Free Math Worksheets

- Recall the cosine value at \(90^\circ\). - What happens to a product when one factor is zero? - Which side is opposite the right angle?

1. \(\cos(90^\circ)=0\). 2. Substituting gives \(c^2=a^2+b^2-2ab(0)=a^2+b^2\), which is the Pythagorean theorem. 3. Since \(c\) is opposite \(\gamma\), when \(\gamma=90^\circ\), side \(c\) is the hypotenuse. The resulting equation is exactly the right-triangle relationship between the hypotenuse and legs.

1. \(\cos(90^\circ)=0\) 2. The Pythagorean theorem results: \(c^2=a^2+b^2\). 3. With \(\gamma=90^\circ\), \(c\) is the hypotenuse, so the formula matches right-triangle geometry.

- The given angle is between sides \(a\) and \(b\). - Use the Law of Cosines to find the side opposite \(\gamma\).

1. Use the Law of Cosines: \(c^2=a^2+b^2-2ab\cos(\gamma)\). 2. Substitute the given values: \(c^2=(8.5)^2+(12.0)^2-2\cdot 8.5\cdot 12.0\cos(42^\circ)\). 3. Thus, \(c=\sqrt{(8.5)^2+(12.0)^2-2\cdot 8.5\cdot 12.0\cos(42^\circ)}\approx 8.04\,\text{cm}\).

\(c\approx 8.04\,\text{cm}\)

- Use the Law of Cosines when all three side lengths are known. - Put the side opposite the target angle by itself on the left side of the formula.

1. Use the Law of Cosines: \(c^2=a^2+b^2-2ab\cos(\gamma)\). 2. Substitute: \(7^2=5^2+8^2-2\cdot5\cdot8\cos(\gamma)\). 3. Simplify: \(49=89-80\cos(\gamma)\), so \(80\cos(\gamma)=40\). 4. Therefore, \(\cos(\gamma)=\frac{1}{2}=0.5\).

\(\cos(\gamma)=0.5\)

- Use the Law of Cosines with the two known sides and their included angle. - Remember that the cosine of an obtuse angle is negative.

1. Use the Law of Cosines: \(x^2=3^2+5^2-2\cdot3\cdot5\cos(120^{\circ})\). 2. Since \(\cos(120^{\circ})=-\frac{1}{2}\), \(x^2=9+25+15=49\). 3. Therefore, \(x=7\,\text{cm}\).

\(7\,\text{cm}\)

- Use the Law of Cosines when all three sides are known. - A negative cosine indicates that the angle is obtuse.

1. By the Law of Cosines, \(c^2=a^2+b^2-2ab\cos(\gamma)\). 2. Solve for the cosine: \(\cos(x)=\frac{3^2+5^2-7^2}{2\cdot3\cdot5}=-\frac{1}{2}\). 3. Therefore, \(x=\cos^{-1}\left(-\frac{1}{2}\right)=120^{\circ}\).

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

- Solve the Law of Cosines for the cosine of the angle opposite side \(c\). - Recall the sign of cosine for acute, right, and obtuse angles. - Compare \(a^2+b^2\) with \(c^2\).

1. Solve the Law of Cosines for \(\cos(\gamma)\): \(\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}\). 2. Substitute: \(\cos(\gamma)=\frac{6^2+9^2-13^2}{2\cdot6\cdot9}=\frac{36+81-169}{108}\). 3. Therefore, \(\cos(\gamma)=\frac{-52}{108}=-\frac{13}{27}\approx-0.4815\). 4. Since \(\cos(\gamma)<0\) for an angle between \(0^\circ\) and \(180^\circ\), \(\gamma\) is obtuse.

\(\cos(\gamma)=-\frac{13}{27}\approx-0.4815\). Because the cosine is negative, \(\gamma\) is obtuse.

- In an isosceles triangle, the two legs have equal lengths. - Substitute \(b=c=s\) into the Law of Cosines before calculating. - Pay attention to the sign of \(\cos(110^\circ)\).

1. Apply the Law of Cosines to the base: \(a^2=s^2+s^2-2s^2\cos(\alpha)\). 2. Factor the expression: \(a^2=2s^2(1-\cos(\alpha))\). 3. Substitute \(s=10\) and \(\alpha=110^\circ\): \(a=\sqrt{2(10)^2(1-\cos(110^\circ))}\approx16.38\,\text{cm}\). 4. In general, the simplified formula is \(a^2=2s^2(1-\cos(\alpha))\), or equivalently, \(a=s\sqrt{2(1-\cos(\alpha))}\).

a) \(a\approx16.38\,\text{cm}\) b) \(a^2=2s^2(1-\cos(\alpha))\), so \(a=s\sqrt{2(1-\cos(\alpha))}\).

- The largest angle is opposite the longest side. - Use the Law of Cosines because all three side lengths are known. - The sign of the cosine can help you classify the angle before finding its measure. - A triangle is obtuse when one interior angle is greater than \(90^\circ\).

1. The largest angle is \(\gamma\) because it is opposite the longest side, \(c=20\,\text{cm}\). 2. Solve the Law of Cosines for \(\cos(\gamma)\): \(\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}\). 3. Substitute: \(\cos(\gamma)=\frac{12^2+15^2-20^2}{2\cdot12\cdot15}=\frac{-31}{360}\). 4. Therefore, \(\gamma=\cos^{-1}\left(-\frac{31}{360}\right)\approx94.94^\circ\). 5. Since the largest angle is greater than \(90^\circ\), the triangle is obtuse.

