Laws of Sines and Cosines and Triangle Area: Precalculus Math Test Answer Key – PDF Download

Precalculus Law Sines Test Answer Key: questions 1–5

Question 1: Area with sine (2 points)

\(\text{Area}=\frac{1}{2}(14)(9)\sin 52^{\circ}=63\sin 52^{\circ}\approx \)\(49.6\text{ cm}^{2}\).

Scoring: 1 point for the correct formula and substitution, 1 point for the correct value.

Question 2: Law of Sines (3 points)

1. \(C=58^{\circ}\)
2. \(b=\frac{15\sin 84^{\circ}}{\sin 38^{\circ}}\approx \)\(24.23\)
3. \(c=\frac{15\sin 58^{\circ}}{\sin 38^{\circ}}\approx \)\(20.66\)

Scoring: 1 point for each part.

Question 3: Ambiguous case (3 points)

1. \(h=20\sin 36^{\circ}\approx 11.76\) and \(11.76< 14< 20\), so two triangles exist
2. \(B_{1}\approx 57.1^{\circ}\) and \(B_{2}\approx 122.9^{\circ}\)
3. \(c_{1}=\frac{14\sin 86.9^{\circ}}{\sin 36^{\circ}}\approx \)\(23.78\) and \(c_{2}=\frac{14\sin 21.1^{\circ}}{\sin 36^{\circ}}\approx \)\(8.58\)

Scoring: 1 point for each part.

Question 4: Law of Cosines: find a side (2 points)

\(c^{2}=13^{2}+19^{2}-2(13)(19)\cos 47^{\circ}=530-494\cos 47^{\circ}\approx 193.09\), so \(c\approx 13.90\).

Scoring: 1 point for the correct substitution, 1 point for the correct value.

Question 5: Law of Cosines: find an angle (2 points)

The largest angle is opposite 15: \(\cos\theta=\frac{10^{2}+13^{2}-15^{2}}{2(10)(13)}=\frac{44}{260}\approx 0.1692\), so \(\theta\approx 80.3^{\circ}\).

Scoring: 1 point for choosing the angle opposite 15 and substituting correctly, 1 point for the correct angle.

Precalculus Law Sines Test Answer Key: questions 6–8

Question 6: Heron's formula (2 points)

\(s=\frac{17+25+28}{2}=35\). \(\text{Area}=\sqrt{35(18)(10)(7)}=\sqrt{44{,}100}=\)210 square units.

Scoring: 1 point for \(s=35\) and the substitution, 1 point for the area.

Question 7: Ships and bearings (3 points)

1. \(145^{\circ}-40^{\circ}=\)\(105^{\circ}\)
2. \(d^{2}=25^{2}+30^{2}-2(25)(30)\cos 105^{\circ}\approx 1913.23\), so \(d\approx 43.74\text{ mi}\)

Scoring: 1 point for the angle; 1 point for the correct Law of Cosines setup and 1 point for the correct distance.

Question 8: Lot area and height (3 points)

1. \(s=\frac{80+110+125}{2}=157.5\). \(\text{Area}=\sqrt{157.5(77.5)(47.5)(32.5)}\approx \)\(4,341\text{ ft}^{2}\)
2. \(\frac{1}{2}(125)h=4,341\), so \(h=\frac{2(4,341)}{125}\approx \)\(69.5\text{ ft}\)

Scoring: part (a) 1 point for \(s\) and the setup, 1 point for the area; part (b) 1 point.

Scoring summary

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