and \(AC\) is adjacent. 2. Thus \(\sin A=\frac{BC}{AB}=\frac{5}{13}\) and \(\cos A=\frac{AC}{AB}=\frac{12}{13}\).

\(\sin A=\frac{BC}{AB}=\frac{5}{13}\); \(\cos A=\frac{AC}{AB}=\frac{12}{13}\).

- Compare the same two types of leg in each triangle. - How do multiplication and division interact when both lengths change by the same factor?

1. The ratios are \(\frac{3}{4}\) and \(\frac{6}{8}=\frac{3}{4}\). 2. Tangent is opposite divided by adjacent, so it is unchanged in these similar triangles. 3. For any scale factor \(k>0\), \(\frac{3k}{4k}=\frac{3}{4}\); multiplying both legs by the same positive factor leaves their ratio unchanged.

Both tangent values are \(\frac{3}{4}\). In general \(\frac{3k}{4k}=\frac{3}{4}\) for \(k>0\), so enlarging or reducing the triangle does not change this tangent.

- How much do both acute angles add to in a right triangle? - Imagine viewing the same legs from the two different acute vertices.

1. The two acute angles sum to \(90^\circ\), so \(B=55^\circ\). 2. The opposite leg for \(35^\circ\) is the adjacent leg for \(B\), giving \(\sin35^\circ=\cos55^\circ\). 3. The other leg switches roles, giving \(\cos35^\circ=\sin55^\circ\).

\(B=55^\circ\); \(\sin35^\circ=\cos55^\circ\) and \(\cos35^\circ=\sin55^\circ\), because the same two legs swap opposite/adjacent roles.

- Which two sides appear in a sine ratio? - Which operation undoes division by the hypotenuse?

1. Sine relates the opposite leg \(x\) to the hypotenuse: \(\sin37^\circ=\frac{x}{14}\). 2. Thus \(x=14\sin37^\circ\approx8.4\,\text{cm}\).

\(\sin37^\circ=\frac{x}{14}\), so \(x\approx8.4\,\text{cm}\).

- Which ratio compares the adjacent leg to the hypotenuse? - If the unknown is in the denominator, what equation will isolate it?

1. Cosine relates the adjacent leg to the hypotenuse: \(\cos47^\circ=\frac{5}{h}\). 2. Hence \(h=\frac{5}{\cos47^\circ}\approx7.33\,\text{in}\).

\(\cos47^\circ=\frac{5}{h}\), so \(h\approx7.33\,\text{in}\).

- Which trigonometric ratio uses only the two legs? - Use the inverse operation to recover the angle from that ratio.

1. The ratio is \(\tan\theta=\frac{7}{9}\). 2. Use inverse tangent: \(\theta=\tan^{-1}\!\left(\frac{7}{9}\right)\approx37.9^\circ\).

\(\theta\approx37.9^\circ\)

- Compare the lengths of a leg and the hypotenuse. - Could opposite divided by hypotenuse exceed \(1\)?

1. In a right triangle, sine is opposite leg divided by hypotenuse. 2. The hypotenuse is longer than either leg, so this quotient is less than \(1\). 3. Therefore \(1.12\) is impossible.

No. For an acute angle, \(0<\sin\theta<1\).

- Comparing just the rises is not enough when the runs differ. - Which ratio expresses the angle's steepness for each ramp?

1. The tangent ratios are \(\frac{7}{12}\approx0.583\) for A and \(\frac{5}{8}=0.625\) for B. 2. Tangent increases as an acute angle increases, so ramp B has the larger incline angle even though its rise is smaller.

Ramp A: \(\frac{7}{12}\approx0.583\); ramp B: \(\frac{5}{8}=0.625\). Ramp B is steeper.

- Identify the two legs relative to the \(28^\circ\) angle. - Which trigonometric ratio relates those two legs? - Solve your equation for \(h\), then round at the end.

The side labeled \(h\) is opposite the \(28^\circ\) angle, and the 45 ft side is adjacent. \(\tan(28^\circ)=\frac{h}{45}\). Thus \(h=45\tan(28^\circ)\approx23.9\,\text{ft}\).

\(23.9\,\text{ft}\)

- Which ratio connects a path's vertical rise to its horizontal run? - Calculate each rise before comparing how the values change. - Do not assume that doubling an angle doubles its tangent.

1. For each path, \(\tan\theta=\frac{h}{10}\), so \(h=10\tan\theta\). 2. The rises are \(10\tan20^\circ\approx3.6\,\text{ft}\), \(10\tan40^\circ\approx8.4\,\text{ft}\), and \(10\tan60^\circ\approx17.3\,\text{ft}\). 3. The angles increase by \(20^\circ\) each time, but the rises increase by about \(4.8\,\text{ft}\) and \(8.9\,\text{ft}\). The increases in rise are not equal.

The rises are about \(3.6\,\text{ft}\), \(8.4\,\text{ft}\), and \(17.3\,\text{ft}\). No: two equal \(20^\circ\) increases cause unequal increases in rise.

- Which ratio connects the rise and the fixed horizontal run? - Compare the two calculated heights before considering even steeper angles. - Why can't a right triangle with a positive horizontal leg have a \(90^\circ\) angle of elevation?

1. With fixed run \(10\,\text{ft}\), \(h=10\tan\theta\). 2. At \(80^\circ\), \(h\approx10\cdot5.6713=56.7\,\text{ft}\), about \(57\,\text{ft}\). 3. At \(85^\circ\), \(h\approx10\cdot11.4301=114.3\,\text{ft}\), about \(114\,\text{ft}\). 4. Near \(90^\circ\), tangent increases without bound because the opposite leg becomes large relative to a fixed adjacent leg; at exactly \(90^\circ\), a fixed positive horizontal run cannot form that angle.

About \(57\,\text{ft}\) and \(114\,\text{ft}\). The height grows very large as the angle approaches \(90^\circ\); there is no finite height for exactly \(90^\circ\) with a positive run.

- Which side of the right triangle represents the sloped surface? - Express both lengths in the same unit. - With the same rise, would a shallower ramp be longer or shorter?

1. Convert the rise to meters: \(45\,\text{cm}=0.45\,\text{m}\). The sloped surface is the hypotenuse \(L\). 2. The shortest ramp uses the largest permitted angle, so \(\sin(3.5^\circ)=\frac{0.45}{L}\). 3. Thus, \(L=\frac{0.45}{\sin(3.5^\circ)}\approx7.37118\,\text{m}\). 4. Round up to the next hundredth to keep the incline at or below the design limit: \(L=7.38\,\text{m}\).

At least \(7.38\,\text{m}\) of sloped ramp surface.

- Model the morning and noon situations as two right triangles with the same height. - Which trigonometric ratio relates a shadow's length to the sun's angle? - Find each shadow length before taking the difference.

1. Let \(s_1\) be the morning shadow length. Then \(\tan(22^\circ)=\frac{42}{s_1}\), so \(s_1=\frac{42}{\tan(22^\circ)}\approx103.95\,\text{m}\). 2. Let \(s_2\) be the noon shadow length. Then \(\tan(58^\circ)=\frac{42}{s_2}\), so \(s_2=\frac{42}{\tan(58^\circ)}\approx26.24\,\text{m}\). 3. Subtract the unrounded values: \(s_1-s_2\approx77.71\,\text{m}\).

\(\approx77.71\,\text{m}\)

- Use tangent to find an incline angle from the rise and horizontal run. - Convert Route B's map length to its actual horizontal run first. - Which side of each right triangle represents the full cable-car route?

1. For Route A, \(\tan(\alpha)=\frac{450}{1200}\), so \(\alpha\approx20.56^\circ\). 2. Its length is \(L_A=\sqrt{1200^2+450^2}\approx1281.60\,\text{m}\). 3. The horizontal run of Route B is \(2.5\,\text{cm}\cdot50{,}000=125{,}000\,\text{cm}=1250\,\text{m}\). 4. For Route B, \(\tan(\beta)=\frac{450}{1250}\), so \(\beta\approx19.80^\circ\). 5. Its length is \(L_B=\sqrt{1250^2+450^2}\approx1328.53\,\text{m}\). 6. Since \(L_B>L_A\), Route B is longer.

a) Route A: \(\approx20.56^\circ\). Route B: \(\approx19.80^\circ\). b) Route B is longer: \(L_B\approx1328.53\,\text{m}\) versus \(L_A\approx1281.60\,\text{m}\).

