Q. 24

# A 12 m high tree

Let AB is the Height of the tree, ED = h is the Height of the broken tree, CE is the Length of broken part.

Given, ECD = 60o, AB = 12.

2CE + CE√3 = 24

CE = 24(2 – √3)

DE = 12 – 48 + 24√3

DE = 12( – 3 + 2√3)

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Cube and Cuboid41 mins
History - Concept and Questions57 mins
Resources and Development Revision28 mins
Sphere and Hemisphere16 mins
Heights and Distances-I45 mins
Corrosion: Types, Pros and Cons41 mins
Right Circular Cylinder45 mins
Right Circular Cone And Frustum43 mins
Revision on Substitution and Elimination Method44 mins
Area Related to Circles- Important Formula and Concepts59 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

On observing fromRajasthan Board Mathematics

The shadow of a tRajasthan Board Mathematics

A tree breaks dueRajasthan Board Mathematics

On one side of a Rajasthan Board Mathematics

The angle of elevRajasthan Board Mathematics

The slope of a hiRajasthan Board Mathematics

From a distance oRajasthan Board Mathematics

From the top of aRajasthan Board Mathematics

The shadow of a vRajasthan Board Mathematics

From a point of tRajasthan Board Mathematics