Question
Two buildings are collinear with the base of a tower and are at a distance 9m and 16m from the base of the tower. The angles of elevation from these two building of the top of the tower are complementary. What is the height of the tower?
More Height and Distance Questions
- At a point 20 metres away from the base of a 20√3, metres high house, the angle of elevation of the top is
- From a point 20 m away from the foot of a pole, the angle of elevation of the top of the pole is 60 degrees. Find the height of the pole.
- The two buildings are 100m apart. From the top of the first building, which is 60m tall, the angle of depression to the top of the second building is 45 de...
- Two buildings are collinear with the base of a tower and are at a distance 9m and 16m from the base of the tower. The angles of elevation from these two bu...
- From a point on the ground, the angle of elevation of the top of a tower is 30 degrees. After moving 20 m towards the tower, the angle becomes 60 degrees. ...
- A pole 21 m high casts a shadow 7√3 m long on the ground. Find the angle of elevation
- The angle of elevation of a tower from a certain point of bus stand is 30°. When a man walks 5m ahead in the direction of the tower, the angle of elevatio...
- Raju is positioned 30 meters away from the base of a tower. From his standing point, the angle of elevation to the top of the tower is 45°. Determine the h...
- If distance between two pillars of length 9 & 4 cm is x cm. If two angle of elevation of their respective top from a point on ground of other are complemen...
- There are two houses of the same height on both sides of a 30-meter wide road. From a point on the road, elevation angles of the houses are 30° and 60° res...
Relevant for Exams:
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)
Let the height of the tower be h And ∠CBD = θ and ∠ DAC = 90 – θ In ∆ BCD tan θ = CD/BC = h/9 …………(i) In ∆ ACD tan(90 - θ) = CD/AC cot θ = h/16 ……………………….(ii) On multiplying Equation (i) and (ii) tan θ × cot θ = h/9 × h/16 1 = h²/144 Now h = 12 m