- For \(\alpha=\angle A\), which leg is opposite \(\alpha\)? - Find the missing leg with the Pythagorean theorem. - Compute the area using the two legs and again using the hypotenuse as the base.

1. Since \(\sin(\alpha)=\frac{a}{c}=\frac{5}{13}\), \(\alpha=\sin^{-1}\left(\frac{5}{13}\right)\approx22.62^\circ\). 2. Find the other leg: \(b=\sqrt{13^2-5^2}=12\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot5\cdot12=30\,\text{cm}^2\). 4. Also, \(A=\frac{1}{2}\cdot13\cdot h_c\). Therefore, \(h_c=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\).

a) \(\alpha\approx22.62^\circ\) b) \(h_c=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\)

- Think of simple side lengths whose ratio of opposite leg to hypotenuse is \(0.6\). - Use the Pythagorean theorem to find the third side. - What is true about right triangles that share the same acute angle?

1. Model \(\sin(\alpha)=0.6=\frac{6}{10}\) with an opposite leg of \(6\) and a hypotenuse of \(10\). 2. By the Pythagorean theorem, the adjacent leg is \(\sqrt{10^2-6^2}=8\). 3. Therefore, \(\cos(\alpha)=\frac{8}{10}=0.8\) and \(\tan(\alpha)=\frac{6}{8}=0.75\). 4. Any two right triangles with the same acute angle \(\alpha\) are similar by AA. Corresponding side ratios are therefore constant.

a) \(\cos(\alpha)=0.8\) and \(\tan(\alpha)=0.75\) b) All right triangles with angle \(\alpha\) are similar by AA, so their corresponding side ratios are equal.

- Use one of the smaller right triangles formed by the altitude. - What geometric-mean relationship connects the altitude and the two parts of the hypotenuse? - Express the needed lengths in terms of \(q\).

1. From the similar right triangles formed by the altitude, \(CD\) is the geometric mean of the two hypotenuse segments: \(CD^2=pq\). 2. Substitute \(p=4q\): \(CD^2=(4q)(q)=4q^2\), so \(CD=2q\). 3. In right triangle \(ACD\), \(\tan(\alpha)=\frac{CD}{AD}=\frac{2q}{4q}=\frac{1}{2}=0.5\).

\(\tan(\alpha)=0.5\)

- Imagine decreasing the acute angle while keeping the hypotenuse fixed. What happens to the height of the triangle? - Which side is always longest in a right triangle? - What is true about a fraction whose numerator is smaller than its denominator?

1. As \(\alpha\) decreases while the hypotenuse stays fixed, the opposite leg \(a\) becomes shorter. 2. As \(\alpha\) approaches \(0^\circ\), \(a\) approaches \(0\), so \(\frac{a}{c}\) approaches \(0\). 3. The hypotenuse is opposite the right angle and is always the longest side of a right triangle. 4. Therefore, \(a<c\), which means \(\frac{a}{c}<1\).

a) The ratio decreases and approaches \(0\). b) The hypotenuse is always longer than either leg, so \(a<c\) and therefore \(\frac{a}{c}<1\).

- Why can the sun's rays be treated as making the same angle with the ground at both poles? - In similar figures, corresponding side lengths have equal ratios. - Identify the opposite and adjacent legs relative to the angle at the ground.

1. The sun's rays are effectively parallel, so both triangles have the same angle of elevation. Each pole is perpendicular to the ground, so both triangles also have a right angle. Therefore, the triangles are similar by AA similarity. 2. Let \(x\) be the shadow length of Pole 2. Corresponding sides are proportional: \(\frac{1.6}{2.0}=\frac{2.4}{x}\). 3. Solve: \(x=\frac{2.4\cdot 2.0}{1.6}=3.0\,\text{m}\). 4. The pole height is opposite the angle of elevation, and the shadow length is adjacent to it. Therefore, \(\frac{\text{pole height}}{\text{shadow length}}=\frac{\text{opposite}}{\text{adjacent}}=\tan(\alpha)\).

a) The triangles share equal angles of elevation and both contain a right angle, so they are similar by AA similarity. b) The shadow of Pole 2 is \(3.0\,\text{m}\) long. c) The ratio is tangent because it compares the opposite leg, the pole height, with the adjacent leg, the shadow length.

- Model the roof's side view as a right triangle. - Which ratio connects the incline angle with the horizontal and vertical legs? - Which side of the triangle is the full length of a rafter?

1. Let \(h\) be the vertical change. Then \(\tan(20^\circ)=\frac{h}{4.50}\), so \(h=4.50\tan(20^\circ)\approx1.64\,\text{m}\). 2. Let \(L\) be the rafter length. Since the rafter is the hypotenuse, \(\cos(20^\circ)=\frac{4.50}{L}\). Thus, \(L=\frac{4.50}{\cos(20^\circ)}\approx4.79\,\text{m}\). 3. The grade is \(100\tan(20^\circ)\%\approx36.40\%\). 4. Since the grade is below \(40\%\), the design meets the project's limit.

a) \(\approx1.64\,\text{m}\) b) \(\approx4.79\,\text{m}\) c) Yes. The grade is \(\approx36.40\%\), below \(40\%\).

- Use the definition of sine in a right triangle. - Choose side lengths whose ratio is exactly \(0.8\). - Use the Pythagorean theorem to find the third side. - Which inverse trigonometric function gives an angle from its sine?

1. Since \(\sin(\alpha)=\frac{\text{opposite leg}}{\text{hypotenuse}}=0.8=\frac{4}{5}\), choose an opposite leg of \(4\,\text{cm}\) and a hypotenuse of \(5\,\text{cm}\). 2. The adjacent leg is \(\sqrt{5^2-4^2}=3\,\text{cm}\), so the side lengths are \(3\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\). 3. Using the inverse sine, \(\alpha=\sin^{-1}(0.8)\approx 53.13^\circ\).

1. One suitable triangle has side lengths \(3\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\), with the \(4\,\text{cm}\) leg opposite \(\alpha\). 2. \(\alpha\approx 53.13^\circ\)

- Use the two legs as the base and height in the area formula. - Once both legs are known, how can you find the hypotenuse? - Match the angle opposite leg \(a\) with the right leg ratio.

1. Use the area formula: \(24=\frac{1}{2}\cdot a\cdot8\), so \(a=6\,\text{cm}\). 2. By the Pythagorean theorem, \(c=\sqrt{6^2+8^2}=10\,\text{cm}\). 3. Since \(\tan(\alpha)=\frac{6}{8}=0.75\), \(\alpha=\tan^{-1}(0.75)\approx36.87^\circ\). 4. Then \(\beta=90^\circ-\alpha\approx53.13^\circ\).

\(c=10\,\text{cm}\), \(\alpha\approx36.87^\circ\), and \(\beta\approx53.13^\circ\)

- What does the map scale tell you about the actual horizontal distance? - Model the trail, horizontal distance, and elevation gain as a right triangle. - Which method gives the trail length, and which ratio finds its incline angle?

1. Convert the map length to the horizontal ground distance: \(5.4\,\text{cm}\cdot10{,}000=54{,}000\,\text{cm}=540\,\text{m}\). 2. The trail length is \(L=\sqrt{540^2+380^2}\approx660.30\,\text{m}\). 3. For the incline, \(\tan(\alpha)=\frac{380}{540}\). 4. Therefore, \(\alpha=\tan^{-1}\left(\frac{380}{540}\right)\approx35.13^\circ\).

Trail length: \(\approx660.30\,\text{m}\). Average incline: \(\approx35.13^\circ\).

- What ratio does percent grade represent? - Which side of the triangle is the actual road length? - Which trigonometric ratio relates the hypotenuse and elevation gain?

1. A \(12\%\) grade means \(\tan(\alpha)=0.12\), so \(\alpha=\tan^{-1}(0.12)\approx6.84^\circ\). 2. The road length is the hypotenuse. Let \(h\) be the elevation gain. Then \(\sin(\alpha)=\frac{h}{850}\). 3. Use the unrounded angle: \(h=850\sin(\tan^{-1}(0.12))\approx101.27\,\text{m}\).

a) \(\approx101.27\,\text{m}\) b) \(\approx6.84^\circ\)

- Use a right-triangle model for each sun angle. - Which length stays the same in both calculations? - How does the shadow change when the sun's angle decreases?

1. Let \(H\) be the tower height. Then \(\tan(52^\circ)=\frac{H}{28}\), so \(H=28\tan(52^\circ)\approx35.84\,\text{m}\). 2. Let \(s\) be the shadow length at \(30^\circ\). Using the unrounded height, \(s=\frac{28\tan(52^\circ)}{\tan(30^\circ)}\approx62.07\,\text{m}\). 3. The increase is \(s-28\approx34.07\,\text{m}\).

