Free Math Worksheets

- 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 \(\sin(\gamma)=\frac{x}{z}\).

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

- Cosine is the ratio of the adjacent leg to the hypotenuse. - Identify the adjacent leg separately for each angle.

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 first. - Decide whether side \(e\) is adjacent to \(\sigma\) or \(\tau\).

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

\(\tau\)

- Sine uses the opposite leg. - Which angle is opposite side \(k\)?

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

\(\zeta\)

- Identify the horizontal leg and the vertical leg in the right triangle. - Remember that the angle is measured from the student's eye level, not from the ground. - Which trigonometric ratio relates the opposite and adjacent legs?

1. Let \(h\) be the vertical distance from the student'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 \(h\): \(h=32\tan(38^\circ)\approx 25.00\,\text{m}\). 3. Add the student's eye height: \(25.00\,\text{m}+1.60\,\text{m}\approx 26.60\,\text{m}\).

The church tower is approximately \(26.60\,\text{m}\) tall.

- Identify the opposite and adjacent legs for each acute angle. - Express \(b\) in terms of \(a\). - Use the definition of tangent as a ratio of the legs.

1. For angle \(\alpha\), \(\tan(\alpha)=\frac{a}{b}\). For angle \(\beta\), \(\tan(\beta)=\frac{b}{a}\). 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 \(\tan(\beta)=2\)

- How is percent grade related to rise over horizontal run? - In a right triangle, which trigonometric ratio compares rise with horizontal run? - Use an inverse trigonometric function to find an angle from a ratio.

1. A \(15\%\) grade means \(\frac{\text{rise}}{\text{run}}=0.15\). Since this ratio is \(\tan(\alpha)\), \(\alpha=\tan^{-1}(0.15)\approx 8.53^\circ\). 2. For a \(10^\circ\) incline, the decimal grade is \(\tan(10^\circ)\approx 0.1763\). 3. Convert to a percent: \(0.1763=17.63\%\), so the grade is approximately \(17.63\%\).

a) The angle of incline is approximately \(8.53^\circ\). b) An angle of \(10^\circ\) corresponds to a grade of approximately \(17.63\%\).

- 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 \(c\approx 14.64\,\text{cm}\)

- 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 \(10\,\text{cm}\).

- 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 \(12.00\,\text{m}^2\).

- 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 hypotenuse first. - For each acute angle, identify the opposite and adjacent legs. - What is the sum of the two acute angles in a right triangle?

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, \(\tan(\phi)=\frac{u}{v}\) and \(\cos(\phi)=\frac{v}{w}\). 4. The two acute angles in a right triangle are complementary, so \(\epsilon+\phi=90^\circ\).

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

- Decide whether each ratio is opposite over hypotenuse, adjacent over hypotenuse, or opposite over adjacent. - Match that ratio to angle \(\delta\) or \(\eta\).

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) \(\sin(\delta)=\frac{d}{f}\)

- Identify the opposite and adjacent legs for each acute angle. - The hypotenuse is opposite the right angle. - The opposite and adjacent legs switch roles when you change acute angles.

1. Relative to \(\alpha\), the opposite leg is \(12\), the adjacent leg is \(35\), and the hypotenuse is \(37\). 2. Thus, \(\sin(\alpha)=\frac{12}{37}\approx 0.324\), \(\cos(\alpha)=\frac{35}{37}\approx 0.946\), and \(\tan(\alpha)=\frac{12}{35}\approx 0.343\). 3. Relative to \(\beta\), the opposite and adjacent legs switch roles. 4. Thus, \(\sin(\beta)=\frac{35}{37}\approx 0.946\), \(\cos(\beta)=\frac{12}{37}\approx 0.324\), and \(\tan(\beta)=\frac{35}{12}\approx 2.917\).

a) \(\sin(\alpha)=\frac{12}{37}\approx 0.324\), \(\cos(\alpha)=\frac{35}{37}\approx 0.946\), and \(\tan(\alpha)=\frac{12}{35}\approx 0.343\) b) \(\sin(\beta)=\frac{35}{37}\approx 0.946\), \(\cos(\beta)=\frac{12}{37}\approx 0.324\), and \(\tan(\beta)=\frac{35}{12}\approx 2.917\)

- Identify the hypotenuse and the legs relative to the given angle. - Choose sine for the opposite leg and cosine for the adjacent leg. - Make sure your calculator is in degree mode.

1. The hypotenuse is \(14\,\text{cm}\), and the given acute angle is \(32^\circ\). 2. Side \(x\) is opposite the angle, so \(x=14\sin(32^\circ)\approx 7.42\,\text{cm}\). 3. Side \(y\) is adjacent to the angle, so \(y=14\cos(32^\circ)\approx 11.87\,\text{cm}\).

\(x\approx 7.42\,\text{cm}\) and \(y\approx 11.87\,\text{cm}\)

- Draw a right triangle and label the string as the hypotenuse. - The height is opposite the given angle. - Use sine.

