Question

for The angle of elevation of the top of a building from the foot of the tower is 30° and to angle of elevation of the top of the tower from the foot of the building is 60°. If the tow is 50 m high, find the height of the building. ​

Answer 1

the hight of building is 90 degree

Answer 2

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

Let assume that,

• AD be the height of the building

• BC be the height of tower

Further assume that, Let

• AD = h metres

Given that,

• BC = 50 metre

• ∠ABD = 30°

• ∠BAC = 60°

Now, In right-angle triangle ABC

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

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

$\rm \implies\:AB = \dfrac{50}{ \sqrt{3} } \: m$

Now, In right angle triangle ABD

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

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

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

$\rm :\longmapsto\:\dfrac{50}{ \sqrt{3} \times \sqrt{3} } \: = \: h$

$\rm \implies\:\boxed{\tt{ \: h \: = \: \frac{50}{3} \: m \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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