Tower height: \(\approx35.84\,\text{m}\). Shadow increase: \(\approx34.07\,\text{m}\).

- Interpret tangent as a ratio of the two legs. - Use the Pythagorean theorem to find a proportional hypotenuse. - Apply the definitions of sine and cosine.

1. Model the opposite and adjacent legs as \(5k\) and \(12k\). 2. By the Pythagorean theorem, the hypotenuse is \(\sqrt{(5k)^2+(12k)^2}=13k\). 3. Thus, \(\sin(\alpha)=\frac{5k}{13k}=\frac{5}{13}\) and \(\cos(\alpha)=\frac{12k}{13k}=\frac{12}{13}\).

\(\sin(\alpha)=\frac{5}{13}\) and \(\cos(\alpha)=\frac{12}{13}\)

- Read the sine ratio directly from the given relationship. - Use the Pythagorean theorem or identity to find the remaining side ratio. - Simplify radicals and rationalize the denominator if needed.

1. Let the opposite leg be \(a\), so the hypotenuse is \(3a\). Then \(\sin(\alpha)=\frac{a}{3a}=\frac{1}{3}\). 2. By the Pythagorean identity, \(\cos^2(\alpha)=1-\frac{1}{9}=\frac{8}{9}\). Since \(\alpha\) is acute, \(\cos(\alpha)=\frac{2\sqrt{2}}{3}\). 3. Therefore, \(\tan(\alpha)=\frac{1/3}{2\sqrt{2}/3}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\).

\(\sin(\alpha)=\frac{1}{3}\), \(\cos(\alpha)=\frac{2\sqrt{2}}{3}\), and \(\tan(\alpha)=\frac{\sqrt{2}}{4}\)

- Use the tangent ratio to choose proportional leg lengths. - Find the hypotenuse with the Pythagorean theorem. - Substitute the resulting sine and cosine values into the identity.

1. Model the opposite and adjacent legs as \(3k\) and \(4k\). 2. The Pythagorean theorem gives a hypotenuse of \(\sqrt{(3k)^2+(4k)^2}=5k\). 3. Therefore, \(\sin(\gamma)=\frac{3}{5}\) and \(\cos(\gamma)=\frac{4}{5}\). 4. Substitution gives \(\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=\frac{9}{25}+\frac{16}{25}=1\).

a) \(\sin(\gamma)=\frac{3}{5}\) and \(\cos(\gamma)=\frac{4}{5}\) b) The identity is verified because \(\frac{9}{25}+\frac{16}{25}=1\).

- Consider the altitude from the vertex. - How does the altitude divide the vertex angle and the base? - Use sine for half the base and cosine for the altitude.

1. The altitude bisects the vertex angle and the base. For the original triangle, each half has a \(20^\circ\) angle at the vertex. 2. Since \(\sin(20^\circ)=\frac{c/2}{12}\), \(c=24\sin(20^\circ)\approx 8.21\,\text{cm}\). 3. Since \(\cos(20^\circ)=\frac{h_c}{12}\), \(h_c=12\cos(20^\circ)\approx 11.28\,\text{cm}\). 4. With vertex angle \(80^\circ\), each half-angle is \(40^\circ\), so the new base is \(24\sin(40^\circ)\approx 15.43\,\text{cm}\).

a) \(c\approx 8.21\,\text{cm}\) b) \(h_c\approx 11.28\,\text{cm}\) c) \(c\approx 15.43\,\text{cm}\)

- Use the vertical cross-section through the face's midpoint and the apex. - What horizontal run belongs to half the square base? - For the seked, compare horizontal run with vertical rise before changing to palms per cubit.

1. In the stated vertical cross-section, the horizontal leg is half the square base's side: \(\frac{230.4}{2}=115.2\,\text{m}\). 2. For the face angle, \(\tan(\alpha)=\frac{146.6}{115.2}\). Thus, \(\alpha=\tan^{-1}\left(\frac{146.6}{115.2}\right)\approx51.84^\circ\). 3. The seked is horizontal run for one cubit of rise. In palms, \(S=\frac{115.2}{146.6}\cdot7\approx5.50068\). 4. Rounded to one decimal place, the seked is \(5.5\) palms.

a) \(\alpha\approx51.84^\circ\) b) \(\approx5.5\) palms

- Convert the percent grade to a vertical-rise to horizontal-run ratio. - What is the distance from the base center to the midpoint of one side? - The cross-sectional right triangle has the support as its longest side.

1. The horizontal run from the center of the square base to a side midpoint is \(\frac{18}{2}=9\,\text{m}\). 2. A \(140\%\) grade means \(\frac{h}{9}=1.40\), so \(h=9\cdot1.40=12.6\,\text{m}\). 3. Since \(\tan(\alpha)=1.40\), \(\alpha=\tan^{-1}(1.40)\approx54.46^\circ\). 4. The support is the hypotenuse in that cross-section: \(L=\sqrt{12.6^2+9^2}\approx15.48\,\text{m}\).

a) \(h=12.6\,\text{m}\) b) \(\alpha\approx54.46^\circ\) c) \(L\approx15.48\,\text{m}\)

- Which trigonometric ratio compares river width with the distance along the bank? - Identify the known adjacent leg and unknown opposite leg. - Consider what happens to tangent close to \(90^\circ\).

1. For \(\gamma=72^\circ\), \(\tan(72^\circ)=\frac{AB}{25}\). Thus, \(AB=25\tan(72^\circ)\approx76.94\,\text{m}\). 2. For \(\gamma=73^\circ\), \(AB=25\tan(73^\circ)\approx81.77\,\text{m}\). 3. Subtract unrounded values to find the change: \(25(\tan(73^\circ)-\tan(72^\circ))\approx4.83\,\text{m}\). 4. As \(\gamma\) approaches \(90^\circ\), \(\tan(\gamma)\) increases without bound, so the calculated river width does too.

a) \(AB\approx76.94\,\text{m}\) b) Increase of \(\approx4.83\,\text{m}\) c) The calculated width increases without bound as \(\gamma\) approaches \(90^\circ\).

- Which ratio is sine, and which ratio is cosine? - Calculate both ratios at each angle. - Divide each new ratio by its original value.

1. At \(10^\circ\), \(\frac{a}{c}=\sin(10^\circ)\approx 0.1736\) and \(\frac{b}{c}=\cos(10^\circ)\approx 0.9848\). 2. At \(20^\circ\), the corresponding ratios are \(\sin(20^\circ)\approx 0.3420\) and \(\cos(20^\circ)\approx 0.9397\). 3. The opposite-leg ratio changes by a factor of \(\frac{0.3420}{0.1736}\approx 1.97\). 4. The adjacent-leg ratio changes by a factor of \(\frac{0.9397}{0.9848}\approx 0.95\).

1. \(\frac{a}{c}\approx 0.1736\) and \(\frac{b}{c}\approx 0.9848\) 2. The ratio \(\frac{a}{c}\) is multiplied by about \(1.97\), while \(\frac{b}{c}\) is multiplied by about \(0.95\).

- Identify the opposite and adjacent legs relative to \(\beta\). - For an angle close to \(90^\circ\), which leg is almost as long as the hypotenuse? - Compare which ratios must be less than or greater than \(1\).

1. Relative to \(\beta\), \(b\) is opposite and \(a\) is adjacent. Thus, \(b=15\sin(82^\circ)\approx 14.85\,\text{cm}\) and \(a=15\cos(82^\circ)\approx 2.09\,\text{cm}\). 2. Because \(82^\circ\) is close to \(90^\circ\), the opposite leg \(b\) is nearly as long as the hypotenuse, while adjacent leg \(a\) is short. 3. Both \(\frac{a}{c}\) and \(\frac{b}{c}\) are less than \(1\), but \(\frac{b}{a}>1\). Therefore, \(\frac{b}{a}\) is greatest.

1. \(a\approx 2.09\,\text{cm}\) and \(b\approx 14.85\,\text{cm}\) 2. \(\frac{b}{a}\) is greatest because \(b\) is much longer than \(a\), while each leg-to-hypotenuse ratio is less than \(1\).

- Use the Pythagorean identity to find sine. - Apply the right-triangle definitions of sine and cosine to find the legs. - Find tangent once from the legs and once from sine and cosine.

