Question
The angle of elevation of the top of a 70 feet tall tower from the initial position of a person on the ground was 30°. She walked towards the tower in a manner that the foot of the tower, her initial position and the final position were all in the same straight line. If she walked 140(√3/3) feet from her initial position, what was the angle of elevation of the top of the tower from her final position?
tan B = 70/AB tan 30˚ = 70/AB => 1/√3 = 70/AB AB = 70√3 Now, => AD = AB - BD => AD = 70√3 - 140√3/3 => AD = 70√3/3 Now, => tan θ = 70/(70√3/3) = 3/√3 => tan θ = √3/1 = tan 60˚More Trigonometry Questions
- Find the value of the given expression. 2 × (cos 60° – tan 45° + sec 60°)
- If sin 2a = (5/9), then find the value of (sin a + cos a).
- Simplify the following trigonometric expression: 10 cos19∘ sec19∘ − 4cot28∘ tan28∘
- In ∆ABC, AB = 5cm, BC = 6cm and AC = 10cm then find out the value of cos A?
- If tanθ – cotθ = a and cosθ + sinθ = b, then (b2 –1)(a2 + 4) = ?
- Find the maximum value of 18 sin A + 7 cos A.
- If (sin 3A sec 6A = 1) , then what will be the value of cos 6A?
- If cot2A = tan(A+6˚ ) , find the value of A? Given that 2A and A+6˚ are acute angles.
- If 3 tan X + cot X = 2√3, then find the value of 6 tan2 X + 2 cot2 X.
- What is the value of (4/3) cot2 (p/6) + 3 cos2 (150o) – 4 cosec2 45o + 8 sin (p/2)?
Hey! Ask a query
Please enter email id
The email must be a valid email address.
Please enter Mobile Number
Please enter valid Mobile Number
Please enter your Doubt
Think You're Ready for RBI Grade B?
RBI Grade B 2026 Phase 1 Memory Based Paper
- 200 Questions with Detailed Solutions
- Section-wise Coverage (GA, English, Quant & Reasoning)