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The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is 32. When the observer moves towards the tower a distance of 100 in, he finds the angle of elevation of the top lobe 63. Find the height of the tower and the distance of the first position from the tower. (Take tan 32 = 0.6248 and tan 63 = 1.9626)

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81.66 m

and

100.7 m

91.66 m

and

146.7 m

1.4 m

and

14.7 m

None of these

answer is B.

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Detailed Solution

Let the height of tower = h Question Image

Let BC = x and CD = 100
From the

Δ ABC

we have,

\Rightarrow \tan C = \frac{AB}{BC}

\Rightarrow \tan 63^{\circ} = \frac{AB}{BC}

\Rightarrow 1.9626 = \frac{h}{x}

\Rightarrow x = \frac{h}{1.9626}

From the

Δ ABD

we have,

\Rightarrow \tan D = \frac{AB}{BC + CD}

\Rightarrow \tan 32^{\circ} = \frac{h}{x + 100}

\Rightarrow 0.6248 = \frac{h}{x + 100}

\Rightarrow x + 100 = \frac{h}{0.6248}

Putting the value of x, we get

\Rightarrow 100 = \frac{h}{0.6248} - \frac{h}{1.9626}

\Rightarrow 100 = \frac{h (1.9626 - 0.6248)}{0.6248 \times 1.9626} \Rightarrow 100 = \frac{h (1.3378)}{0.6248 \times 1.9626} \Rightarrow 100 \times 0.6248 \times 1.9626 = h \times 1.3378

\Rightarrow h = \frac{100 \times 0.6248 \times 1.9626}{1.3378}

\Rightarrow h = \frac{122.6232}{1.3378} \Rightarrow h = 91.66

Since,

x = \frac{h}{1.9626}

\Rightarrow x = \frac{91.66}{1.9626} \Rightarrow x = 46.7

Therefore, the distance of the first position from the tower

= 100 + 46.7 = 146.7 m

Hence, the height of tower and the distance of the first position from the tower are

91.66 m

and

146.7 m

respectively.

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