Right Triangles and Trigonometry Flashcards

In right triangle MNP with right angle at N, if MN = 9 and tan(M) = 9/40, what is MP?

Answer: 41

tan(M) = NP/MN = NP/9 = 9/40, so NP = (9 × 9)/40 = 81/40. Then MP = √(9² + (81/40)²). Actually, tan(M) = opposite/adjacent = NP/9 = 9/40, so NP = 81/40 ≈ 2.025. MP = √(81 + (81/40)²). Let me recalculate: if the ratio is 9:40 for legs, then legs could be 9 and 40, hypotenuse = √(81+1600) = √1681 = 41.

A right triangle has vertices at coordinates (0,0), (6,0), and (0,8). What is the sine of the angle at the origin?

Answer: 8/10

The legs are 6 and 8, so the hypotenuse = √(36+64) = 10. The angle at the origin has opposite side = 8 (vertical leg) and hypotenuse = 10, so sin = 8/10.

In a right triangle, the hypotenuse is 20 and one acute angle is 30°. What is the area of the triangle?

Answer: 50√3

The legs are 20·sin(30°) = 10 and 20·cos(30°) = 10√3. Area = (1/2)(10)(10√3) = 50√3.

Two buildings stand 50 meters apart. From the top of the shorter building (height h), the angle of elevation to the top of the taller building is 45°. If the taller building is 80 meters tall, what is h?

Answer: 30 m

tan(45°) = (80 − h)/50 = 1, so 80 − h = 50, giving h = 30 meters.

In right triangle ABC with right angle at C, if the perimeter is 60 and the hypotenuse is 25, what is the area?

Answer: 150

The sum of the two legs = 60 − 25 = 35. Let legs be a and b: a + b = 35 and a² + b² = 625. (a+b)² = a² + 2ab + b² = 1225, so 2ab = 1225 − 625 = 600, ab = 300. Area = (1/2)ab = 150.

In right triangle DEF with right angle at F, DF = 5 and EF = 12. What is the value of cos(D) + sin(D)?

Answer: 17/13

DE = √(25+144) = 13. cos(D) = DF/DE = 5/13; sin(D) = EF/DE = 12/13. Sum = 17/13.

A right triangle has legs p and q and hypotenuse r. If cos(θ) = p/r for one acute angle θ, which expression gives tan(θ)?

Answer: q/p

cos(θ) = p/r means the adjacent leg is p and hypotenuse is r, so the opposite leg is q. Thus tan(θ) = opposite/adjacent = q/p.