1. From the Pythagorean identity, \(\sin(\alpha)=\sqrt{1-\left(\frac{5}{13}\right)^2}=\frac{12}{13}\). 2. Since \(\cos(\alpha)=\frac{b}{c}\), \(b=13\cdot\frac{5}{13}=5\,\text{cm}\). Since \(\sin(\alpha)=\frac{a}{c}\), \(a=13\cdot\frac{12}{13}=12\,\text{cm}\). 3. Using side lengths, \(\tan(\alpha)=\frac{a}{b}=\frac{12}{5}=2.4\). 4. Using sine and cosine, \(\tan(\alpha)=\frac{12/13}{5/13}=\frac{12}{5}=2.4\).

a) \(\sin(\alpha)=\frac{12}{13}\) b) \(a=12\,\text{cm}\) and \(b=5\,\text{cm}\) c) \(\tan(\alpha)=\frac{12}{5}=2.4\)

- Percent grade compares rise with horizontal run. - Which trigonometric ratio represents that comparison? - Convert the ramp rise to meters before using the grade. - Test the doubled angle without rounding the first angle too early.

1. A \(6\%\) grade means \(\tan(\alpha)=0.06\), so \(\alpha=\tan^{-1}(0.06)\approx3.43^\circ\). 2. The rise is \(42\,\text{cm}=0.42\,\text{m}\). Since \(0.06=\frac{0.42}{L}\), the horizontal run is \(L=\frac{0.42}{0.06}=7\,\text{m}\). 3. Doubling the unrounded angle gives \(2\alpha=2\tan^{-1}(0.06)\approx6.87^\circ\). 4. The new grade is \(100\tan(2\tan^{-1}(0.06))\%\approx12.04\%\). 5. Because \(12.04\%\neq12\%\), the claim is not exactly true.

a) \(\alpha\approx3.43^\circ\) b) Total horizontal run: \(7\,\text{m}\) c) False. Doubling the angle gives a grade of \(\approx12.04\%\), not exactly \(12\%\).

- Relate each positive line slope to the tangent of its inclination angle. - Subtract the unrounded inclination angles to find the smaller angle between the lines. - Halve the incline of line \(h\), then turn that angle into a slope.

1. For a line with positive slope \(m\), \(\tan(\alpha)=m\). Thus, \(\alpha_g=\tan^{-1}(0.5)\approx26.57^\circ\) and \(\alpha_h=\tan^{-1}(2)\approx63.43^\circ\). 2. Subtract the unrounded inclination angles: \(\gamma=\tan^{-1}(2)-\tan^{-1}(0.5)\approx36.87^\circ\). 3. The bisecting line has angle \(\alpha_k=\frac{1}{2}\tan^{-1}(2)\approx31.72^\circ\). 4. Therefore, \(m_k=\tan\left(\frac{1}{2}\tan^{-1}(2)\right)\approx0.62\).

a) \(\alpha_g\approx26.57^\circ\) and \(\alpha_h\approx63.43^\circ\) b) \(\gamma\approx36.87^\circ\) c) \(m_k\approx0.62\)

- Read the two legs of the right-triangle cross-section from the diagram. - Relative to \(\alpha\), which leg is opposite and which is adjacent? - Choose the trigonometric ratio that uses those two legs.

1. From the diagram, the face diagonal is \(4\sqrt{2}\,\text{cm}\), and the perpendicular edge is \(4\,\text{cm}\). 2. Relative to \(\alpha\), the edge is opposite and the face diagonal is adjacent. 3. Therefore, \(\tan(\alpha)=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}\). 4. Thus, \(\alpha=\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)\approx35.26^\circ\).

\(\alpha\approx35.26^\circ\)

- Compare the corresponding lengths in the two panels. - Express a cube's space diagonal in terms of its edge length. - Which ratio compares the adjacent edge with the space diagonal?

1. For a cube with edge length \(s\), the space diagonal is \(d_s=s\sqrt{3}\). 2. If \(\gamma\) is the angle between an edge and the space diagonal, then \(\cos(\gamma)=\frac{s}{s\sqrt{3}}=\frac{1}{\sqrt{3}}\). 3. The edge length cancels, so the angle is independent of the cube's size. 4. \(\gamma=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx54.7^\circ\).

The angle is unchanged by the enlargement: \(\gamma\approx54.7^\circ\) in both cubes.

- Use \(b\) and \(c\) to find the face-diagonal leg of the displayed right triangle. - Then find the space diagonal or use a trig ratio directly from the two legs. - Relative to \(\beta\), identify the adjacent side and hypotenuse.

1. The diagonal perpendicular to edge \(a\) in the cross-section is \(d_f=\sqrt{8^2+6^2}=10\,\text{cm}\). 2. The space diagonal is \(d_s=\sqrt{15^2+10^2}=\sqrt{325}\,\text{cm}\). 3. Relative to \(\beta\), edge \(a\) is adjacent and \(d_s\) is the hypotenuse, so \(\cos(\beta)=\frac{15}{\sqrt{325}}\). 4. Therefore, \(\beta=\cos^{-1}\left(\frac{15}{\sqrt{325}}\right)\approx33.69^\circ\).

\(\beta\approx33.69^\circ\)

- Find the diagonal of the square base from its side length. - Read the marked angle from the diagram and identify the opposite and adjacent legs. - Use tangent to solve for the height.

1. The diagonal of the square base is \(d_f=10\sqrt{2}\,\text{cm}\). 2. The diagram marks a \(40^\circ\) angle between the space diagonal and the base diagonal. In the right triangle, \(d_f\) is adjacent and height \(c\) is opposite. 3. Thus, \(\tan(40^\circ)=\frac{c}{10\sqrt{2}}\). 4. Therefore, \(c=10\sqrt{2}\tan(40^\circ)\approx11.87\,\text{cm}\).

\(c\approx11.87\,\text{cm}\)

- Read the two leg lengths and the marked angles from the diagram. - First find the missing hypotenuse with the Pythagorean theorem. - For each acute angle, identify the opposite and adjacent legs before choosing a ratio.

1. Read the leg lengths from the diagram. By the Pythagorean theorem, \(b=\sqrt{2.4^2+0.7^2}=\sqrt{6.25}=2.5\,\text{cm}\). 2. Relative to \(\alpha\) at \(A\), the opposite leg is \(2.4\,\text{cm}\), so \(\sin(\alpha)=\frac{2.4}{2.5}=0.96\). 3. Relative to \(\gamma\) at \(C\), the adjacent leg is \(2.4\,\text{cm}\), so \(\cos(\gamma)=\frac{2.4}{2.5}=0.96\). 4. Also, \(\tan(\alpha)=\frac{2.4}{0.7}\approx3.43\).

a) \(b=2.5\,\text{cm}\) b) \(\sin(\alpha)=0.96\), \(\cos(\gamma)=0.96\), and \(\tan(\alpha)\approx3.43\)

- Read the two leg lengths and the marked angle from the diagram. - Use the Pythagorean theorem before evaluating the trig ratios. - Relative to \(\rho\), identify the opposite and adjacent legs carefully.

1. Read the two leg lengths from the diagram. By the Pythagorean theorem, \(q=\sqrt{20^2+21^2}=\sqrt{841}=29\,\text{cm}\). 2. Relative to \(\rho\) at \(R\), the opposite leg is \(21\), the adjacent leg is \(20\), and the hypotenuse is \(29\). 3. Therefore, \(\sin(\rho)=\frac{21}{29}\approx0.724\), \(\cos(\rho)=\frac{20}{29}\approx0.690\), and \(\tan(\rho)=\frac{21}{20}=1.05\).

a) \(q=29\,\text{cm}\) b) \(\sin(\rho)=\frac{21}{29}\approx0.724\), \(\cos(\rho)=\frac{20}{29}\approx0.690\), and \(\tan(\rho)=\frac{21}{20}=1.05\)

- Read the two side lengths from the diagram and use them to find the diagonal. - For \(\sin(\alpha)\), identify the side opposite \(\alpha\) and the hypotenuse. - After finding one acute angle, use complementarity for the other.

1. By the Pythagorean theorem, \(d=\sqrt{24^2+10^2}=\sqrt{676}=26\,\text{cm}\). 2. Relative to \(\alpha\), the opposite side is \(10\,\text{cm}\), so \(\sin(\alpha)=\frac{10}{26}=\frac{5}{13}\). 3. Then \(\alpha=\sin^{-1}\left(\frac{5}{13}\right)\approx 22.62^\circ\). 4. Since the acute angles are complementary, \(\beta=90^\circ-22.62^\circ\approx 67.38^\circ\).

a) \(d=26\,\text{cm}\) b) \(\sin(\alpha)=\frac{5}{13}\) c) \(\alpha\approx 22.62^\circ\) and \(\beta\approx 67.38^\circ\)

- Angle \(\alpha\) appears in both the smaller left triangle and the large triangle. - Identify the hypotenuse, opposite leg, and adjacent leg separately in each triangle. - Altitude \(h\) is perpendicular to side \(c\).

