Question

An observer in a boat found the angle of
elevation of the top of the hill to be 60°. As he moved 20 m away
from the hill, the angle of elevation changed to 45°. What is the
height of the hill?
(a) 10(+3+3) m
(b) 10/3 m
(c) 10 m
(d) None of these





please its urgent answer with method​

Answer 1

Answer:

answer is ; _ (10/3)

...

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Answer 2

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

Let assume that AB be the hill and let observer be at point C in the boat so that angle of elevation of top of hill is 60°,

Now, the observer retreats 20 m away from hill AB, let this position is at D and angle of elevation of top of the hill is 45°.

Let assume that,

• AB = h m

• BC = x m

Given that,

• CD = 20 m.

Consider,

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

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

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

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

Now,

$\rm :\longmapsto\:In \: \triangle \: ABD$

$\rm :\longmapsto\:tan45 \degree \: = \: \dfrac{AB}{BD}$

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

$\rm :\longmapsto\:x + 20 = h$

$\rm :\longmapsto\:\dfrac{h}{ \sqrt{3} } + 20 = h$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: using \: (1)\bigg \}}$

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

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

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

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

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

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

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

$\bf\implies \:h = 10(3 + \sqrt{3}) \: m$

Hence,

$\: \: \: \: \: \: \: \: \: \underbrace{ \boxed{\bf \:Height \: of \: hill \: = 10(3 + \sqrt{3}) \: m }}$

Additional Information :-

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