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The angular elevation of a tower from a point is $30^{\circ} .$ As we more $100$ m nearer to the base of the tower the angle of elevation becomes $60^{\circ}$ . What is the height of the tower and the distance of the first point from the base of the tower?

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86.6m,

150 m

77.6m,

90 m

80m,

120 m

66m,

180 m

answer is A.

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Detailed Solution

Given that,

PQ = 100

m
The angles of elevation at

P

and

Q

30^{\circ}

and

60^{\circ}

AB is representing the tower,

P, Q

are the points of observation,
We have the formula,

tanθ = \frac{height}{base}

Considering the required height of the tower

AB = h

metres and

BQ = x

metres.
We have to find the height,
In

ΔBAQ,

\frac{h}{x} = \tan 60^{\circ}

\Rightarrow h = x tan 60^{\circ}

\Rightarrow h = x \sqrt{3} . . . . . . . . . . (i)

ΔABP,

\frac{h}{100 + x} = \tan 30^{\circ}

\Rightarrow \frac{h}{100 + x} = \frac{1}{\sqrt{3}}

\Rightarrow h \sqrt{3} = 100 + x

\Rightarrow x \sqrt{3} \times \sqrt{3} = 100 + x

\Rightarrow 3 x = 100

\Rightarrow 2 x = 100

\Rightarrow x = 50

From (i) we get,

h = 50 \sqrt{3 m}

\Rightarrow h = 50 \times 1.732 m

\Rightarrow h = 86.6 m

Then, the distance of the first point from the base will be,

AP = (x + 100)

\Rightarrow AP

(50 + 100)

\Rightarrow AP

150

m
Thus, the height of the tower is 86.6m and the distance is 150m.
Therefore the correct option is 1.

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