Question

From the top of a cliff the angle of depression of the top and bottom of a tower are observed to be 45 and 60 respectively if the height of the tower is 20m find the height of the cliff and the distance between the cliff and the tower​

Answer 1

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

Let height of the cliff, AB = 'h' meter.

Let CD be height of tower = 20 m.

Let distance between cliff and tower, BC = 'x' meter.

Now,

• AE = AB - BE = AB - CD = h - 20 m.

Now,

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

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

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

$\bf\implies \:h = \sqrt{3} x - - - (1)$

Now,

$\rm :\longmapsto\:In \: \triangle \: AED$

$\rm :\longmapsto\:tan45 \degree \: = \dfrac{AE}{ED}$

$\rm :\longmapsto\:1 = \dfrac{h - 20}{x}$

$\rm :\longmapsto\:1 = \dfrac{ \sqrt{3}x - 20}{x} \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:x = \sqrt{3}x - 20$

$\rm :\longmapsto \: \sqrt{3}x - x = 20$

$\rm :\longmapsto\:x( \sqrt{3} - 1) = 20$

$\rm :\longmapsto\:x = \dfrac{20}{ \sqrt{3} - 1}$

$\rm :\longmapsto\:x = \dfrac{20}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1}{ \sqrt{3} + 1}$

$\rm :\longmapsto\:x = \dfrac{20( \sqrt{3} + 1)}{3 - 1}$

$\rm :\longmapsto\:x = \dfrac{20( \sqrt{3} + 1)}{2}$

$\rm :\longmapsto\:x = 10( \sqrt{3} + 1) \: m - - - (2)$

$\rm :\longmapsto\:x = 10(1.732 + 1)$

$\rm :\longmapsto\:x = 10 \times 2.732$

$\bf\implies \:x = 27.32 \: m$

On substituting the value of x from equation (2) , in equation (1), we get

$\rm :\longmapsto\:h = 10 \sqrt{3}( \sqrt{3} + 1)$

$\rm :\longmapsto\:h = 10(3 + \sqrt{3})$

$\rm :\longmapsto\:h = 10(3 + 1.732)$

$\rm :\longmapsto\:h = 10 \times 4.732$

$\rm :\longmapsto\:h = 47.32 \: m$

Hence,

• Height of the cliff = 47.32 m

and

• Distance between cliff and tower = 27.32 m

Additional Information :-

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

Answer 2

The above answer is the correct one