Question

If the angle of elevation of the top of a coconut tree from a point on the ground is 60° and the point is 20 metre away from the foot of the tree, let us find the height of the tree​

Answer 1

Your answer:

Let AB be the height of the Tree

And angle ACB be the incidence angle

And BC be the distance of the point from tree

Given:-

• $\tt \angle ACB= 60^{o}$

• $\tt BC = 20m$

Solution:-

$\tt In \ \ \triangle \ \ ABC \\\\ \tt \tan 60^o = \dfrac{Perpendicular}{Base} \\\\ \tt \Rightarrow \tan 60^{0} = \dfrac{AB}{BC} \\\\ \tt \Rightarrow \sqrt3 = \dfrac{AB}{20m} \\\\ \tt \Rightarrow 20\sqrt3m = AB$

So, Height of Coconut tree is $\tt 20\sqrt3m$

So, Converting it into decimal

20(1.732m) = 34.64m

To Answer this Question We must know:

• Trigonometric ratios.
• Trigonometric values for different angles.
• Concept of Incidence angle.

Answer 2

$\mathcal{\huge{\fbox{\red{Question\:?}}}}$

✴ If the angle of elevation of the top of a coconut tree from a point on the ground is 60° and the point is 20 metre away from the foot of the tree. Find the height of the tree ?

$\mathcal{\huge{\fbox{\green{Answer:-}}}}$

✒ The height of the coconut tree is 20√3 meters.

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

Let's ,C be the point where the person is standing.

According to the question,

• AB is the coconut tree we have to find its height.

• CB is 20 meters distance.

◼ The angle of elevation at the top of the coconut tree from the base is 60° .

Now,

We know that, tan θ = perpendicular/base

➡ tan 60°= AB/BC

✨ tan 60° = √3

➡ √3=AB/BC

Where, BC = 20 meters

➡ √3 = AB / 20

➡ 20√3 = AB

Hence, AB = 20√3

The following solution also attached with diagram.

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