ICSE Class X Math 2015 Solved Paper

Question : 40 of 52

Marks: +1, -0

The horizontal distance between two towers is

120 {\m}

. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is

30^{∘}

and

24^{∘}

respectively.

Find the height of the two towers. Give your answer correct to 3 significant figures.

Solution:

Let

A B

and

C D

are two towers.

{\BD}=120 {\m}= {\EC}, ∠ {\ACE}=30^{∘}, ∠ {\CBD}=24^{∘} \;\text ". "\;

In right angle

△ {\CBD}

\tan 24^{∘}\;=\;{C D}/{B D}=\;{C D}/{120}

C D\;=120 × 0.4452

\;=53.42 {\m} .

In right angle

△ {\ACE}

\tan 30^{∘}\;=\;{ {\AE}}/{ {\EC}}=\;{ {\AE}}/{120}

{\AE}\;=120 × 0.5773

\;=69.28 {\m}

{\AB}\;= {\AE}+ {\EB}

\;= {\AE}+ {\CD} \;\; [ {\EB}= {\CD}]

\;=69 .28+53 .42

\;=122 .70 {\m} .

Height of 1st tower is

122.70 {\m}

and 2nd tower is

53.42 {\m}

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