Question

The angle of elevation of the top of a tower, from a point on the ground and at a distance of 150m from its foot, is 30° . Find the height of the tower correct to one place of decimal.​

Answer 1

tan30°=x/150

tan30°=x/1501/√3×150=x

tan30°=x/1501/√3×150=xx=50√3

tan30°=x/1501/√3×150=xx=50√3h=50√3

tan30°=x/1501/√3×150=xx=50√3h=50√3height of the tower is 50√3

tan30°=x/1501/√3×150=xx=50√3h=50√3height of the tower is 50√3in decimal=50×1.73

tan30°=x/1501/√3×150=xx=50√3h=50√3height of the tower is 50√3in decimal=50×1.73 =86.50

Answer 2

$\sf\large\underline\blue{Given:-}$

• The angle of elevation of a tower , from a point on the ground and at a distance of 150m from its foot, is 30°

$\sf\large\underline\blue{To\: Find :-}$

• What's the height of the tower?

$\sf\large\underline\blue{Solution:-}$

$\small{\underline{\sf{\blue{Let-}}}}$

The height of the tower be x m

Here, we need to consider the trigonometric ratio " tan theta " to solve this problem.

We know,

$\sf\implies\: \: \tan\theta =\dfrac{height}{base}$

Now, put the given values

$\sf\implies \tan30\degree = \dfrac{x}{150m} \\\\ \sf\implies \dfrac{1}{\sqrt{3}}=\dfrac{x}{150m}\\\\ \sf\implies x = \dfrac{150m}{\sqrt{3}} \\\\ \sf\implies x = 50\sqrt{3} m \\\\ \sf\implies x = 86.5m$

$\underline{\boxed{\bf{\green{\therefore The\:height\:of\:the\:tower\:is\:86.5 meters.}}}}$

Extra Info :-

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$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

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