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
- The length of the shadow of a vertical tower increases by 9 m on horizontal ground when the height of the sun changes from 45° to 30 ° , then find the heig...
- The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds flight the elevation changes to 30°, if the aeroplane is flying ...
- A man whose height is 1.5 metres is standing 24√3 metres away from a light pole. The angle of elevation from his head to the top of the light pole is 30 ° ...
- From the top of an upright pole 30√3 feet high, the angle of elevation to the top of an upright tower was 60°. If the foot of the pole was 55 feet away fro...
- A pole 21 m high casts a shadow 7√3 m long on the ground. Find the angle of elevation
- From a point situated at some distance from the base of a pillar, the angle of elevation of the top of the pillar is found to be 45 degrees. After walking ...
- A man 1.5 m tall is 22.5 m away from a tower 24 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
- A man 3 m tall is 23 m away from a tower 26 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
- The angle of depression of a point on the ground as seen from the top of a tower, 20 feet high, is 45°. Find the distance of the point on the ground from t...
- 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)
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