1. The string is the hypotenuse of a right triangle, and the kite’s height \(h\) is opposite the \(55^{\circ}\) angle. 2. Use sine: \(\sin(55^{\circ}) = \frac{h}{100}\). 3. Therefore, \(h = 100\sin(55^{\circ}) \approx 81.92\,\text{ft}\).

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

- Identify the opposite and adjacent legs relative to \(40^\circ\). - Rearrange the sine equation to find the hypotenuse. - Use tangent or the Pythagorean theorem to find the adjacent leg.

1. The \(9\,\text{cm}\) leg is opposite the \(40^\circ\) angle. 2. Since \(\sin(40^\circ)=\frac{9}{u}\), \(u=\frac{9}{\sin(40^\circ)}\approx 14.00\,\text{cm}\). 3. Since \(\tan(40^\circ)=\frac{9}{v}\), \(v=\frac{9}{\tan(40^\circ)}\approx 10.73\,\text{cm}\).

\(u\approx 14.00\,\text{cm}\) and \(v\approx 10.73\,\text{cm}\)

- The ramp surface is the hypotenuse. - The vertical rise is opposite the angle. - Use sine and isolate \(l\).

1. The ramp surface is the hypotenuse, and the \(4\)-foot rise is opposite the \(4^{\circ}\) angle. 2. Use sine: \(\sin(4^{\circ}) = \frac{4}{l}\). 3. Solve for \(l\): \(l = \frac{4}{\sin(4^{\circ})} \approx 57.34\,\text{ft}\).

\(l \approx 57.34\,\text{ft}\)

- Identify the hypotenuse first. - Determine which leg is opposite and which is adjacent to \(50^\circ\). - Choose the matching trigonometric ratios.

1. The hypotenuse is \(10\,\text{cm}\), and the marked angle is \(50^\circ\). 2. Side \(m\) is adjacent to the angle, so \(m=10\cos(50^\circ)\approx 6.43\,\text{cm}\). 3. Side \(k\) is opposite the angle, so \(k=10\sin(50^\circ)\approx 7.66\,\text{cm}\).

\(k\approx 7.66\,\text{cm}\) and \(m\approx 6.43\,\text{cm}\)

- Draw the pole, shadow, and sunlight as a right triangle. - Identify the opposite and adjacent legs relative to the angle. - Use tangent.

1. The shadow is adjacent to the \(35^{\circ}\) angle, and the pole’s height is opposite. 2. Use tangent: \(\tan(35^{\circ}) = \frac{h}{25}\). 3. Therefore, \(h = 25\tan(35^{\circ}) \approx 17.51\,\text{ft}\).

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

- Identify the opposite and adjacent sides relative to \(\alpha\). - Use inverse tangent to find \(\alpha\). - The two acute angles in a right triangle are complementary.

1. In the right triangle formed by the diagonal, the height is opposite \(\alpha\) and the width is adjacent. 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 blank requires an angle or a trigonometric function.

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 \(b=12\cos(28^\circ)\approx 10.60\,\text{cm}\).

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

- Identify the two given sides in the right triangle formed by the diagonal. - Which trigonometric ratio uses the adjacent side and hypotenuse?

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

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

- The opposite and adjacent legs depend on which acute angle you use. - Notice the cofunction relationship between \(\sin(\beta)\) and \(\cos(\alpha)\).

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 \(\cos(\alpha)\approx 0.923\)

- Identify the hypotenuse from the right angle. - For each requested ratio, identify the opposite and adjacent legs relative to its angle.

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 adjacent leg and the hypotenuse relative to \(\angle A\). - Which trigonometric ratio relates the adjacent leg to the hypotenuse?

1. Relative to \(\angle A\), \(AB\) is the adjacent leg and \(AC\) is the hypotenuse. 2. Therefore, \(\cos(30^\circ)=\frac{AB}{AC}=\frac{5}{x}\). 3. Solving for \(x\), \(x=\frac{5}{\cos(30^\circ)}=\frac{5}{\frac{\sqrt{3}}{2}}=\frac{10\sqrt{3}}{3}\,\text{cm}\approx 5.77\,\text{cm}\).

\(x=\frac{10\sqrt{3}}{3}\,\text{cm}\approx 5.77\,\text{cm}\)

- The altitude divides the original triangle into two smaller right triangles. - Which smaller triangle contains all the given information?

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 right triangle \(ADC\). Which side is its hypotenuse? - Which side is adjacent to \(\angle A\)?

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

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

- Which side of the right triangle represents the sloped ramp surface? - Express both lengths in the same unit before calculating. - To make the angle smaller for the same rise, must the ramp be longer or shorter?

1. Convert the rise to meters: \(45\,\text{cm}=0.45\,\text{m}\). Let \(L\) be the length of the sloped surface, which is the hypotenuse. 2. Use sine: \(\sin(3.5^\circ)=\frac{0.45}{L}\). 3. Solve for \(L\): \(L=\frac{0.45}{\sin(3.5^\circ)}\approx 7.3712\,\text{m}\). 4. Because the angle may not exceed \(3.5^\circ\), round up to the next hundredth: \(L=7.38\,\text{m}\).

The sloped ramp surface must be at least \(7.38\,\text{m}\) long.

