The angle of elevation of the rope of a tree from a point A on the ground is $60^{\circ}$ . On walking 20m away from its base to a point B, the angle of elevation changes to $30^{\circ}$ . Find the height of the tree.

10√3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

20√3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30√3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40√3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
10√3

In triangle ACD

tanɵ = $\frac{D C}{C A}$ = $\frac{h}{x}$

tan $60^{\circ}$ = $\sqrt{3}$ = $\frac{h}{x}$ ..................... 1

in triangle CDB

tan $30^{\circ}$ = $\frac{C D}{C B}$

$\frac{1}{\sqrt{3}}$ = $\frac{h}{x}$ + 20.......................2

Solving 1 and 2, we get h= 10 $\sqrt{3}$

Suggest Corrections

Similar questions

The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower is $30^{\circ}$ . Find the height of the tower.

Q. Find the height of a tree when it is found that on walking away from it 20 m. in a horizontal line through its base, the elevation of its top changes from

60^{0}

30^{0}

Join BYJU'S Learning Program

The angle of elevation of the rope of a tree from a point A on the ground is 60∘ . On walking 20m away from its base to a point B, the angle of elevation changes to 30∘. Find the height of the tree.

The correct option is A 10√3 In triangle ACD tanɵ = DCCA = hx tan60∘ = √3 = hx ..................... 1 in triangle CDB tan30∘ = CDCB 1√3 = hx + 20.......................2 Solving 1 and 2, we get h= 10√3

The angle of elevation of the rope of a tree from a point A on the ground is $60^{\circ}$ . On walking 20m away from its base to a point B, the angle of elevation changes to $30^{\circ}$ . Find the height of the tree.