A vertical pole is fixed on a tower which is 30 m high. From a point on the ground, the angles of elevation of the bottom and the top of the pole are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the pole.

30 m

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15√3 m

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30(√3-1) m

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60 m

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Solution

The correct option is C
30(√3-1) m

BC = 30m

Let CD = x

In $Δ$ ABC

AB = BC = 30m

In $Δ$ ABD

BD = AB $\sqrt{3}$ = 30 $\sqrt{3}$

Hence, height of pole = CD = BD – BC = 30( $\sqrt{3}$ -1)m

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A vertical pole is fixed on a tower which is 30 m high. From a point on the ground, the angles of elevation of the bottom and the top of the pole are 45∘ and 60∘ respectively. Find the height of the pole.

The correct option is C 30(√3-1) m BC = 30m Let CD = x In ΔABC AB = BC = 30m In ΔABD BD = AB√3 = 30√3 Hence, height of pole = CD = BD – BC = 30(√3-1)m

A vertical pole is fixed on a tower which is 30 m high. From a point on the ground, the angles of elevation of the bottom and the top of the pole are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the pole.

The correct option is C
30(√3-1) m

BC = 30m

Let CD = x

In $Δ$ ABC

AB = BC = 30m

In $Δ$ ABD

BD = AB $\sqrt{3}$ = 30 $\sqrt{3}$

Hence, height of pole = CD = BD – BC = 30( $\sqrt{3}$ -1)m