Trigonometry: Aptitude Questions with Answers

Q1. If tanθ + secθ = 2, find sinθ.

A. 4/5
B. 3/5
C. 12/13
D. 5/13

Answer: B

Solution:
Let k = tanθ + secθ = 2.
Use identity: secθ − tanθ = 1 / (tanθ + secθ) = 1/2.
Now, add: (tanθ + secθ) + (secθ − tanθ) = 2 + 1/2 = 5/2 = 2secθ ⇒ secθ = 5/4
So cosθ = 4/5 ⇒ sinθ = √(1 − cos²θ) = √(1 − 16/25) = √(9/25) = 3/5

Q2. Solve for θ ∈ (0°, 90°): sin(θ + 15°) = cos(2θ − 15°)

A. 20°
B. 25°
C. 30°
D. 35°

Answer: C

Solution:
cos(2θ − 15°) = sin(90° − (2θ − 15°)) = sin(105° − 2θ)
So, sin(θ + 15°) = sin(105° − 2θ)
⇒ θ + 15° = 105° − 2θ ⇒ 3θ = 90° ⇒ θ = 30°

Q3. If (1 + sinθ)/(1 − sinθ) = 9, find cosθ (acute θ).

A. 4/5
B. 3/5
C. 5/13
D. 12/13

Answer: B

Solution:
Cross-multiply: 1 + sinθ = 9 − 9sinθ ⇒ 10sinθ = 8 ⇒ sinθ = 4/5
Then cosθ = √(1 − sin²θ) = √(1 − 16/25) = √(9/25) = 3/5

Q4. If sinθ + cosθ = 7/5, find sin³θ + cos³θ.

A. 91/125
B. 83/125
C. 7/25
D. 49/125

Answer: A

Solution:
Use identity: a³ + b³ = (a + b)³ − 3ab(a + b)
We know a + b = 7/5,
Also ab = (a + b)² − (a² + b²) = (49/25 − 1)/2 = 12/25
Then: (7/5)³ − 3×12/25×7/5 = 343/125 − 252/125 = 91/125

Q5. Solve 2sin²θ + 3cosθ = 0 for 0° ≤ θ ≤ 360°

A. 60°, 300°
B. 120°, 240°
C. 90°, 270°
D. 135°, 225°

Answer: B

Solution:
Use sin²θ = 1 − cos²θ ⇒
2(1 − cos²θ) + 3cosθ = 0 ⇒ 2 − 2cos²θ + 3cosθ = 0
⇒ 2cos²θ − 3cosθ − 2 = 0
Solving: cosθ = −1/2 ⇒ θ = 120°, 240°

Q6. If tanθ = 3/4, evaluate (sinθ·cosθ)/(sin²θ − cos²θ)

A. −12/7
B. 12/7
C. −7/12
D. 7/12

Answer: A

Solution:
tanθ = 3/4 ⇒ sinθ = 3/5, cosθ = 4/5
Then sinθ·cosθ = 12/25, sin²θ − cos²θ = (9 − 16)/25 = −7/25
⇒ Expression = (12/25) / (−7/25) = −12/7

Q7. From top of a tower, angles of depression of A and B are 30° and 45°. If AB = 40 m, find height of tower.

A. 20(√3 + 1) m
B. 10(√3 + 1) m
C. 20(√3 − 1) m
D. 40 m

Answer: A

Solution:
Let height of tower = h
Distance to A = h√3, to B = h
So AB = h(√3 − 1) = 40 ⇒ h = 40 / (√3 − 1) = 20(√3 + 1) after rationalization

Q8. From point A, elevation angle to top of tower is 60°, from B (20 m farther), it’s 30°. Find height of tower.

A. 10√3 m
B. 15√3 m
C. 20 m
D. 5√3 m

Answer: A

Solution:
Let distance from A be x ⇒ h = √3·x
From B, distance = x + 20 ⇒ h = (x + 20)/√3
Equating: √3·x = (x + 20)/√3 ⇒ 3x = x + 20 ⇒ x = 10 ⇒ h = 10√3

Q9. Two ships are on opposite sides of a lighthouse. Elevations are 30° and 45°, and separation = 100(√3 + 1). Find height.

A. 100 m
B. 80 m
C. 120 m
D. 50(√3 − 1) m

Answer: A

Solution:
Let height = h
Then distances = h√3 and h ⇒ total = h(√3 + 1) = 100(√3 + 1) ⇒ h = 100

Q10. Man sees building at 30°, walks 40 m toward it, angle becomes 60°. Find height of building.

A. 20√3 m
B. 10√3 m
C. 30 m
D. 40 m

Answer: A

Solution:
Let initial distance = d ⇒ h = d/√3 ⇒ d = √3·h
After walking 40 m, distance = d − 40 ⇒ h = √3(d − 40)
Now equate: d/√3 = √3(d − 40) ⇒ d = 60 ⇒ h = 60/√3 = 20√3