Right Triangle Trigonometry: Geometry Math Test Answer Key – PDF Download

Geometry Right Triangle Test Answer Key: questions 1–5

Question 1: Ratios in a right triangle (2 points)

1. \(\sin A=\frac{12}{37},\ \cos A=\frac{35}{37},\ \tan A=\frac{12}{35}\)
2. \(\sin B=\frac{35}{37},\ \cos B=\frac{12}{37}\)

Scoring: part (a) 1 point for all three correct ratios (accept 1 point only if all three are correct); part (b) 1 point for both correct.

Question 2: Similar triangles and sine (2 points)

1. Since \(\sin D=\sin A\), \(\frac{EF}{39}=\frac{10}{26}\). Then \(EF=\frac{10\cdot 39}{26}=15\). \(EF=15\)
2. The triangles are similar by AA, so corresponding sides are proportional. Then \(\frac{BC}{AB}=\frac{EF}{DE}\), and these are the sine of angles \(A\) and \(D\).

Scoring: (a) 1 point for a correct proportion and the value 15; (b) 1 point for naming similarity and proportional sides.

Question 3: Complementary angles (2 points)

1. The complement of \(17^{\circ}\) is \(73^{\circ}\). \(\sin 17^{\circ}=\cos 73^{\circ}\)
2. The angles are complementary: \((4x-6)+(x+11)=90\), so \(5x+5=90\) and \(x=17\). \(x=17\)

Scoring: (a) 1 point; (b) 1 point for the complementary equation and the correct value.

Question 4: Special right triangles (2 points)

1. The hypotenuse is the leg times \(\sqrt{2}\), so each leg is \(\frac{10\sqrt{2}}{\sqrt{2}}=10\). 10 in
2. The hypotenuse is \(2(8)=16\) and the longer leg is \(8\sqrt{3}\). \(16\text{ cm},\ 8\sqrt{3}\text{ cm}\)

Scoring: (a) 1 point; (b) 1 point for both lengths (accept 13.9 cm as an approximation of the long leg).

Question 5: Finding sides (2 points)

\(x\) is opposite and 14 is adjacent: \(x=14\tan 27^{\circ}\approx 7.1\).

\(y\) is the hypotenuse: \(y=\frac{14}{\cos 27^{\circ}}\approx 15.7\).

\(x\approx 7.1,\ y\approx 15.7\)

Scoring: 1 point for each correct side. Award 1 point total if a correct setup has a rounding or calculator-mode slip.

Geometry Right Triangle Test Answer Key: questions 6–9

Question 6: Finding angles (2 points)

1. \(\sin P=\frac{9}{13}\), so \(m\angle P=\sin^{-1}(\frac{9}{13})\approx 43.8^{\circ}\). \(43.8^{\circ}\)
2. \(\tan Q=\frac{15}{6}=2.5\), so \(m\angle Q=\tan^{-1}(2.5)\approx 68.2^{\circ}\). \(68.2^{\circ}\)

Scoring: 1 point for each angle, 1 point if the correct inverse function is set up but the answer is off by rounding.

Question 7: Solve a right triangle (3 points)

Hypotenuse: \(AB=\sqrt{14^{2}+9^{2}}=\sqrt{277}\approx 16.6\).

\(\tan A=\frac{14}{9}\), so \(m\angle A\approx 57.3^{\circ}\).

\(m\angle B=90^{\circ}-57.3^{\circ}=32.7^{\circ}\).

\(AB\approx 16.6,\ m\angle A\approx 57.3^{\circ},\ m\angle B\approx 32.7^{\circ}\)

Scoring: 1 point for each correct part. Accept a correct method with a rounding difference of 0.1.

Question 8: Angle of depression (3 points)

The angle of depression equals the angle of elevation from the boat (alternate interior angles), so the angle at the boat is \(21^{\circ}\).

\(\tan 21^{\circ}=\frac{90}{d}\), so \(d=\frac{90}{\tan 21^{\circ}}\approx 234.5\).

\(d\approx 234.5\text{ ft}\)

Scoring: 1 point for the angle at the boat (21 degrees); 1 point for the tangent equation; 1 point for the correct distance.

Question 9: Zip line (2 points)

Cable (hypotenuse): \(\sin 25^{\circ}=\frac{32}{L}\), so \(L=\frac{32}{\sin 25^{\circ}}\approx 75.7\).

Horizontal distance (adjacent): \(\tan 25^{\circ}=\frac{32}{x}\), so \(x=\frac{32}{\tan 25^{\circ}}\approx 68.6\).

\(\text{cable}\approx 75.7\text{ ft},\ \text{horizontal}\approx 68.6\text{ ft}\)

Scoring: 1 point for each correct measurement.

Scoring summary

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