Question

A boat is 40m away from the base of a cliff. Given that the angle of depression of the
boat from the top of the cliff is 60°, find the height of the cliff. (Take√ =1.732)

Answer 1

Given :

A boat is 40m away from the base of a cliff. The angle of depression of the boat from the top of the cliff is 60°.

To Find :

The height of the cliff.

Solution :

Let A be the top of the cliff.

So, AB = h metres

Let C be the position of boat 40 metres away from the cliff of angle of depression 60°,

• ∠ACB = 60°

• ∠ABC = 90°

The base is given to us. We have to find the height.

We know that,

From right angled ∆ABC,

$\boxed{\bf\tan\ 60^{\circ}=\dfrac{Height}{Base}}$

$\\ \implies\sf\tan\ 60^{\circ}=\dfrac{AB}{BC}$

where,

• tan 60° = √3
• AB = h m
• BC = 40 m

Substituting the values,

$\\ \implies\sf\sqrt{3}=\dfrac{h}{40}$

By cross multiplying,

$\\ \implies\sf\sqrt{3}\times40=h$

Taking √3 = 1.732,

$\\ \implies\sf1.732\times40=h$

$\\ \implies\sf69.28=h$

$\\ \therefore\boxed{\bf Height=69.28\ m.}$

The height of the cliff is 69.28 m.

Explore More :

Trigonometric Ratio :

$\\ \sf\bullet\ \sin\theta=\dfrac{Height}{Hypotenuse}$

$\\ \sf\bullet\ \cos\theta=\dfrac{Base}{Hypotenuse}$

$\\ \sf\bullet\ \tan\theta=\dfrac{Height}{Base}$

$\\ \sf\bullet\ \cot\theta=\dfrac{Base}{Height}$

$\\ \sf\bullet\ \sec\theta=\dfrac{Hypotenuse}{Base}$

$\\ \sf\bullet\ \cosec\theta=\dfrac{Hypotenuse}{Height}$

Trigonometric Table :

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

Answer 2

Answer:

Given:

A boat is 40m away from the base of a cliff.

The angle of depression of the boat from the

top of the cliff is 60°.

To Find

The height of the cliff.

Solution:

Let A be the top of the cliff.

So, AB =h metres

Let C be the position of boat 40 metres away

from the cliff of angle of depression 60,

LACB= 60°

LABC = 90°

The base is given to us. We have to find the

height.

We know that,

From right angled AABC,

Height

tan 60 =

Base

AB

tan 60= BO

BC

where,

tan 60° = V3

AB h m

BC 40 m

Substituting the values,

Step-by-step explanation:

hope it's helpful