Question
From a watch tower of 205 m height, the angles of depression of two cliffs in a horizontal line through the base of the tower are 45° and 30°. Find the distance between the cliffs if they are on the same side.
More Height and Distance Questions
- A tree breaks due to a storm and the top touches the ground 30 meters away from its base, making a 30-degree angle with the ground. What was the height of ...
- From a point on the ground, the angle of elevation of the top of a tower is 30 degrees. After moving 10√3 m towards the tower, the angle becomes 60 degrees...
- A man 3 m tall is 19 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
- The angle of elevation of the top of a building from a point on the ground is 30° and moving 40 meters towards the building it becomes 60°. The height of t...
- A pole 9√3 m high casts a shadow 27 m long on the ground. Find the angle of elevation
- At the foot of the Lighthouse, the elevation of its top is 45⁰. A Bird after flying 8 km towards the building up a slope of 30⁰ inclinations, the elevation...
- From a point on the ground, the angle of elevation of the top of a tower is 45 degrees. After moving 20 m away from the tower, the angle becomes 30 degree...
- From the top of a lighthouse, the angle of depression to the top and bottom of a tower is 30° and 45°, respectively. If the lighthouse is 210 metres high, ...
- A shadow of a tower standing on level ground is found to be 40√3 meters longer when the Sun's altitude is 30° than when it is 60°. The height of the tower ...
- A pole 21 m high casts a shadow 7√3 m long on the ground. Find the angle of elevation
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)
In triangle ABC- Tan45 =205/AC AC =205m Now – Tan30 =205/AD 1/ √3 =205/AD AD=205√3. Distance between the cliffs = AD=AC+CD 205 √3 =205+CD CD=205√3-205 =205(√3-1) m