Question
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 the tree before it broke?
More Height and Distance Questions
- Two ships are on opposite sides in front of a lighthouse in such a way that all three of them are in line. The angles of depression of two ships from the t...
- 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...
- 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 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 ...
- In the village of Hazelton, there was a totem pole 50 m high of which the lower 24 m was carved. According to a clue the treasure was hidden at a point on ...
- A tree of height 'h' meters is partially bent at a certain point above the ground, causing its top to touch the ground at a distance of 12 meters from its ...
- The length of the shadow of a vertical tower increases by 8 m on horizontal ground when the height of the sun changes from 45° to 30 ° , then find the heig...
- 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 hei...
- The angle of depression of a point on the ground as seen from the top of a tower, 35 feet high, is 45°. Find the distance of the point on the ground from t...
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
The height of the tree before it broke =AB+AC trigonometry (tan) = opposite/adjacent): tan 30° = AB/30 Since tan 30° = 1/√3 1 /√3 =AB/30 AB=30/√3 =10√3m AB = 10√3 Cos 30° =BC/AC √3 /2 =30 /AC AC =60 /√3 AC =20√3. Now the height of the tree=AB+AC =10√3 +20√3 =30√3