1. In right triangle \(BDC\), \(a\) is the hypotenuse and \(h\) is opposite \(\beta\). Therefore, \(\sin(\beta)=\frac{h}{a}\). 2. In right triangle \(ADC\), \(h\) is opposite \(\alpha\) and \(p\) is adjacent. Therefore, \(\tan(\alpha)=\frac{h}{p}\). 3. In the large triangle, \(c\) is the hypotenuse and \(b\) is adjacent to \(\alpha\), so \(\cos(\alpha)=\frac{b}{c}\). 4. In the large triangle, \(a\) is opposite \(\alpha\), so \(\sin(\alpha)=\frac{a}{c}\).

a) \(\sin(\beta)=\frac{h}{a}\) b) \(\tan(\alpha)=\frac{h}{p}\) c) \(\cos(\alpha)=\frac{b}{c}\) d) \(\sin(\alpha)=\frac{a}{c}\)

- Convert all lengths to the same unit before forming a ratio. - Identify which angle is opposite the given leg. - Check that the angle sizes are reasonable for the diagram.

1. Convert the leg length: \(5.4\,\text{dm}=54\,\text{cm}\). 2. The hypotenuse is \(185\,\text{cm}\), and the leg opposite \(\beta\) is \(54\,\text{cm}\). 3. Thus, \(\sin(\beta)=\frac{54}{185}\), so \(\beta=\sin^{-1}\left(\frac{54}{185}\right)\approx 17.0^\circ\). 4. Therefore, \(\alpha=90^\circ-\beta\approx 73.0^\circ\).

\(\alpha\approx 73.0^\circ\) and \(\beta\approx 17.0^\circ\)

- Read the given leg and acute angle from the diagram. - Use the complementary-angle relationship to find the other acute angle. - Choose ratios that pair the given leg with each unknown side.

1. The acute angles are complementary, so \(\alpha=90^\circ-50^\circ=40^\circ\). 2. Since \(\cos(50^\circ)=\frac{5}{c}\), \(c=\frac{5}{\cos(50^\circ)}\approx 7.78\,\text{cm}\). 3. Since \(\tan(50^\circ)=\frac{b}{5}\), \(b=5\tan(50^\circ)\approx 5.96\,\text{cm}\).

\(b\approx 5.96\,\text{cm}\) \(c\approx 7.78\,\text{cm}\) \(\alpha=40^\circ\)

- The right-angle mark tells you which side is the hypotenuse. - Use the known leg and hypotenuse to find the missing leg. - Which ratio connects the side adjacent to \(\beta\) with the hypotenuse?

1. The marked right angle is at \(C\), so the \(10\,\text{cm}\) side is the hypotenuse. 2. By the Pythagorean theorem, \(b=\sqrt{10^2-4^2}=\sqrt{84}\approx9.17\,\text{cm}\). 3. Relative to \(\beta\), the \(4\,\text{cm}\) leg is adjacent and the \(10\,\text{cm}\) side is the hypotenuse, so \(\cos(\beta)=\frac{4}{10}\). 4. Thus, \(\beta=\cos^{-1}(0.4)\approx66.4^\circ\).

\(b\approx9.17\,\text{cm}\) \(\beta\approx66.4^\circ\)

- Locate the right angle and its opposite side. - For the marked acute angle, decide which unknown leg is opposite and which is adjacent. - Which ratios relate those legs to the hypotenuse?

1. The right-angle mark at \(C\) identifies the \(8\,\text{cm}\) side as the hypotenuse. 2. Relative to the \(40^\circ\) angle, \(a\) is opposite and \(b\) is adjacent. 3. Use sine: \(a=8\sin(40^\circ)\,\text{cm}\approx5.14\,\text{cm}\). 4. Use cosine: \(b=8\cos(40^\circ)\,\text{cm}\approx6.13\,\text{cm}\).

\(a\approx5.14\,\text{cm}\) \(b\approx6.13\,\text{cm}\)

- Read the slanted-side length and acute angle from the figure. - Use the right triangle containing the height to relate that height to the slanted side. - Use the angle relationship for adjacent angles of a parallelogram.

1. In the right triangle formed by the \(4\,\text{cm}\) side and the height, \(\sin(50^\circ)=\frac{h}{4}\). 2. Therefore, \(h=4\sin(50^\circ)\approx 3.06\,\text{cm}\). 3. Adjacent angles in a parallelogram are supplementary, so \(\beta=180^\circ-50^\circ=130^\circ\).

\(h\approx 3.06\,\text{cm}\) \(\beta=130^\circ\)

- Read the side length and height from the diagram. - Use the right triangle formed by the altitude to relate \(\alpha\) to those lengths. - Adjacent angles of a rhombus have the same supplementary relationship as in any parallelogram.

1. In the right triangle formed by the height and a side of the rhombus, \(\sin(\alpha)=\frac{4}{6}=\frac{2}{3}\). 2. Therefore, \(\alpha=\sin^{-1}\left(\frac{2}{3}\right)\approx 41.81^\circ\). 3. Adjacent angles in a rhombus are supplementary, so \(\beta=180^\circ-41.81^\circ\approx 138.19^\circ\).

\(\alpha\approx 41.81^\circ\) \(\beta\approx 138.19^\circ\)

- Read the equal-side length and base angle from the diagram. - Use the altitude to split the isosceles triangle into two congruent right triangles. - In one right triangle, identify the height and half of the base relative to the marked base angle.

1. The altitude divides the isosceles triangle into two congruent right triangles. Using one of them, \(\sin(70^\circ)=\frac{h}{8}\), so \(h=8\sin(70^\circ)\approx 7.52\,\text{cm}\). 2. Half the base is adjacent to the \(70^\circ\) angle. Thus, \(\cos(70^\circ)=\frac{b/2}{8}\), so \(b=16\cos(70^\circ)\approx 5.47\,\text{cm}\).

Height: \(h\approx 7.52\,\text{cm}\) Base length: \(b\approx 5.47\,\text{cm}\)

- Read the base and height from the diagram. - The altitude splits the isosceles triangle into two congruent right triangles. - Find one congruent side and half the vertex angle from one of those right triangles.

1. The altitude bisects the \(10\,\text{cm}\) base, so each right triangle has legs \(5\,\text{cm}\) and \(4\,\text{cm}\). 2. By the Pythagorean theorem, \(s=\sqrt{5^2+4^2}=\sqrt{41}\approx 6.40\,\text{cm}\). 3. The altitude bisects the vertex angle, so \(\tan\left(\frac{\gamma}{2}\right)=\frac{5}{4}\). 4. Therefore, \(\gamma=2\tan^{-1}\left(\frac{5}{4}\right)\approx 102.68^\circ\).

\(s\approx 6.40\,\text{cm}\) \(\gamma\approx 102.68^\circ\)

- Read the two base lengths and leg length from the figure. - Imagine dropping altitudes from the endpoints of the shorter base; how is the difference of the bases split? - Use a right-triangle ratio for \(\alpha\), then use the parallel bases for \(\gamma\).

1. Dropping the two altitudes leaves a horizontal leg of length \(\frac{9-4}{2}=2.5\,\text{cm}\) in each side right triangle. 2. For the lower base angle, \(\cos(\alpha)=\frac{2.5}{4}=0.625\), so \(\alpha=\cos^{-1}(0.625)\approx 51.32^\circ\). 3. Because the bases are parallel, consecutive interior angles along a leg are supplementary. Thus, \(\gamma=180^\circ-51.32^\circ\approx 128.68^\circ\).

\(\alpha\approx 51.32^\circ\) \(\gamma\approx 128.68^\circ\)

- Read the lower base, leg length, and lower base angle from the diagram. - Drop altitudes from the endpoints of the upper base to form two congruent right triangles. - Use one right triangle to find the height and one horizontal offset, then relate the two bases.

