Problem 1: Area from two sides and an angle
The angle is between the two given sides, so \(\text{Area}=\frac{1}{2}ab\sin C\).
\(\text{Area}=\frac{1}{2}(6)(9)\sin 30^{\circ}=27(0.5)=13.5\).
13.5 square units
Problem 2: Area, decimal answer
\(\text{Area}=\frac{1}{2}(12)(15)\sin 50^{\circ}=90\sin 50^{\circ}\approx 68.9\).
\(\approx 68.9\text{ cm}^{2}\)
Problem 3: Area with an obtuse angle
The formula works for obtuse angles. \(\text{Area}=\frac{1}{2}(7.5)(10)\sin 115^{\circ}=37.5\sin 115^{\circ}\approx 34.0\).
\(\approx 34.0\text{ in}^{2}\)
Problem 4: Angle from the area
\(60=\frac{1}{2}(10)(15)\sin C=75\sin C\), so \(\sin C=\frac{60}{75}=0.8\).
\(C=\sin^{-1}(0.8)\approx 53.1^{\circ}\).
\(m\angle C\approx 53.1^{\circ}\)
Problem 5: Law of Sines for a side
\(\frac{b}{\sin 62^{\circ}}=\frac{10}{\sin 48^{\circ}}\), so \(b=\frac{10\sin 62^{\circ}}{\sin 48^{\circ}}\approx 11.9\).
\(b\approx 11.9\)
Problem 6: Third angle, then two sides
\(A=180^{\circ}-35^{\circ}-80^{\circ}=65^{\circ}\).
\(a=\frac{24\sin 65^{\circ}}{\sin 80^{\circ}}\approx 22.1\) and \(b=\frac{24\sin 35^{\circ}}{\sin 80^{\circ}}\approx 14.0\).
\(A=65^{\circ},\ a\approx 22.1,\ b\approx 14.0\)
Problem 7: Law of Sines for an angle
\(\frac{\sin B}{12}=\frac{\sin 70^{\circ}}{15}\), so \(\sin B=\frac{12\sin 70^{\circ}}{15}\approx 0.7518\).
\(B=\sin^{-1}(0.7518)\approx 48.7^{\circ}\).
\(m\angle B\approx 48.7^{\circ}\)
Problem 8: Law of Cosines, SAS
\(c^{2}=7^{2}+12^{2}-2(7)(12)\cos 60^{\circ}=49+144-168(0.5)=109\).
\(c=\sqrt{109}\approx 10.4\).
\(c\approx 10.4\)
Problem 9: Law of Cosines, obtuse angle
\(a^{2}=13^{2}+9^{2}-2(13)(9)\cos 112^{\circ}\). Since \(\cos 112^{\circ}\approx -0.3746\) is negative, subtracting it adds: \(a^{2}\approx 250+87.66\approx 337.7\).
\(a\approx 18.4\).
\(a\approx 18.4\)
Problem 10: Law of Cosines, SSS
The largest angle is opposite the longest side, 12.
\(\cos C=\frac{7^{2}+9^{2}-12^{2}}{2(7)(9)}=\frac{-14}{126}\approx -0.1111\).
\(C=\cos^{-1}(-0.1111)\approx 96.4^{\circ}\).
\(\approx 96.4^{\circ}\)
Problem 11: All three angles
Let \(a=13\), \(b=14\) and \(c=15\).
\(\cos A=\frac{14^{2}+15^{2}-13^{2}}{2(14)(15)}=\frac{252}{420}=0.6\), so \(A\approx 53.1^{\circ}\).
\(\cos B=\frac{13^{2}+15^{2}-14^{2}}{2(13)(15)}=\frac{198}{390}\), so \(B\approx 59.5^{\circ}\).
\(C=180^{\circ}-53.1^{\circ}-59.5^{\circ}\approx 67.4^{\circ}\) (check with \(\cos C=\frac{140}{364}\)).
\(A\approx 53.1^{\circ},\ B\approx 59.5^{\circ},\ C\approx 67.4^{\circ}\)
Problem 12: Choose the law
- Two sides and the included angle (SAS): Law of Cosines, to find \(c\).
- Two angles and a side (AAS): Law of Sines, since \(a\) and \(A\) are a known pair.
- Three sides (SSS): Law of Cosines, solved for an angle.
- Two sides and a non-included angle (SSA): Law of Sines, and check whether two triangles exist.
