Bluebook SAT Test Right Triangles and Trigonometry 6

Question 1

In right triangle ABC, the right angle is at C. The altitude from C to hypotenuse AB meets AB at point D. If AD = 4 and DB = 9, what is the perimeter of triangle ABC?

Accepted Answer

13 + 5√13

Answer

By the geometric mean (altitude-on-hypotenuse) theorem: CD = √(AD · DB) = √(4 · 9) = 6. Each leg equals the geometric mean of the hypotenuse and the adjacent segment: AC = √(AD · AB) = √(4 · 13) = 2√13, and BC = √(DB · AB) = √(9 · 13) = 3√13. AB = AD + DB = 13. Perimeter = 13 + 2√13 + 3√13 = 13 + 5√13. Choice D is a trap: √117 = 3√13, so 13 + √117 = 13 + 3√13, which is missing the AC term.

Question 2

In right triangle PQR with the right angle at R, sin(P) = 5/13. What is the value of cos(2P) − cos(2Q)?

Accepted Answer

238/169

Answer

Since the right angle is at R, angles P and Q are complementary: P + Q = 90°, so Q = 90° − P. Using the double-angle formula: cos(2P) = 1 − 2sin²(P) = 1 − 2(25/169) = 119/169. For cos(2Q): since 2Q = 180° − 2P, we get cos(2Q) = cos(180° − 2P) = −cos(2P) = −119/169. Therefore cos(2P) − cos(2Q) = 119/169 − (−119/169) = 238/169. Choice A (0) is the trap for students who assume complementary angles produce equal cosines of double angles.

Question 3

Square ABCD has vertices labeled in order (A, B, C, D consecutive) with side length 10. Point E is the midpoint of side BC. What is sin(∠DAE), where ∠DAE is the angle at vertex A in triangle DAE?

Accepted Answer

2√5/5

Answer

Place coordinates: A = (0, 10), B = (10, 10), C = (10, 0), D = (0, 0), E = (10, 5). Vector AD = (0, −10), |AD| = 10. Vector AE = (10, −5), |AE| = √(100 + 25) = 5√5. The area of triangle DAE by the cross-product method: (1/2)|0·(−5) − (−10)·10| = (1/2)(100) = 50. Then sin(∠DAE) = 2 · Area / (|AD| · |AE|) = 100 / (10 · 5√5) = 2/√5 = 2√5/5. Choice A (√5/5) is the cosine of the angle, a common reversal error.

Question 4

In right triangle XYZ with the right angle at Z and hypotenuse XY = 10, the sides satisfy 3 sin(X) = 4 cos(X). What is the length of side YZ?

Accepted Answer

8

Answer

The equation 3 sin(X) = 4 cos(X) gives tan(X) = 4/3, so this is a 3-4-5 scaled right triangle with sin(X) = 4/5 and cos(X) = 3/5. Side YZ is opposite angle X, so YZ = XY · sin(X) = 10 · (4/5) = 8. Side XZ = 10 · cos(X) = 6. Verify: 6² + 8² = 36 + 64 = 100 = 10². Choice A (6) is XZ, the leg adjacent to X — a common mix-up between opposite and adjacent.

Question 5

In right triangle ABC with the right angle at C, tan(A) = 2/3. If the area of triangle ABC is 54, what is the length of hypotenuse AB?

Accepted Answer

3√26

Answer

Let BC = 2k (opposite A) and AC = 3k (adjacent to A), since tan(A) = BC/AC = 2/3. Area = (1/2)(BC)(AC) = (1/2)(2k)(3k) = 3k² = 54, so k² = 18 and k = 3√2. Thus BC = 6√2 and AC = 9√2. By the Pythagorean theorem: AB² = BC² + AC² = (6√2)² + (9√2)² = 72 + 162 = 234. AB = √234 = √(9 · 26) = 3√26. Choice C (9√2) is the longer leg, not the hypotenuse.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Right Triangles and Trigonometry 6