- Use a right-triangle model for each time. What stays the same, and what changes? - How are the tower height, shadow length, and sun angle related? - Find both shadow lengths before subtracting.

1. Let \(s_1\) be the morning shadow length. Then \(\tan(22^\circ)=\frac{42}{s_1}\), so \(s_1=\frac{42}{\tan(22^\circ)}\approx 103.95\,\text{m}\). 2. Let \(s_2\) be the noon shadow length. Then \(\tan(58^\circ)=\frac{42}{s_2}\), so \(s_2=\frac{42}{\tan(58^\circ)}\approx 26.24\,\text{m}\). 3. Find the decrease: \(s_1-s_2\approx 103.95-26.24=77.71\,\text{m}\).

The tower's shadow shortens by approximately \(77.71\,\text{m}\).

- Which ratio relates the side opposite \(\alpha\) to the hypotenuse? - Find the missing leg with the Pythagorean theorem. - Compute the area using the legs and again using the hypotenuse as the base.

1. Since \(\sin(\alpha)=\frac{a}{c}=\frac{5}{13}\), \(\alpha=\sin^{-1}\left(\frac{5}{13}\right)\approx 22.62^\circ\). 2. Find the other leg: \(b=\sqrt{13^2-5^2}=12\,\text{cm}\). 3. The area is \(A=\frac{1}{2}\cdot 5\cdot 12=30\,\text{cm}^2\). 4. Also, \(A=\frac{1}{2}\cdot 13\cdot h_c\). Therefore, \(h_c=\frac{60}{13}\,\text{cm}\approx 4.62\,\text{cm}\).

a) \(\alpha\approx 22.62^\circ\) b) \(h_c=\frac{60}{13}\,\text{cm}\approx 4.62\,\text{cm}\)

- 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.

- 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 \(\frac{a}{c}<1\).

- Model the side view as a right triangle and identify the known quantities. - Which trigonometric ratio relates the angle to the horizontal and vertical legs? - Which side of the triangle represents a rafter?

1. Let \(h\) be the vertical change. Then \(\tan(20^\circ)=\frac{h}{4.50}\), so \(h=4.50\tan(20^\circ)\approx 1.64\,\text{m}\). 2. Let \(L\) be the rafter length. Since the rafter is the hypotenuse, \(\cos(20^\circ)=\frac{4.50}{L}\). Thus, \(L=\frac{4.50}{\cos(20^\circ)}\approx 4.79\,\text{m}\). 3. Since \(\tan(20^\circ)\approx 0.3640\), the percent grade is approximately \(36.40\%\). 4. Since \(36.40\%\leq 40\%\), the design meets the requirement.

a) The vertical change is approximately \(1.64\,\text{m}\). b) Each rafter must be approximately \(4.79\,\text{m}\) long. c) Yes. The roof grade is approximately \(36.40\%\), which is below the \(40\%\) limit.

- 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. \(\alpha\approx 53.13^\circ\)

- Express the given ratio as a fraction. - Which trigonometric ratio compares the two legs? - How are the two acute angles in a right triangle related?

1. The ratio \(a:b=2:3\) means \(\frac{a}{b}=\frac{2}{3}\). 2. Since \(\tan(\alpha)=\frac{a}{b}\), \(\alpha=\tan^{-1}\left(\frac{2}{3}\right)\approx 33.69^\circ\). 3. The acute angles are complementary, so \(\beta=90^\circ-33.69^\circ\approx 56.31^\circ\).

\(\alpha\approx 33.69^\circ\) and \(\beta\approx 56.31^\circ\)

- Use the legs as the base and height in the triangle area formula. - Once both legs are known, use the Pythagorean theorem. - Which inverse trigonometric function finds an angle from the ratio of the legs?

1. Use the area formula: \(24=\frac{1}{2}\cdot a\cdot 8\), so \(a=6\,\text{cm}\). 2. By the Pythagorean theorem, \(c=\sqrt{6^2+8^2}=10\,\text{cm}\). 3. Since \(\tan(\alpha)=\frac{6}{8}=0.75\), \(\alpha=\tan^{-1}(0.75)\approx 36.87^\circ\). 4. Then \(\beta=90^\circ-36.87^\circ\approx 53.13^\circ\).

\(c=10\,\text{cm}\), \(\alpha\approx 36.87^\circ\), and \(\beta\approx 53.13^\circ\)

- 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 theorem gives the hypotenuse? - Which trigonometric ratio compares the elevation gain with the horizontal distance?

1. Convert the map length to the horizontal ground distance: \(5.4\,\text{cm}\cdot 10{,}000=54{,}000\,\text{cm}=540\,\text{m}\). 2. Use the Pythagorean theorem to find the trail length: \(L=\sqrt{540^2+380^2}\approx 660.30\,\text{m}\). 3. Use tangent to find the angle: \(\tan(\alpha)=\frac{380}{540}\). 4. Therefore, \(\alpha=\tan^{-1}\left(\frac{380}{540}\right)\approx 35.13^\circ\).

The trail is approximately \(660.30\,\text{m}\) long and has an average angle of incline of approximately \(35.13^\circ\).

