Question

Watching from the top of a building and from a
window 10m
below the top. The angles of
depressions of car on the ground are found to be 60°
and 45° respectively.
Find the distance of the
car from the building.​

Answer 1

Answer:

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Answer 2

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

Let AB be the building and AC be the window 10 m below the top of building.

and

Let car is at the point D when angle of depression from top of building B and window C is 60° and 45°.

Let distance between car and building, AD = 'x' meters.

Let AC = 'h' meter.

So, we have

• AD = x meter

• AC = h meter

• BC = 10 meter.

Now,

$\rm :\longmapsto\:In \: \triangle \: ADC \:$

$\rm :\longmapsto\:tan45 \degree \: = \: \dfrac{AC}{AD}$

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

$\bf\implies \:h \: = \: x - - - (1)$

Now,

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

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

$\rm :\longmapsto\: \sqrt{3} = \dfrac{h + 10}{x}$

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

$\rm :\longmapsto\: \sqrt{3}x - x = 10$

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

$\rm :\longmapsto\:x = \dfrac{10}{ \sqrt{3} - 1}$

$\rm :\longmapsto\:x = \dfrac{10}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1}{ \sqrt{3} + 1}$

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

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

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

$\rm :\longmapsto\:x = 5( \sqrt{3} + 1) \:$

$\rm :\longmapsto\:x = 5(1.732 + 1)$

$\rm :\longmapsto\:x = 5 \times 2.732$

$\bf\implies \:x \: = \: 13.66 \: m$

Hence,

• Distance between car and building is 13.66 meters

Additional Information :-

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