Question

From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is?​

Answer 1

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Answer 2

Refer to attachment for diagram.

Now in triangle PQR.

$\qquad \mathtt{ \frac{ Opposite }{ Adjacent } = \tan \theta } \\ \\</p><p></p><p>\rightarrow \quad \mathtt{ \frac{QR}{QP} = \tan 30^{\circ} } </p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \mathtt{ \frac{100}{QP} = \frac{1}{\sqrt{3}} } \qquad \left( \because \tan 30^{\circ} = \frac{1}{\sqrt{3}} \right)</p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \mathtt{ QP = 100 \times \sqrt{3} } </p><p></p><p>\\ \\ \rightarrow \quad \mathtt{QP = 100 \times 1.73 } \qquad \left( \mathtt{ Take \; \sqrt{3} = 1.73 } \right) </p><p></p><p>\\ \\ </p><p></p><p>\rightarrow \quad \mathtt{QP = 173} </p><p></p><p>\\ \\</p><p> \mathtt{\bold{Distance \; of \; P \; from \; down \; of \; the \; tower \; is \; 173 \: m} }</p><p></p><p>$