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From a point $p$ on a level ground the angle of elevation of the top of a tower is $30^{0} .$ If the tower is $100 m$ high the distance of point $p$ from the foot of the tower is

[use $\sqrt{3}$ =1.73]

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149 m

156 m

173 m

200 m

answer is C.

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Detailed Solution

Given, angle of elevation of the top of a tower is

30^{0} .

We have to find the foot of the tower.
Let

AB

be the tower then

∠ APB = 30^{0}

and

AB = 100 m

We know that tan

θ = \frac{perpendicular}{base}

.
Then,

\frac{AB}{AP} = \tan 30^{0}

\Rightarrow \frac{100}{AP} = \frac{1}{\sqrt{3}}

[since

\tan 30^{0}

\frac{1}{\sqrt{3}}

]

\Rightarrow AP = 100 \sqrt{3} m

\Rightarrow AP = (100 \times 1.73) m

\Rightarrow AP = 173 m

Hence, the foot of the tower is

173 m

.
Therefore, the correct option is 3.

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