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

1. A \(12\%\) grade means \(\tan(\alpha)=0.12\), so \(\alpha=\tan^{-1}(0.12)\approx 6.84^\circ\). 2. The road length is the hypotenuse. Let \(h\) be the elevation gain. Then \(\sin(\alpha)=\frac{h}{850}\). 3. Solve: \(h=850\sin(6.8428^\circ)\approx 101.27\,\text{m}\).

a) The elevation gain is approximately \(101.27\,\text{m}\). b) The angle of incline is approximately \(6.84^\circ\).

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

1. Let \(H\) be the tower height. Then \(\tan(52^\circ)=\frac{H}{28}\), so \(H=28\tan(52^\circ)\approx 35.84\,\text{m}\). 2. Let \(s\) be the shadow length when the angle is \(30^\circ\). Using the unrounded height, \(\tan(30^\circ)=\frac{H}{s}\), so \(s=\frac{28\tan(52^\circ)}{\tan(30^\circ)}\approx 62.07\,\text{m}\). 3. The increase is \(62.07-28=34.07\,\text{m}\).

The tower is approximately \(35.84\,\text{m}\) tall, and the shadow becomes approximately \(34.07\,\text{m}\) longer.

- 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 a vertical cross-section through the center of the pyramid. - Identify the vertical height and the horizontal distance from the center of the base to the midpoint of an edge.

1. Consider a vertical cross-section through the apex and the midpoints of two opposite base edges. The horizontal leg is half the base side: \(\frac{230.4}{2}=115.2\,\text{m}\). 2. For the face angle, \(\tan(\alpha)=\frac{146.6}{115.2}\). Thus, \(\alpha=\tan^{-1}\left(\frac{146.6}{115.2}\right)\approx 51.84^\circ\). 3. The seked is the horizontal run per cubit of vertical rise. In palms, \(S=\frac{115.2}{146.6}\cdot 7\approx 5.50\). 4. Therefore, the seked is approximately \(5.5\) palms.

a) The face angle is approximately \(51.84^\circ\). b) The seked is approximately \(5.5\) palms.

- Convert the percent grade to a decimal ratio. - What horizontal distance runs from the center of the square base to the midpoint of an edge? - Which theorem finds the slanted support length in the cross-sectional right triangle?

1. The horizontal run from the center of the base to the midpoint of an edge is \(\frac{18}{2}=9\,\text{m}\). 2. A \(140\%\) grade means \(\frac{h}{9}=1.40\), so \(h=9(1.40)=12.6\,\text{m}\). 3. Since \(\tan(\alpha)=1.40\), \(\alpha=\tan^{-1}(1.40)\approx 54.46^\circ\). 4. The support is the hypotenuse of the cross-sectional right triangle: \(L=\sqrt{12.6^2+9^2}\approx 15.48\,\text{m}\).

a) The pyramid's height is \(12.6\,\text{m}\). b) The face angle is approximately \(54.46^\circ\). c) The glass panel support must be approximately \(15.48\,\text{m}\) long.

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

1. For \(\gamma=72^\circ\), \(\tan(72^\circ)=\frac{AB}{25}\). Thus, \(AB=25\tan(72^\circ)\approx 76.94\,\text{m}\). 2. For \(\gamma=73^\circ\), \(AB=25\tan(73^\circ)\approx 81.77\,\text{m}\). 3. The change is \(81.77-76.94=4.83\,\text{m}\). 4. As \(\gamma\) approaches \(90^\circ\), \(\tan(\gamma)\) increases without bound. Therefore, the calculated width also increases without bound, and small angle errors cause increasingly large distance errors.

a) The river is approximately \(76.94\,\text{m}\) wide. b) The calculated width changes by approximately \(4.83\,\text{m}\). c) The calculated width increases without bound as \(\gamma\) approaches \(90^\circ\).

- 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 is rise divided by horizontal run, expressed as a percent. - Which trigonometric ratio represents rise over horizontal run? - Convert centimeters and meters to consistent units. - Use your angle from part a) to test the claim.

1. A \(6\%\) grade means \(\tan(\alpha)=0.06\). Therefore, \(\alpha=\tan^{-1}(0.06)\approx 3.43^\circ\). 2. Convert the rise: \(42\,\text{cm}=0.42\,\text{m}\). Since \(0.06=\frac{0.42}{L}\), \(L=\frac{0.42}{0.06}=7\,\text{m}\). 3. Doubling the angle gives \(2\alpha=2\tan^{-1}(0.06)\approx 6.87^\circ\). 4. Using the unrounded doubled angle, \(\tan(2\alpha)\approx 0.12043\), so the new percent grade is approximately \(12.04\%\). 5. Because \(12.04\%\neq 12\%\), the claim is not exactly true, although it is a close approximation for this small angle.

a) \(\alpha\approx 3.43^\circ\) b) The sloped ramp sections require a total horizontal length of \(7\,\text{m}\). c) The claim is false. Doubling the angle gives a grade of approximately \(12.04\%\), not exactly \(12\%\).