Problem 13: Cosines and the Pythagorean theorem
Since \(\cos 90^{\circ}=0\), \(c^{2}=a^{2}+b^{2}-2ab(0)=a^{2}+b^{2}\).
With \(a=6\) and \(b=8\): \(c^{2}=36+64=100\), so \(c=10\).
\(c^{2}=a^{2}+b^{2}\)
Problem 14: Proof of the area formula
The reasons are shown below.
| Step | Statements | Reasons |
|---|---|---|
| 1 | Draw altitude \(\overline{AD}\) to \(\overline{BC}\), with \(AD=h\) | Through a point not on a line, there is exactly one perpendicular to the line |
| 2 | \(\text{Area}=\frac{1}{2}ah\) | Area of a triangle is one half base times height |
| 3 | \(\sin C=\frac{h}{b}\) | Definition of sine in right triangle ADC |
| 4 | \(h=b\sin C\) | Multiplication property of equality (multiply both sides by \(b\)) |
| 5 | \(\text{Area}=\frac{1}{2}ab\sin C\) | Substitution (replace \(h\) in step 2 with \(b\sin C\)) |
Problem 15: Area after the Law of Sines
\(C=180^{\circ}-40^{\circ}-75^{\circ}=65^{\circ}\). Find \(b\) with the Law of Sines: \(b=\frac{12\sin 75^{\circ}}{\sin 40^{\circ}}\approx 18.03\).
Sides \(a\) and \(b\) have included angle \(C\): \(\text{Area}=\frac{1}{2}(12)(18.03)\sin 65^{\circ}\approx 98.1\).
\(\approx 98.1\) square units
Problem 16: Area of a parallelogram
Each triangle has area \(\frac{1}{2}(14)(9)\sin 65^{\circ}\). The parallelogram is two triangles: \(2\cdot\frac{1}{2}(14)(9)\sin 65^{\circ}=126\sin 65^{\circ}\approx 114.2\).
\(\approx 114.2\) square units
Problem 17: Distance across a pond
Two sides and the included angle are known, so use the Law of Cosines.
\(AB^{2}=420^{2}+350^{2}-2(420)(350)\cos 70^{\circ}\approx 298{,}900-100553.9\approx 198346\).
\(AB\approx 445\).
About 445 ft
Problem 18: Two boats leave a dock
The paths are two sides and the included angle of a triangle.
\(d^{2}=18^{2}+24^{2}-2(18)(24)\cos 70^{\circ}\approx 900-295.5\approx 604.5\).
\(d\approx 24.6\).
\(\approx 24.6\text{ mi}\)
Problem 19: Sod for a triangular lot
\(\text{Area}=\frac{1}{2}(120)(85)\sin 74^{\circ}=5100\sin 74^{\circ}\approx 4902.4\), which rounds to 4902 square feet.
Cost: \(4902\times 0.45=2205.90\).
4902 ft²; $2,205.90
Problem 20: Locating a fire
The angle at the fire is \(C=180^{\circ}-52^{\circ}-71^{\circ}=57^{\circ}\). Side \(c=12\) is opposite \(C\).
Distance from A (side \(b\), opposite \(B\)): \(b=\frac{12\sin 71^{\circ}}{\sin 57^{\circ}}\approx 13.5\).
Distance from B (side \(a\)): \(a=\frac{12\sin 52^{\circ}}{\sin 57^{\circ}}\approx 11.3\).
\(\approx 13.5\text{ mi from A},\ \approx 11.3\text{ mi from B}\)
Problem 21: Plane with a turn
The turn of \(40^{\circ}\) leaves an interior angle of \(180^{\circ}-40^{\circ}=140^{\circ}\) in the triangle.
\(d^{2}=150^{2}+90^{2}-2(150)(90)\cos 140^{\circ}=30{,}600+20683.2\approx 51283.2\).
\(d\approx 226.5\).
\(\approx 226.5\text{ mi}\)
Problem 22: Garden angle and area
The largest angle is opposite 22 ft: \(\cos C=\frac{14^{2}+18^{2}-22^{2}}{2(14)(18)}=\frac{36}{504}\approx 0.0714\), so \(C\approx 85.9^{\circ}\).
The sides 14 and 18 enclose \(C\): \(\text{Area}=\frac{1}{2}(14)(18)\sin 85.9^{\circ}=126\sin 85.9^{\circ}\approx 125.7\).
\(C\approx 85.9^{\circ},\ \text{area}\approx 125.7\text{ ft}^{2}\)
See every chapter on the Geometry math page.