Question
From a watch tower of 150 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
- From the top of a tower, the angle of depression of a car on the ground is 30°. After the car moves 40 m towards the tower in a straight line, the angle of...
- From a point A on the ground, the angle of elevation of the top of a tower is 30°. From another point B, which is 20 m closer to the tower along the same s...
- 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 pole 21 m high casts a shadow 7√3 m long on the ground. Find the angle of elevation
- A long metallic cable is stretched from a point P on the ground to the top of a cliff, forming an angle of 30 ° with the horizontal ground. The total lengt...
- Find the area of maximum side of square that can be inscribed in a right angled triangle of side 15, 20 and 25 cm.
- There are two houses of the same height on both sides of a 15-meter wide road. From a point on the road, elevation angles of the houses are 30° and 60° res...
- A boat is being rowed away from a cliff 270 m high. From the top of the cliff, the angle of depression of the boat changes from 60° to 45° in 2 minutes. Wh...
- 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 ...
- The angle of elevation of an aeroplane from a point on the ground is 60°. After 12 seconds flight the elevation changes to 30°, if the aeroplane is flying ...
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 150/AC AC =150m Now – Tan30 =150/AD 1/√3 =150/AD AD=150√3. Distance between the cliffs = AD=AC+CD 150√3 =150+CD CD=150√3-150 =150(√3-1) m