- Relate a line’s slope to the tangent of its inclination angle. - Subtract the two inclination angles to find the smaller angle between the lines. - Halve the angle for line \(h\), then convert the resulting angle back to a slope.

1. For a line with slope \(m\), \(\tan(\alpha)=m\). Thus, \(\alpha_g=\tan^{-1}(0.5)\approx 26.57^\circ\) and \(\alpha_h=\tan^{-1}(2)\approx 63.43^\circ\). 2. The smaller angle between the lines is \(\gamma=63.43^\circ-26.57^\circ\approx 36.87^\circ\). 3. The bisecting line has angle \(\alpha_k=\frac{63.4349^\circ}{2}\approx 31.7175^\circ\). 4. Therefore, \(m_k=\tan(31.7175^\circ)\approx 0.62\).

a) \(\alpha_g\approx 26.57^\circ\) and \(\alpha_h\approx 63.43^\circ\) b) \(\gamma\approx 36.87^\circ\) c) \(m_k\approx 0.62\)

- Identify the right triangle formed by the face diagonal, edge, and space diagonal. - Express the face diagonal using the edge length. - Choose a trigonometric ratio using the opposite and adjacent sides.

1. The face diagonal is \(d_f = 4\sqrt{2}\,\text{cm}\). 2. In the right triangle, the side opposite \(\alpha\) is \(4\,\text{cm}\), and the adjacent side is \(4\sqrt{2}\,\text{cm}\). 3. Therefore, \(\tan(\alpha) = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}\). 4. \(\alpha = \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \approx 35.26^{\circ}\).

\(\alpha \approx 35.26^{\circ}\)

- Express the space diagonal in terms of a general edge length \(s\). - Use cosine with the adjacent edge and the space diagonal. - Check what happens to the ratio when every length is multiplied by the same scale factor.

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. \(\gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 54.7^{\circ}\).

The angle is the same in both cubes and is approximately \(54.7^{\circ}\).

- Find the space diagonal using the three edge lengths. - In the right triangle containing edge \(a\) and the space diagonal, identify the adjacent side and hypotenuse. - Use cosine.

1. The space diagonal is \(d_s = \sqrt{15^2 + 8^2 + 6^2} = \sqrt{325}\,\text{cm}\). 2. In the relevant right triangle, edge \(a = 15\,\text{cm}\) is adjacent to \(\beta\), and the space diagonal is the hypotenuse. 3. Therefore, \(\cos(\beta) = \frac{15}{\sqrt{325}}\). 4. \(\beta = \cos^{-1}\left(\frac{15}{\sqrt{325}}\right) \approx 33.69^{\circ}\).

\(\beta \approx 33.69^{\circ}\)

- Find the diagonal of the square base. - The angle is between the space diagonal and the base diagonal. - Use tangent with the height as the opposite side.

1. The diagonal of the square base is \(d_f = 10\sqrt{2}\,\text{cm}\). 2. In the right triangle, \(d_f\) is adjacent to the \(40^{\circ}\) angle, and height \(c\) is opposite. 3. Thus, \(\tan(40^{\circ}) = \frac{c}{10\sqrt{2}}\). 4. Therefore, \(c = 10\sqrt{2}\tan(40^{\circ}) \approx 11.87\,\text{cm}\).

\(c \approx 11.87\,\text{cm}\)

- First find the missing side with the Pythagorean theorem. - For each angle, identify the opposite leg, adjacent leg, and hypotenuse. - Pay attention to which acute angle is being used.

1. 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 \(a\), so \(\sin(\alpha)=\frac{2.4}{2.5}=0.96\). 3. Relative to \(\gamma\) at \(C\), the adjacent leg is \(a\), so \(\cos(\gamma)=\frac{2.4}{2.5}=0.96\). 4. Also, \(\tan(\alpha)=\frac{a}{c}=\frac{2.4}{0.7}\approx 3.43\).

a) \(b=2.5\,\text{cm}\) b) \(\sin(\alpha)=0.96\), \(\cos(\gamma)=0.96\), and \(\tan(\alpha)\approx 3.43\)

- Use the Pythagorean theorem to find the hypotenuse. - Identify the opposite and adjacent legs relative to \(\rho\). - Substitute the side lengths into the sine, cosine, and tangent ratios.

1. 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 \(r=21\), the adjacent leg is \(p=20\), and the hypotenuse is \(q=29\). 3. Therefore, \(\sin(\rho)=\frac{21}{29}\approx 0.724\), \(\cos(\rho)=\frac{20}{29}\approx 0.690\), and \(\tan(\rho)=\frac{21}{20}=1.05\).

a) \(q=29\,\text{cm}\) b) \(\sin(\rho)=\frac{21}{29}\approx 0.724\), \(\cos(\rho)=\frac{20}{29}\approx 0.690\), and \(\tan(\rho)=\frac{21}{20}=1.05\)

- Use the Pythagorean theorem for the diagonal. - Identify the side opposite \(\alpha\) and the hypotenuse. - Use an inverse trigonometric function, then find the complementary angle.

