Question

the angle of elevation of the top of a tree from a point A on the grounds is 60°. on walking 20m away from it's base, to a point B, the angle of elevation changes to 30°. find the height of the tree​

Answer 1

SOLUTION:-

Given:

•The angle of elevation of the top 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°.

To find:

The height of the tree.

Explanation:

Let the height of the tower CD= h m

Now,

In right angled triangle ACD & BCD.

Using trigonometric ratio of the angle theta:

cotangent theta:

$\frac{Base}{Perpendicular} = cot \theta$

In ∆ACD, we get

$\frac{AC}{h} = cot30 \degree = \sqrt{3} \\ \\ \frac{AC}{h} = \sqrt{3} \\ \\ AC = \sqrt{3} h$

&

In ∆BCD, we get;

$\frac{BC}{h} = cot60 \degree = \frac{1}{ \sqrt{3} } \\ \\ \frac{bc}{h} = \frac{1}{ \sqrt{3} } \\ [Cross \: multiplication] \\ h = \sqrt{3} BC \\ \\ BC = \frac{h}{ \sqrt{3} }$

We have,

20m away from it's base.

AB= 20m = AC - BC

So,

$20 = \sqrt{3} h \: - \: \frac{h}{ \sqrt{3} } \\ \\ 20 = h( \sqrt{3} - \frac{1}{ \sqrt{3} } ) \\ \\ 20 = h( \frac{ \sqrt{3} \times \sqrt{3} - 1}{ \sqrt{3} } ) \\ \\ 20 = h( \frac{3 - 1}{ \sqrt{3} } ) \\ \\ 20m = h \times \frac{2}{ \sqrt{3} } \\ \\ 20m = \frac{2h}{ \sqrt{3} } \\ \\ 20 \sqrt{3} = 2h \\ \\ h = \frac{20 \sqrt{3} }{2} \\ \\ h = 10 \sqrt{3} m$

Thus,

The height of the tree is 10√3 m.

Answer 2

Answer:

Given:

•The angle of elevation of the top 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°.

To find:

The height of the tree.

Explanation:

Let the height of the tower CD= h m

Now,

In right angled triangle ACD & BCD.

Using trigonometric ratio of the angle theta:

cotangent theta:

In ∆ACD, we get

&

In ∆BCD, we get;

We have,

20m away from it's base.

AB= 20m = AC - BC

So,

Thus,

The height of the tree is 10√3 m.

Step-by-step explanation: