Question

the angle of elevation of the top of a tree from a point, which is 20m away from the foot of the tree,is 60°.Find the height of the tree.(take √3=1.73) answer in detail​

Answer 1

Answer:

height of the tree is 34.60m.

Answer 2

Given:-

• Angle of elevation of the top of a tree from a point on ground is 60°
• Distance of that point from the foot of the tree = 20 m

To Find:-

• Height of the tree

Note:-

• Refer to the attachment for the figure.

Solution:-

Here in this scenario a right-angled triangle is being formed in which,

AB = 20 m

∠ABC = 60°

We know,

$\sf{TanB = \dfrac{Perpendicular}{Base}}$

Hence,

$\sf{Tan60^\circ = \dfrac{AC}{AB}}$

From Trigonometric Table we have,

Tan60° = √3

Hence,

$\sf{\sqrt{3} = \dfrac{AC}{20}}$

$\sf{\implies AC = 20\sqrt{3}}$

Taking √3 as 1.73

AC = 20 × 1.73

=> AC = 34.6 m

Therefore the height of the tree is 34.6 m

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Explore more:-

Trigonometric table is very important for solving these kind of sums.

Here is the trigonometric table:-

$Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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