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 \(\beta\approx 17.0^\circ\)

- Identify the hypotenuse. - Find the other acute angle from the complementary-angle relationship. - Choose ratios that use the given leg and 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 \(\tan(50^\circ)=\frac{b}{5}\), \(b=5\tan(50^\circ)\approx 5.96\,\text{cm}\).

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

- An angle inscribed in a semicircle is a right angle. - Use the Pythagorean theorem to find the missing leg. - Use a trigonometric ratio involving the adjacent side and hypotenuse.

1. The angle opposite the diameter is \(90^{\circ}\), so the diameter is the hypotenuse. 2. By the Pythagorean theorem, \(b = \sqrt{10^2 - 4^2} = \sqrt{84} \approx 9.17\,\text{cm}\). 3. Relative to \(\beta\), the \(4\,\text{cm}\) leg is adjacent and the \(10\,\text{cm}\) side is the hypotenuse. Thus, \(\cos(\beta) = \frac{4}{10} = 0.4\). 4. Therefore, \(\beta = \cos^{-1}(0.4) \approx 66.4^{\circ}\).

\(b \approx 9.17\,\text{cm}\) \(\beta \approx 66.4^{\circ}\)

- What type of triangle is formed when one side is a circle's diameter? - Use sine and cosine with the hypotenuse to find the two legs.

1. An angle inscribed in a semicircle is a right angle, so the diameter is the triangle's hypotenuse. 2. Relative to the \(40^\circ\) angle, \(a\) is opposite and \(b\) is adjacent. 3. Use sine: \(a=8\sin(40^\circ)\,\text{cm}\approx 5.14\,\text{cm}\). 4. Use cosine: \(b=8\cos(40^\circ)\,\text{cm}\approx 6.13\,\text{cm}\).

\(a\approx 5.14\,\text{cm}\) \(b\approx 6.13\,\text{cm}\)

- Identify the right triangle containing the height and side \(b\). - Use sine to find the opposite leg. - Adjacent angles in a parallelogram are supplementary.

1. In the right triangle formed by side \(b\) 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 \(\beta = 180^{\circ} - 50^{\circ} = 130^{\circ}\).

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

- A rhombus is a parallelogram with four congruent sides. - Use the right triangle formed by the height and a side. - Adjacent angles in a parallelogram are supplementary.

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 \(\beta = 180^{\circ} - 41.81^{\circ} \approx 138.19^{\circ}\).

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

- Draw the altitude from the vertex to the base. - What special triangles does the altitude create? - In one right triangle, identify the side that is half the base. - Use sine for the height and cosine for the base.

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: \(b \approx 5.47\,\text{cm}\)

- The altitude creates two congruent right triangles. What are their leg lengths? - Which theorem finds the hypotenuse? - Find half the vertex angle first.

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. Thus, \(\tan\left(\frac{\gamma}{2}\right)=\frac{5}{4}\). 4. Therefore, \(\gamma=2\tan^{-1}\left(\frac{5}{4}\right)\approx 102.68^\circ\).

The congruent side length is \(s\approx 6.40\,\text{cm}\), and the vertex angle is \(\gamma\approx 102.68^\circ\).

- Draw two altitudes to divide the trapezoid into a rectangle and two right triangles. - Find the short horizontal leg of either right triangle. - Which trigonometric ratio uses the adjacent leg and hypotenuse? - Use the relationship between consecutive interior angles formed by parallel lines.

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

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

- Draw the two altitudes. - Use sine and cosine to find the height and one short horizontal leg. - Relate the two bases to the two equal horizontal legs.

1. In either right triangle along a side, \(\sin(50^{\circ}) = \frac{h}{5}\). Therefore, \(h = 5\sin(50^{\circ}) \approx 3.83\,\text{cm}\). 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: \(h \approx 3.83\,\text{cm}\)

- Use the observer, the balloon's center, and a tangent point to identify a right triangle. - Which part of the visual angle appears in one right triangle? - Which trigonometric ratio relates the radius and the center distance? - Use degree mode.

1. Half the visual angle is \(\frac{1.5^\circ}{2}=0.75^\circ\). 2. A tangent sight line and the radius to the tangent point form a right triangle. The radius is opposite the half-angle, and the center distance \(d\) is the hypotenuse. 3. Thus, \(\sin(0.75^\circ)=\frac{12}{d}\). 4. Solve: \(d=\frac{12}{\sin(0.75^\circ)}\approx 916.76\,\text{m}\).

The observer is approximately \(916.76\,\text{m}\) from the balloon's center.

- Use sine and cosine in the smaller right triangle formed by the altitude. - Which segment is adjacent to \(\angle A\)? - Find the full hypotenuse, then subtract 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, \(c=\frac{6}{\cos(40^\circ)}\,\text{cm}\approx 7.83\,\text{cm}\). 4. Therefore, using unrounded values, \(p=c-q\approx 3.24\,\text{cm}\).

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

- First find \(\angle B\). - Apply sine and cosine in right triangle \(BCD\). - Use tangent in right triangle \(ACD\) to find 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 \(h\approx 7.37\,\text{cm}\)

- Use the hypotenuse and \(\angle B\) to find the adjacent leg. - In right triangle \(BCD\), use cosine to find \(p\) and sine to find \(h\). - The two hypotenuse segments add to \(c\).