1. In either side right triangle, \(\sin(50^\circ)=\frac{h}{5}\), so \(h=5\sin(50^\circ)\approx 3.83\,\text{cm}\). 2. Let \(x\) be the horizontal leg of either right triangle. Then \(\cos(50^\circ)=\frac{x}{5}\), so \(x=5\cos(50^\circ)\). 3. The two horizontal legs account for the difference between the bases. Thus, \(c=12-2x=12-10\cos(50^\circ)\approx 5.57\,\text{cm}\).

Upper base: \(c\approx 5.57\,\text{cm}\) Height: \(h\approx 3.83\,\text{cm}\)

- Read the known leg and acute angle from the diagram. - Use the smaller right triangle containing \(q\) and \(h\) first. - Find the full hypotenuse before subtracting to obtain the remaining segment.

1. In right triangle \(ACD\), \(q\) is adjacent to the \(40^\circ\) angle, so \(q=6\cos(40^\circ)\,\text{cm}\approx 4.60\,\text{cm}\). 2. The altitude is opposite the \(40^\circ\) angle, so \(h=6\sin(40^\circ)\,\text{cm}\approx 3.86\,\text{cm}\). 3. In the full triangle, \(AB=\frac{6}{\cos(40^\circ)}\,\text{cm}\approx 7.83\,\text{cm}\). 4. Therefore, using unrounded values, \(p=AB-q\approx 3.24\,\text{cm}\).

\(p\approx 3.24\,\text{cm}\), \(q\approx 4.60\,\text{cm}\), and \(h\approx 3.86\,\text{cm}\)

- Read the given leg and angle from the figure, then find the other acute angle. - Work first in the smaller right triangle containing the given leg. - Use the altitude in the other small right triangle to recover the remaining hypotenuse segment.

1. The other acute angle is \(\angle B=90^\circ-35^\circ=55^\circ\). 2. In right triangle \(BCD\), \(p\) is adjacent to \(55^\circ\), so \(p=9\cos(55^\circ)\,\text{cm}\approx 5.16\,\text{cm}\). 3. The altitude is opposite \(55^\circ\), so \(h=9\sin(55^\circ)\,\text{cm}\approx 7.37\,\text{cm}\). 4. In right triangle \(ACD\), \(\tan(35^\circ)=\frac{h}{q}\), so \(q=\frac{9\sin(55^\circ)}{\tan(35^\circ)}\,\text{cm}\approx 10.53\,\text{cm}\).

\(p\approx 5.16\,\text{cm}\), \(q\approx 10.53\,\text{cm}\), and \(h\approx 7.37\,\text{cm}\)

- Read the hypotenuse length and acute angle from the diagram. - First find the leg adjacent to the marked angle in the large triangle. - Work in the smaller right triangle at that endpoint, then use the two hypotenuse segments' sum.

1. The leg adjacent to \(50^\circ\) is \(a=15\cos(50^\circ)\,\text{cm}\approx 9.64\,\text{cm}\). 2. In right triangle \(BCD\), \(p\) is adjacent to \(50^\circ\), so \(p=a\cos(50^\circ)=15\cos^2(50^\circ)\,\text{cm}\approx 6.20\,\text{cm}\). 3. The altitude is opposite \(50^\circ\), so \(h=a\sin(50^\circ)=15\cos(50^\circ)\sin(50^\circ)\,\text{cm}\approx 7.39\,\text{cm}\). 4. Since \(p+q=15\), \(q=15-p\approx 8.80\,\text{cm}\).

\(p\approx 6.20\,\text{cm}\), \(q\approx 8.80\,\text{cm}\), and \(h\approx 7.39\,\text{cm}\)

- Use the equal legs to split the difference of the two bases evenly between the two ends. - Form a right triangle to determine the trapezoid's height. - A diagonal, the height, and its horizontal run form another right triangle. - Use symmetry to combine the two equal diagonal inclinations into the marked obtuse angle.

1. The difference of the base lengths is \(18-10=8\,\text{cm}\). Because the trapezoid is isosceles, each horizontal overhang is \(4\,\text{cm}\). 2. Using either leg, the height is \(h=\sqrt{7^2-4^2}=\sqrt{33}\,\text{cm}\). 3. From \(A\) to \(C\), the horizontal run is \(4+10=14\,\text{cm}\). If \(\theta\) is the acute angle diagonal \(AC\) makes with the longer base, then \(\tan\theta=\frac{\sqrt{33}}{14}\), so \(\theta\approx22.31^\circ\). 4. By symmetry, diagonal \(BD\) makes the same acute angle with the base on the other side. Therefore, \(\epsilon=180^\circ-2\theta\approx135.38^\circ\). 5. Rounded to the nearest tenth, \(\epsilon\approx135.4^\circ\).

\(\epsilon\approx135.4^\circ\)

- Read the diagonal length and its angle with the side from the figure. - A diagonal divides the rectangle into two congruent right triangles. - Use sine and cosine to recover the rectangle's two side lengths before finding area.

1. The diagonal is the hypotenuse of a right triangle. The side adjacent to the \(38^\circ\) angle is \(9.4\cos(38^\circ)\approx 7.407\,\text{cm}\). 2. The opposite side is \(9.4\sin(38^\circ)\approx 5.787\,\text{cm}\). 3. Multiply the side lengths: \(A=(9.4\cos(38^\circ))(9.4\sin(38^\circ))\approx 42.8677\,\text{cm}^2\). 4. Therefore, \(A\approx 42.87\,\text{cm}^2\).

The area is approximately \(42.87\,\text{cm}^2\).

- Read the two leg lengths from the diagram. - Find the hypotenuse before using a side ratio for one acute angle. - Use the complementary relationship for the remaining acute angle.

1. Use the Pythagorean theorem: \(c=\sqrt{4^2+7^2}=\sqrt{65}\,\text{cm}\approx 8.06\,\text{cm}\). 2. For angle \(\alpha\), the opposite leg is \(4\) and the adjacent leg is \(7\), so \(\tan(\alpha)=\frac{4}{7}\). 3. Therefore, \(\alpha=\tan^{-1}\left(\frac{4}{7}\right)\approx 29.7^\circ\). 4. The acute angles are complementary, so \(\beta=90^\circ-29.7^\circ\approx 60.3^\circ\).

\(c\approx 8.06\,\text{cm}\), \(\alpha\approx 29.7^\circ\), and \(\beta\approx 60.3^\circ\)

- Read the upper base, slanted side, and marked angle from the diagram. - Drop a perpendicular from the upper-right vertex to split the trapezoid into a rectangle and a right triangle. - Find the horizontal leg of that right triangle and add it to the upper-base length.

1. Drop a perpendicular from \(C\) to \(AD\) at \(E\). Then \(ABCE\) is a rectangle, so \(AE=BC=6\,\text{cm}\). 2. In right triangle \(CED\), \(ED\) is adjacent to \(50^\circ\) and \(CD=8\,\text{cm}\) is the hypotenuse. 3. Thus, \(ED=8\cos(50^\circ)\,\text{cm}\approx 5.14\,\text{cm}\). 4. Therefore, \(AD=AE+ED\approx 6+5.14=11.14\,\text{cm}\).

\(x\approx 11.14\,\text{cm}\)

- Start with the right triangle whose given side and angle let you find the shared side \(AB\). - Then switch to the second right triangle, where \(AB\) becomes a known leg. - In each triangle, identify the hypotenuse before choosing a trig ratio.

1. In right triangle \(ABC\), \(BC\) is opposite the marked \(30^\circ\) angle and \(AB\) is the hypotenuse. Thus, \(\sin(30^\circ)=\frac{5}{AB}\), so \(AB=10\,\text{cm}\). 2. In right triangle \(ABD\), \(AB\) is opposite the marked \(40^\circ\) angle and \(AD\) is the hypotenuse. Thus, \(\sin(40^\circ)=\frac{10}{AD}\), so \(x=\frac{10}{\sin(40^\circ)}\approx 15.56\,\text{cm}\).

\(x\approx 15.56\,\text{cm}\)

- Read the three given lengths from the diagram and use the perpendicular segment to identify two right triangles. - First find the shared altitude \(AC\) from the left triangle. - Then use the right triangle containing \(\alpha\) to form a tangent ratio.

1. In right triangle \(ABC\), \(AC=\sqrt{13^2-5^2}=\sqrt{144}=12\,\text{cm}\). 2. In right triangle \(ACD\), \(\tan(\alpha)=\frac{9}{12}=0.75\). 3. Therefore, \(\alpha=\tan^{-1}(0.75)\approx 36.87^\circ\).

\(\alpha\approx 36.87^\circ\)

- Find the two horizontal lengths separately. - What is the full angle \(\angle CAB\)? - Use tangent in both right triangles that share side \(AC\).

