Geometry Right Triangle Trigonometry Test Solutions and Scoring (PDF)
This right triangles test answers for Geometry covers right triangle trigonometry (Common Core G-SRT.C.6, G-SRT.C.7, G-SRT.C.8), with a free PDF to print.
Geometry · Common Core State Standards: G-SRT.C.6, G-SRT.C.7, G-SRT.C.8
Right triangles test answers: worked solutions
Question 1 – Find a side [2 pts]
\(\text{sin}\,40^{\circ} = \frac{x}{18}\)
\(x = 18 \cdot 0.6428 = 11.6\)
x = 11.6
Scoring: 1 point for the ratio, 1 point for x.
Question 2 – Find an angle [2 pts]
\(\text{tan}\,A = \frac{9}{14}\)
\(A = \text{tan}^{-1}\left( \frac{9}{14} \right) = 32.7^{\circ}\)
32.7°
Scoring: 1 point for the ratio, 1 point for the angle.
Question 3 – Three ratios [3 pts]
(a) \(\sqrt{20^{2} + 21^{2}} = \sqrt{841} = 29\).
(b) \(\text{sin}\,A = \frac{20}{29}\), \(\text{cos}\,A = \frac{21}{29}\), \(\text{tan}\,A = \frac{20}{21}\).
Scoring: 1 point for the hypotenuse, 2 points for the three ratios (deduct 1 for each error).
Question 4 – Complementary [2 pts]
\(\text{cos}\,(90^{\circ} – A) = \text{sin}\,A = \frac{8}{17}\).
The adjacent side is \(\sqrt{17^{2} – 8^{2}} = 15\), so \(\text{cos}\,A = \frac{15}{17}\).
Scoring: 1 point each.
Question 5 – Cliff height [2 pts]
\(\text{tan}\,35^{\circ} = \frac{h}{120}\)
\(h = 120 \cdot 0.7002 = 84\)
About 84 feet
Scoring: 1 point for the ratio, 1 point for the height.
Question 6 – Solve a right triangle [3 pts]
Other angle: \(90^{\circ} – 65^{\circ} = 25^{\circ}\).
Opposite the 65° angle: \(30 \cdot \text{sin}\,65^{\circ} = 27.2\).
Adjacent: \(30 \cdot \text{cos}\,65^{\circ} = 12.7\).
25°; legs 27.2 and 12.7
Scoring: 1 point for the angle, 1 point for each leg.
Question 7 – Special triangle [2 pts]
\(\text{leg} \cdot \sqrt{2} = 16\), so \(\text{leg} = \frac{16}{\sqrt{2}} = 8\sqrt{2}\).
8√2
Scoring: 1 point for the setup, 1 point for the radical.
Question 8 – Depression [2 pts]
\(\text{tan}\,8^{\circ} = \frac{60}{d}\), so \(d = \frac{60}{0.1405} = 427\).
About 427 feet
Scoring: 1 point for the setup, 1 point for the distance.
Question 9 – Ramp [2 pts]
\(\text{sin}\,A = \frac{4}{15}\), so \(A = \text{sin}^{-1}\left( \frac{4}{15} \right) = 15.5^{\circ}\).
15.5°
Scoring: 1 point for the ratio, 1 point for the angle.
Total score
20 points
Key facts behind this right triangles test answers
- sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent.
- sin A = cos (90 degrees minus A) for the acute angles of a right triangle.
- Use an inverse function (sin^-1, cos^-1, tan^-1) to find an angle.
- Solving a right triangle means finding all sides and acute angles.
- Angles of elevation and depression are measured from the horizontal.
How to use this right triangles test answers
Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.
All Geometry chapters: Geometry math.
