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The angles of elevation of the top of a tower observed from two points on the ground are $30^{o}$ and $45^{o}$ . If the distance between the two points is $20$ metres, then find the height of the tower.

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27.3 m

17.3 m

54.6 m

60.2 m

answer is A.

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Detailed Solution

It is given that the angles of elevation of the top of a tower observed from two points on the ground are

30^{o}

and

45^{o}

.
The distance between the two points is

20

m.
Let the height of the tower AB be

y

.
Let C and D be the two points and BC =

x .

△ ADB

\tan 30^{0} = \frac{y}{x + 20}

....(1)
In

△ ACB,

\tan 45^{0} = \frac{y}{x}

⇒

x = y

Putting the value of

x

in (1), we get,
⇒

\frac{1}{\sqrt{3}} = \frac{y}{y + 20}

⇒

y + 20 = \sqrt{3} y

⇒

y = \frac{20}{\sqrt{3} - 1}

⇒

y = \frac{20}{0.732}

… (since

\sqrt{3} = 1.732

).
⇒

y = 27.3

m
Hence, the correct option is 1.

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