1. In right triangle \(ACD\), \(CD=12\tan(15^\circ)\,\text{cm}\approx 3.22\,\text{cm}\). 2. The full angle at \(A\) is \(15^\circ+30^\circ=45^\circ\). 3. In right triangle \(ABC\), \(BC=12\tan(45^\circ)=12\,\text{cm}\). 4. Therefore, \(BD=BC-CD\approx 12-3.22=8.78\,\text{cm}\).

\(x\approx 8.78\,\text{cm}\)

- The altitude splits the total base into two separate right-triangle base segments. - For each right triangle, relate the vertical leg to its horizontal base segment using the marked acute angle. - Add the two horizontal segments only after finding each one.

1. In right triangle \(ABC\), \(\tan(34^\circ)=\frac{BC}{8}\), so \(BC=8\tan(34^\circ)\,\text{cm}\). 2. In right triangle \(ACD\), \(\tan(52^\circ)=\frac{CD}{8}\), so \(CD=8\tan(52^\circ)\,\text{cm}\). 3. Therefore, \(BD=BC+CD=8\bigl(\tan(34^\circ)+\tan(52^\circ)\bigr)\,\text{cm}\approx15.64\,\text{cm}\).

\(x\approx15.64\,\text{cm}\)

- Express \(x\) as the difference of two angles at \(A\). - Find the angle at \(A\) in each nested right triangle. - Determine the full length \(CB\) first.

1. In right triangle \(ACD\), \(\angle CAD=\tan^{-1}\left(\frac{4}{10}\right)\approx 21.80^\circ\). 2. The full base is \(CB=4+8=12\,\text{cm}\). In right triangle \(ABC\), \(\angle CAB=\tan^{-1}\left(\frac{12}{10}\right)\approx 50.19^\circ\). 3. Therefore, \(x=\angle CAB-\angle CAD\approx 50.19^\circ-21.80^\circ=28.39^\circ\).

\(x\approx 28.39^\circ\)

- Divide the regular octagon into eight congruent isosceles triangles. - Find the central angle, then bisect one triangle. - Use right-triangle trigonometry in the half-triangle.

1. The central angle is \(\frac{360^\circ}{8}=45^\circ\). Bisecting one central triangle gives an angle of \(22.5^\circ\) and a half-side of \(2\,\text{cm}\). 2. For the apothem, \(\tan(22.5^\circ)=\frac{2}{r}\), so \(r=\frac{2}{\tan(22.5^\circ)}\approx 4.83\,\text{cm}\). 3. For the circumradius, \(\sin(22.5^\circ)=\frac{2}{R}\), so \(R=\frac{2}{\sin(22.5^\circ)}\approx 5.23\,\text{cm}\).

The circumradius is \(R\approx 5.23\,\text{cm}\), and the apothem is \(r\approx 4.83\,\text{cm}\).

- How large is the angle at the center of one of five equal central triangles? - Which part of the side length belongs to either right triangle? - In a right half-triangle, compare half the side and the apothem with the radius.

1. The central angle of a regular pentagon is \(\frac{360^\circ}{5}=72^\circ\). Its perpendicular bisector divides the central triangle into two right triangles with a \(36^\circ\) angle at the center. 2. For half the side, \(\sin(36^\circ)=\frac{a/2}{10}\). Thus, \(a=20\sin(36^\circ)\approx11.76\,\text{cm}\). 3. For the apothem, \(\cos(36^\circ)=\frac{r}{10}\). Thus, \(r=10\cos(36^\circ)\approx8.09\,\text{cm}\).

Side length: \(a\approx11.76\,\text{cm}\) Apothem: \(r\approx8.09\,\text{cm}\)

- Identify the two legs from the hypotenuse and the marked acute angle. - Use sine and cosine to find those legs. - Then use the triangle area formula with the two perpendicular legs.

1. The leg opposite \(30^\circ\) is \(10\sin(30^\circ)=5\,\text{cm}\). 2. The adjacent leg is \(10\cos(30^\circ)=5\sqrt{3}\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot 5\cdot 5\sqrt{3}=12.5\sqrt{3}\,\text{cm}^2\approx 21.65\,\text{cm}^2\).

\(A\approx 21.65\,\text{cm}^2\)

- Which two sides form the perpendicular base and height for the area? - Which trigonometric ratio relates the marked adjacent leg to the unknown opposite leg? - Find the missing leg before using the area formula.

1. The opposite leg is \(4\tan(60^\circ)=4\sqrt{3}\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\cdot 4\cdot 4\sqrt{3}=8\sqrt{3}\,\text{cm}^2\approx 13.86\,\text{cm}^2\).

\(A\approx 13.86\,\text{cm}^2\)

- Which angle is complementary to the marked \(53^\circ\) angle? - How are the sine of an angle and cosine of its complement related? - Which leg is adjacent to \(53^\circ\), and which side is the hypotenuse?

1. The two acute angles are complementary: \(37^\circ+53^\circ=90^\circ\). 2. Therefore, \(\cos(53^\circ)=\sin(37^\circ)\approx0.6018\). 3. Segment \(AC\) is adjacent to \(53^\circ\), and \(AB=20\) is the hypotenuse, so \(\frac{AC}{20}=\cos(53^\circ)\). 4. Hence, \(AC\approx20\cdot0.6018=12.036\approx12.0\) to the nearest tenth.

\(AC\approx12.0\)

- Classify the triangle using its angle marks. - Identify the positions of the known and unknown sides relative to the acute angle. - Which method uses those two sides without first finding another value?

1. The triangle is right. The known side \(AC\) is adjacent to the \(35^\circ\) angle, and \(BC\) is opposite it. 2. Right-triangle trigonometry is most direct, using \(\tan(35^\circ)=\frac{BC}{14}\). 3. Solving gives \(BC=14\tan(35^\circ)\approx9.8029\). 4. To the nearest tenth, \(BC\approx9.8\).

Right-triangle trigonometry: \(\tan(35^\circ)=\frac{BC}{14}\), so \(BC\approx9.8\).

- How does the downward sight line form a matching acute angle with the horizontal at ground level? - Which side is the platform height, and which side is the required horizontal distance? - Check whether the student's ratio uses those two sides.

1. The angle of depression equals the angle of elevation from the car, \(40^\circ\). 2. The \(30\,\text{ft}\) height is opposite that angle and the horizontal distance \(x\) is adjacent. Sine compares the opposite side with the hypotenuse, not the adjacent side. 3. \(\tan40^\circ=\frac{30}{x}\), so \(x=\frac{30}{\tan40^\circ}\approx36\,\text{ft}\).

The student used sine instead of tangent. The horizontal distance is about \(36\,\text{ft}\).

- How many equal central angles make a full turn? - What do the matching length marks tell you about the triangle's symmetry? - Once you know the angle at \(O\) in one half, which ratio connects \(OM\) and \(OA\)?

1. All eight central angles are equal, so \(\angle AOB=\frac{360^\circ}{8}=45^\circ\). 2. Since \(OA=OB\) and the tick marks give \(AM=MB\), the midpoint segment \(OM\) is perpendicular to \(AB\) and bisects \(\angle AOB\). 3. Thus \(\angle AOM=22.5^\circ\). In right triangle \(OAM\), \(\cos22.5^\circ=\frac{OM}{OA}=\frac{OM}{8}\). 4. Therefore \(OM=8\cos22.5^\circ\approx7.39\,\text{cm}\).

\(\angle AOM=\frac{360^\circ}{8\cdot2}=22.5^\circ\); \(OM=8\cos22.5^\circ\approx7.39\,\text{cm}\).

- First express the rise and run in the same unit. - What happens to the angle when the horizontal run decreases but the rise stays fixed? - The percent grade compares vertical rise with horizontal run, not ramp length.

1. Convert the rise: \(72\,\text{cm}=0.72\,\text{m}\). With horizontal distance \(x\), \(\tan(3.5^\circ)=\frac{0.72}{x}\), so \(x=\frac{0.72}{\tan(3.5^\circ)}\approx11.77190\,\text{m}\). Rounding upward gives a minimum of \(11.78\,\text{m}\) to the requested precision. 2. The percent grade at the limit is \(100\tan(3.5^\circ)\%\approx6.12\%\). 3. With a \(10\,\text{m}\) horizontal run, \(\tan(\theta)=\frac{0.72}{10}=0.072\). 4. Hence \(\theta=\tan^{-1}(0.072)\approx4.12^\circ\). Since \(4.12^\circ>3.5^\circ\), the shorter design exceeds the project's limit.

a) At least \(11.78\,\text{m}\) of horizontal distance. b) \(\approx6.12\%\) c) \(\theta\approx4.12^\circ\). No; it exceeds the \(3.5^\circ\) design limit.