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 \(p+q=c\), \(q=15-p\approx 8.80\,\text{cm}\).

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

- A diagonal divides a rectangle into two right triangles. - Use sine and cosine to find the two side lengths from the hypotenuse. - Multiply the side lengths to find the rectangle’s 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 \(42.87\,\text{cm}^2\).

- Use the Pythagorean theorem to find the third side. - Which trigonometric ratio uses only the two legs? - Use an inverse trigonometric function to find an angle from a side ratio.

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 \(\beta\approx 60.3^\circ\)

- Divide the trapezoid into a rectangle and a right triangle by drawing an altitude. - Which part of the lower base is equal to the upper base?

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, \(AD=AE+ED\approx 6+5.14=11.14\,\text{cm}\).

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

- Which right triangle has enough information to find a side first? - Identify the hypotenuse in each right triangle. - Choose the trigonometric ratio that relates each given angle to the known and unknown sides.

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

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

- Use the altitude to separate the figure into two right triangles. - First find \(AC\) from the left triangle. - Which side ratio in the right triangle \(ACD\) gives the tangent of \(\alpha\)?

1. In right triangle \(ABC\), \(AC=\sqrt{13^2-5^2}=\sqrt{144}=12\,\text{cm}\). 2. In right triangle \(ACD\), \(\tan(\alpha)=\frac{CD}{AC}=\frac{9}{12}=0.75\). 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 divides the figure into two right triangles. - Find each section of the base separately. - Use tangent to relate the altitude to each base segment.

1. In right triangle \(ABC\), \(BC=8\tan(30^\circ)\,\text{cm}\approx 4.62\,\text{cm}\). 2. In right triangle \(ACD\), \(CD=8\tan(45^\circ)=8\,\text{cm}\). 3. Therefore, \(BD=BC+CD\approx 4.62+8=12.62\,\text{cm}\).

\(x\approx 12.62\,\text{cm}\)

- 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 \(r\approx 4.83\,\text{cm}\).

- Find the central angle of a regular pentagon. - Bisect one of the isosceles central triangles to form a right triangle. - Use that right triangle to find half the side and the apothem.

1. The central angle of a regular pentagon is \(\frac{360^{\circ}}{5} = 72^{\circ}\). Bisecting the central triangle gives an angle of \(36^{\circ}\). 2. For half the side, \(\sin(36^{\circ}) = \frac{a/2}{10}\). Thus, \(a = 20\sin(36^{\circ}) \approx 11.76\,\text{cm}\). 3. For the apothem, \(\cos(36^{\circ}) = \frac{r}{10}\). Thus, \(r = 10\cos(36^{\circ}) \approx 8.09\,\text{cm}\).

Side length: \(a \approx 11.76\,\text{cm}\) Apothem: \(r \approx 8.09\,\text{cm}\)

- Use sine and cosine to find the two legs before calculating the area.

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 trigonometric ratio relates the two legs of a right triangle?

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

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

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

1. For Route A, \(\tan(\alpha)=\frac{450}{1200}\), so \(\alpha\approx 20.56^\circ\). 2. Its length is \(L_A=\sqrt{1200^2+450^2}\approx 1281.60\,\text{m}\). 3. The horizontal run of Route B is \(2.5\,\text{cm}\cdot 50{,}000=125{,}000\,\text{cm}=1250\,\text{m}\). 4. For Route B, \(\tan(\beta)=\frac{450}{1250}\), so \(\beta\approx 19.80^\circ\). 5. Its length is \(L_B=\sqrt{1250^2+450^2}\approx 1328.53\,\text{m}\). 6. Since \(1328.53>1281.60\), Route B is longer.

a) Route A has an angle of incline of approximately \(20.56^\circ\), and Route B has an angle of incline of approximately \(19.80^\circ\). b) Route B is longer. Its length is approximately \(1328.53\,\text{m}\), compared with approximately \(1281.60\,\text{m}\) for Route A.

- 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 \(ACD\), \(\tan(\alpha)=\frac{CD}{AD}=\frac{2q}{4q}=\frac{1}{2}=0.5\).

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

- 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.

- Convert all lengths to the same unit before calculating. - What happens to the angle when the horizontal run decreases but the rise stays fixed? - Use rise over horizontal run to determine the angle.

1. Convert the rise: \(72\,\text{cm}=0.72\,\text{m}\). Let \(x\) be the horizontal distance. Then \(\tan(3.5^\circ)=\frac{0.72}{x}\), so \(x=\frac{0.72}{\tan(3.5^\circ)}\approx 11.77\,\text{m}\). 2. Since \(\tan(3.5^\circ)\approx 0.06116\), the percent grade is approximately \(6.12\%\). 3. With a \(10\,\text{m}\) horizontal run, \(\tan(\theta)=\frac{0.72}{10}=0.072\). 4. Therefore, \(\theta=\tan^{-1}(0.072)\approx 4.12^\circ\). Since \(4.12^\circ>3.5^\circ\), the design is not allowed.

a) The minimum horizontal distance is approximately \(11.77\,\text{m}\). b) The ramp grade is approximately \(6.12\%\). c) The angle would be approximately \(4.12^\circ\). This design is not allowed because it exceeds \(3.5^\circ\).