The largest angle is \(\gamma\approx94.94^\circ\), so the triangle is obtuse.

- Use the distance formula to find the three side lengths. - Apply the Law of Cosines to the side opposite angle \(\alpha\). - Solve the resulting equation for \(\cos(\alpha)\).

1. Find the side lengths: \(AB=5\), \(AC=\sqrt{3^2+4^2}=5\), and \(BC=\sqrt{(3-5)^2+(4-0)^2}=\sqrt{20}\). 2. Apply the Law of Cosines with \(BC\) opposite \(\alpha\): \((\sqrt{20})^2=5^2+5^2-2\cdot5\cdot5\cos(\alpha)\). 3. Simplify: \(20=50-50\cos(\alpha)\), so \(50\cos(\alpha)=30\). 4. Therefore, \(\cos(\alpha)=\frac{30}{50}=0.6\).

\(\cos(\alpha)=0.6\)

- Draw the shorter diagonal and identify the triangle it forms. - Use the Law of Cosines with the two side lengths and their included angle. - Recall the exact value of \(\cos(60^{\circ})\).

1. The shorter diagonal forms a triangle with the two side lengths and the included \(60^{\circ}\) angle. 2. By the Law of Cosines, \(f^2=8^2+5^2-2\cdot8\cdot5\cos(60^{\circ})\). 3. Since \(\cos(60^{\circ})=\frac{1}{2}\), \(f^2=64+25-40=49\). 4. Therefore, \(f=7\,\text{cm}\).

\(7\,\text{cm}\)

- Use the Law of Cosines for the side opposite the given angle. - Recall that \((\sqrt{2})^2=2\). - Use the exact value of \(\cos(45^\circ)\).

1. Apply the Law of Cosines to side \(b\): \(b^2=a^2+c^2-2ac\cos(\beta)\). 2. Substitute: \(x^2=(4\sqrt{2})^2+7^2-2\cdot4\sqrt{2}\cdot7\cos(45^\circ)\). 3. Since \(\cos(45^\circ)=\frac{\sqrt{2}}{2}\), \(x^2=32+49-56\sqrt{2}\left(\frac{\sqrt{2}}{2}\right)=81-56=25\). 4. Because a side length is positive, \(x=5\).

\(x=5\)

- Find side \(b\) first with the Law of Cosines. - Compare the resulting side length with side \(a\) and use the isosceles triangle theorem.

1. Use the Law of Cosines to find \(b\): \(b^2=3^2+(3\sqrt{3})^2-2\cdot3\cdot3\sqrt{3}\cos(30^{\circ})\). 2. Since \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), \(b^2=9+27-27=9\), so \(b=3\). 3. Because \(a=b\), the opposite angles are congruent. Therefore, \(\alpha=\beta=30^{\circ}\).

\(x=30^{\circ}\)

- Use the diagonal to divide the quadrilateral into two triangles with three known side lengths. - Apply the Law of Cosines in each triangle. - The angles at \(B\) and \(D\) are sums of two smaller triangle angles.

1. In \(\triangle ABD\), the Law of Cosines gives \(\cos(\alpha)=\frac{15^2+13^2-14^2}{2\cdot15\cdot13}\), so \(\alpha\approx 59.5^{\circ}\). 2. In \(\triangle BCD\), \(\cos(\gamma)=\frac{12^2+10^2-14^2}{2\cdot12\cdot10}\), so \(\gamma\approx 78.5^{\circ}\). 3. At \(B\), \(\angle ABD\approx 53.1^{\circ}\) and \(\angle DBC\approx 44.4^{\circ}\). Therefore, \(\beta\approx 53.1^{\circ}+44.4^{\circ}=97.5^{\circ}\). 4. At \(D\), the two triangle angles are approximately \(67.4^{\circ}\) and \(57.1^{\circ}\). Therefore, \(\delta\approx 124.5^{\circ}\). 5. The angles total \(360.0^{\circ}\), as required for a quadrilateral.

\(\alpha\approx 59.5^{\circ}\) \(\beta\approx 97.5^{\circ}\) \(\gamma\approx 78.5^{\circ}\) \(\delta\approx 124.5^{\circ}\)

- The Law of Cosines leads to a quadratic equation in the unknown side. - Solve the quadratic, then use the given inequality to select the correct solution.

1. Apply the Law of Cosines: \(7^2=x^2+8^2-2\cdot x\cdot8\cos(60^{\circ})\). 2. Simplify: \(49=x^2+64-8x\), so \(x^2-8x+15=0\). 3. Factor: \((x-3)(x-5)=0\), giving \(x=3\) or \(x=5\). 4. Because \(x>4\), the required length is \(x=5\).

\(x=5\)

- Find the third side with the Law of Cosines. - Then use the Law of Sines to find the sine of the target angle. - Compare the opposite side lengths to choose the correct inverse-sine solution.

1. First use the Law of Cosines to find \(c\): \(c^2=2^2+(1+\sqrt{3})^2-2\cdot2(1+\sqrt{3})\cos(60^{\circ})=6\). Thus, \(c=\sqrt{6}\). 2. By the Law of Sines, \(\frac{\sin(\alpha)}{2}=\frac{\sin(60^{\circ})}{\sqrt{6}}\), so \(\sin(\alpha)=\frac{1}{\sqrt{2}}\). 3. This gives \(\alpha=45^{\circ}\) or \(135^{\circ}\). Since \(a=2<c=\sqrt{6}\), the opposite angles satisfy \(\alpha<\gamma=60^{\circ}\). Therefore, \(\alpha=45^{\circ}\).

\(x=45^{\circ}\)