Question

two boys on the either side of the school building of 25m high observes his top and its angle of elevation 30 and 60 respectively find the distance between the boys

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Answer 1

Let suppose that AB be the height of school building and Let the two boys be at C and D from the base of school building B.

Let the angle of elevation from C of the top of school building is 60° and from D is 30°.

Let distance between two boys be 'x' meters.

and

Let the boy at C is at a distance of 'y' meter from the base of building.

Now, we have following information with us :

• Height of school building, h = 25 m

• Distance, BC = 'y' m

• Distance, CD = 'x' m.

• ∠ACB = 60°

• ∠ADB = 30°

Now,

$\rm :\longmapsto\:In \: \triangle \: ABC$

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{AB}{BC}$

$\rm :\longmapsto\: \sqrt{3} = \dfrac{25}{y}$

$\rm :\longmapsto\:y = \dfrac{25}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\bf\implies \:y = \dfrac{25 \sqrt{3} }{3} \: m$

Now,

$\rm :\longmapsto\:In \: \triangle \: ADB$

$\rm :\longmapsto\:tan30 \degree \: = \: \dfrac{AB}{DB}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} } = \dfrac{25}{x + y}$

$\rm :\longmapsto\:x + y = 25 \sqrt{3}$

$\rm :\longmapsto\:x + \dfrac{25 \sqrt{3} }{3} = 25 \sqrt{3}$

$\rm :\longmapsto\:x = 25 \sqrt{3} - \dfrac{25 \sqrt{3} }{3}$

$\rm :\longmapsto\:x = \dfrac{75 \sqrt{3} - 25 \sqrt{3} }{3}$

$\rm :\longmapsto\:x = \dfrac{50 \sqrt{3} }{3} \: m$

$\rm :\longmapsto\:x = \dfrac{50 \times 1.732 }{3} \: m$

$\bf\implies \:x = 28.87 \: m \: (approx.)$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$