- 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 determine \(\alpha\) and \(\gamma\). - Draw the altitude to create a right triangle. - Use cosine in part a and tangent in part b.

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, \(s=\frac{5}{\cos 72^\circ}\,\text{cm}\approx 16.18\,\text{cm}\). 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 \(\alpha=\arctan 2\approx 63.43^\circ\). Since \(2\alpha<180^\circ\), such a triangle exists.

a) \(s=\frac{5}{\cos 72^\circ}\,\text{cm}\approx 16.18\,\text{cm}\) b) Yes; \(\alpha=\arctan 2\approx 63.43^\circ\).

- Recall that tangent is opposite over adjacent. - In the seked ratio, identify the horizontal and vertical quantities. - What happens to a fraction when its denominator doubles and its numerator stays fixed?

1. Let \(x\) be the horizontal run and \(y\) the vertical rise, measured in the same units. The seked is \(S=\frac{x}{y}\cdot 7\). 2. The tangent of the face angle is \(\tan(\alpha)=\frac{y}{x}\). 3. Rearranging the seked equation 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, \(\alpha=\tan^{-1}\left(\frac{4}{3}\right)\approx 53.13^\circ\). 5. With fixed base dimensions, the horizontal run is constant. Doubling the height doubles the denominator in \(S=\frac{x}{y}\cdot 7\), so the seked is cut in half.

a) Since \(S=\frac{x}{y}\cdot 7\), rearranging gives \(\frac{y}{x}=\frac{7}{S}\). Because \(\tan(\alpha)=\frac{y}{x}\), \(\tan(\alpha)=\frac{7}{S}\). b) \(\alpha\approx 53.13^\circ\) c) The seked is halved because it is inversely proportional to the height when the base remains fixed.

- Identify the sides of the right triangle relative to \(\beta\). - Which side is the hypotenuse when the right angle is at \(A\)? - Rewrite \(d=50s\) as a ratio of side 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 for \(d\) gives \(d=\frac{s}{\cos(\beta)}\). 3. For \(s=10\,\text{km}\) and \(\beta=88^\circ\), \(d=\frac{10}{\cos(88^\circ)}\approx 286.54\,\text{km}\). 4. If \(d=50s\), then \(\cos(\beta)=\frac{s}{50s}=\frac{1}{50}\). 5. Therefore, \(\beta=\cos^{-1}\left(\frac{1}{50}\right)\approx 88.85^\circ\).

a) \(d=\frac{s}{\cos(\beta)}\) b) \(d\approx 286.54\,\text{km}\) c) \(\beta\approx 88.85^\circ\)

- Express \(\frac{c}{b}\) as the reciprocal of a trigonometric ratio. - Evaluate the expression at both angles and compare by division. - Consider what happens to cosine near \(90^\circ\).

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.

- Model a cross section of the ball as a circle. - Draw a line of sight tangent to the circle and identify the right triangle. - The two tangent lines split the visual angle into two equal angles. - Relate the radius, the center-to-eye distance, and half the visual angle.

1. The basketball has radius \(r = \frac{9.5}{2} = 4.75\,\text{in.}\). 2. A line of sight tangent to the circular cross section forms a right triangle with the player’s eye and the center of the ball. The angle at the eye is half of \(\alpha\). 3. Therefore, \(\sin\left(\frac{\alpha}{2}\right) = \frac{4.75}{32}\). 4. Solve for the full angle: \(\alpha = 2\sin^{-1}\left(\frac{4.75}{32}\right) \approx 17.07^{\circ}\). 5. To the nearest tenth, \(\alpha \approx 17.1^{\circ}\).

The visual angle is approximately \(17.1^{\circ}\).

- Divide the trapezoid into a rectangle and two congruent right triangles. - Use the stated side relationships to find the base angle of an outer triangle. - Find the acute angle that each diagonal makes with the longer base. - Use symmetry and the triangle angle sum.

1. Let the shorter base and each leg have length \(c\). Then the longer base has length \(2c\), and each overhang along the longer base has length \(\frac{c}{2}\). 2. In either outer right triangle, \(\cos(\alpha) = \frac{c/2}{c} = \frac{1}{2}\), so \(\alpha = 60^{\circ}\). The height is \(h = c\sin(60^{\circ}) = \frac{\sqrt{3}}{2}c\). 3. A diagonal has horizontal run \(\frac{3}{2}c\), so its acute angle with the base satisfies \(\tan(\theta) = \frac{(\sqrt{3}/2)c}{(3/2)c} = \frac{\sqrt{3}}{3}\). Thus, \(\theta = 30^{\circ}\). 4. By symmetry, the two diagonals make equal \(30^{\circ}\) angles with the base. The marked obtuse angle is \(\epsilon = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}\).

\(\epsilon = 120^{\circ}\)