- Compare the denominators in the sine and tangent ratios. - For an acute angle, how does \(\cos(\theta)\) compare with \(1\)? - How does sine change as an acute angle increases?

1. For an acute angle, \(\tan(\beta)=\frac{\sin(\beta)}{\cos(\beta)}\). Because \(0<\cos(\beta)<1\), it follows that \(\sin(\beta)=\tan(\beta)\cos(\beta)<0.3\). 2. Since sine increases on acute angles and \(\sin(\alpha)=0.3\), \(\alpha>\beta\). 3. Calculate \(\alpha=\sin^{-1}(0.3)\approx 17.46^\circ\) and \(\beta=\tan^{-1}(0.3)\approx 16.70^\circ\), confirming that \(\alpha>\beta\).

1. \(\alpha\) is larger than \(\beta\). 2. \(\alpha\approx 17.46^\circ\) and \(\beta\approx 16.70^\circ\).

- Compare the denominators in the sine and tangent ratios. - Which denominator is smaller in a right triangle? - As the angle becomes nearly \(90^\circ\), what happens to the adjacent leg?

1. Let \(O\) be the opposite leg, \(A\) the adjacent leg, and \(H\) the hypotenuse. Then \(\sin(\alpha)=\frac{O}{H}\) and \(\tan(\alpha)=\frac{O}{A}\). 2. Since the hypotenuse is the longest side, \(H>A\). With the same positive numerator, the ratio with the smaller denominator is larger. Thus, \(\tan(\alpha)>\sin(\alpha)\). 3. As \(\alpha\) approaches \(90^\circ\), \(\sin(\alpha)\) approaches \(1\), while the adjacent leg approaches \(0\), so \(\tan(\alpha)\) increases without bound. 4. Therefore, the difference does not always decrease; it becomes arbitrarily large near \(90^\circ\).

a) Since the adjacent leg is shorter than the hypotenuse, \(\frac{O}{A}>\frac{O}{H}\), so \(\tan(\alpha)>\sin(\alpha)\). b) The claim is false. As \(\alpha\) approaches \(90^\circ\), \(\tan(\alpha)\) increases without bound while \(\sin(\alpha)\) approaches \(1\).

- Use the triangle angle sum to relate the base and vertex angles. - Split the isosceles triangle into two right triangles using its altitude. - For each part, decide which leg lengths or ratios the base angle controls.

1. For part a), \(2\alpha+\gamma=180^\circ\) and \(\alpha=2\gamma\). Thus, \(5\gamma=180^\circ\), so \(\gamma=36^\circ\) and \(\alpha=72^\circ\). 2. The altitude bisects the base, so \(\cos(72^\circ)=\frac{5}{s}\). Therefore, \(s=\frac{5}{\cos(72^\circ)}\,\text{cm}\approx16.18\,\text{cm}\). 3. For part b), if \(h_c=c\), then in one right half of the triangle \(\tan(\alpha)=\frac{h_c}{c/2}=2\). 4. Thus, \(\alpha=\tan^{-1}(2)\approx63.43^\circ\). This gives a valid isosceles triangle because two such base angles total less than \(180^\circ\).

a) \(s=\frac{5}{\cos(72^\circ)}\,\text{cm}\approx16.18\,\text{cm}\) b) Yes. \(\alpha=\tan^{-1}(2)\approx63.43^\circ\).

- Recall which two lengths tangent compares. - In the seked ratio, identify the horizontal and vertical quantities. - What happens to a fraction when its denominator doubles and numerator stays fixed?

1. Let \(x\) be the horizontal run and \(y\) the vertical rise, measured in the same units. The seked is \(S=\frac{x}{y}\cdot7\). 2. The tangent of the face angle is \(\tan(\alpha)=\frac{y}{x}\). 3. Rearranging gives \(\frac{y}{x}=\frac{7}{S}\), so \(\tan(\alpha)=\frac{7}{S}\). 4. For \(S=5.25\), \(\tan(\alpha)=\frac{7}{5.25}=\frac{4}{3}\). Therefore, \(\alpha=\tan^{-1}\left(\frac{4}{3}\right)\approx53.13^\circ\). 5. With fixed base dimensions, the horizontal run is constant. Doubling the height doubles the denominator in \(S=\frac{x}{y}\cdot7\), so the seked is cut in half.

a) \(S=\frac{7x}{y}\) and \(\tan(\alpha)=\frac{y}{x}\), so \(\tan(\alpha)=\frac{7}{S}\). b) \(\alpha\approx53.13^\circ\) c) The seked is halved because its horizontal run stays fixed while its height doubles.

- Identify the sides of the right triangle relative to \(\beta\). - Which side is the hypotenuse when the right angle is at \(A\)? - In part c), rewrite \(d=50s\) as a ratio of corresponding lengths.

1. In right triangle \(ABP\), \(AB=s\) is adjacent to \(\beta\), and \(BP=d\) is the hypotenuse. Thus, \(\cos(\beta)=\frac{s}{d}\). 2. Solving gives \(d=\frac{s}{\cos(\beta)}\). 3. With \(s=10\,\text{km}\) and \(\beta=88^\circ\), \(d=\frac{10}{\cos(88^\circ)}\approx286.54\,\text{km}\). 4. If \(d=50s\), then \(\cos(\beta)=\frac{1}{50}\). 5. Therefore, \(\beta=\cos^{-1}\left(\frac{1}{50}\right)\approx88.85^\circ\).

a) \(d=\frac{s}{\cos(\beta)}\) b) \(d\approx286.54\,\text{km}\) c) \(\beta\approx88.85^\circ\)

- Relate \(\frac{c}{b}\) to a right-triangle trigonometric ratio. - Compare the size of \(V\) at the two given angles using the same expression. - Think about what happens to the denominator as \(\alpha\) gets close to \(90^\circ\).

1. Since \(\cos(\alpha)=\frac{b}{c}\), \(V=\frac{1}{\cos(\alpha)}\). 2. If \(V=2\), then \(\cos(\alpha)=\frac{1}{2}\), so \(\alpha=60^\circ\). 3. \(V(80^\circ)=\frac{1}{\cos(80^\circ)}\approx 5.7588\), and \(V(85^\circ)=\frac{1}{\cos(85^\circ)}\approx 11.4737\). 4. The factor is \(\frac{11.4737}{5.7588}\approx 1.99\), so the claim is accurate. 5. As \(\alpha\) approaches \(90^\circ\), \(\cos(\alpha)\) approaches \(0\). Its reciprocal therefore increases without bound.

a) \(\alpha=60^\circ\) b) The claim is accurate because \(\frac{V(85^\circ)}{V(80^\circ)}\approx 1.99\). c) Since \(\cos(\alpha)\) approaches \(0\), \(V=\frac{1}{\cos(\alpha)}\) increases without bound.

- After fixing the side and angle, describe the locus imposed by \(AC=7\,\text{cm}\). - Compare the circle radius with the perpendicular distance from \(A\) to the angle ray. - Two valid intersections need not give congruent triangles; compare the resulting third sides.

1. Fix \(AB=10\,\text{cm}\) and draw the ray from \(B\) that makes a \(35^\circ\) angle with \(\overrightarrow{BA}\). 2. Point \(C\) must lie on the circle centered at \(A\) with radius \(7\,\text{cm}\). 3. The perpendicular distance from \(A\) to the line containing the ray is \(10\sin35^\circ\approx5.74\,\text{cm}<7\,\text{cm}\), so the line meets the circle twice. 4. The distance from \(B\) to the perpendicular foot is \(10\cos35^\circ\approx8.19\,\text{cm}\). The two circle intersections are \(\sqrt{7^2-(10\sin35^\circ)^2}\approx4.01\,\text{cm}\) from that foot along the line. 5. Both intersections lie on the forward ray, giving \(BC\approx8.19-4.01=4.18\,\text{cm}\) or \(BC\approx8.19+4.01=12.20\,\text{cm}\). 6. Because the possible third-side lengths differ, the two triangles are not congruent.

The angle ray meets the circle centered at \(A\) with radius \(7\,\text{cm}\) at two forward points. The two possible values are \(BC\approx4.18\,\text{cm}\) and \(BC\approx12.20\,\text{cm}\), so the SSA data determine two noncongruent triangles.