Question

131m from the foot of a cliff on level of ground,
the angle of elevation of the top of cliffis 60'
Find the height of cliff (Take v3 1732)
226.892
224641
226 817
225.895​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

${\huge\tt\bf{GIVEN}}\begin{cases}\sf{131\:m\:far\:from\:cliff}\\\sf{Angle\:of\:elevation=60^{\circ}}\\ \sf{\sqrt {3}=1.732}\end{cases}$

$\huge\tt\bf {TO\:FIND:-}$

• Height of the cliff = x.

$\huge\tt\bf {SOLUTION:-}$

Considering the figure attached.

We know that, $\huge {\boxed {\tt {tan\theta =\dfrac {Perpendicular} {Base}}}}$

So,

$\leadsto \tt {\dfrac {x}{131}=tan 60^{\circ}}$

We know that,

$\huge {\boxed {\tt {tan 60^{\circ}=\sqrt {3}}}}$

$\leadsto \tt {x=131 \times \sqrt {3}}$

$\leadsto \tt {x=131 \times 1.732}$

$\leadsto \tt {x=226.892\:m}$

$\huge {\boxed {\tt \red{Height\:of\:the\:cliff = x=226.892\:m}}}$

$\rule {200}2$

Answer 2

Given:

• The distance between the foot of the cliff and the point of elevation is 131 m.
• The angle of elevation is 60°.

To find: The height of the cliff.

Answer:

(Diagram for reference attached below.)

$\sf In\ triangle\ ABC,\\\\\\tan\ \theta\ =\ \dfrac{Opposite}{Adjacent}\\\\\\tan\ \theta\ =\ \dfrac{AB}{BC}\\\\\\tan\ {60}^{\circ}\ =\ \dfrac{AB}{131}\\\\\\We\ know\ that\ tan\ {60}^{\circ}\ =\ \sqrt{3}.\\ \\\\\sqrt{3}\ =\ \dfrac{AB}{131}$

$\sf 131\ \times\ \sqrt{3}\ =\ AB\\ \\\bf (\sqrt{3}\ =\ 1.732)\\\\\sf 131\ \times\ 1.732\ =\ 226.892\\\\\\Therefore,\ the\ height\ of\ the\ cliff\ is\ 226.892\ m.$