SSC CPO SI and ASI 2013 Paper

Question : 109 of 200

Marks: +1, -0

From the top of a hill 200 m high, the angle of depression to the top and the bottom of a tower are observed to be 30° and 60°, respectively. The height of the tower (in m) is

${400 √3}/ 3$
$166 2/ 3$
$133 1/ 3$
$200 √3$

Validate

Solution:

(3)

Height of the hill is JK = 200 m
∠JMN =30° and∠JLK= 60°
Let MN = LK = a meter
From ΔJKL

\tan 60°= { {JK}}/{ {KL}}= {200}/{ {a}}

and

{a}= {200}/{√{3}} {meter}

Now from ΔJNM

\tan 30°= {J N}/{N M}= {J N}/{ {200}/{√{3}}}

⇒{1}/{√{3}}= {\√{3} J N}/{200}

Therefore, JN

= {200}/{3} {m}

Height of tower is LM = 200 – JN =

200- {200}/{3}= {400}/{3}

=133 {1}/{3} {m}

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