Question

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

Answer 1

EXPLANATION.

$\sf : \implies \: \: angle \: of \: elevation \: from \: top \: of \: building \: from \: foot \: of \: tower \: = 30 \degree \\ \\ \sf : \implies \: \: angle \: of \: elevation \: of \: top \: of \: tower \: from \: foot \: of \: building \: = 60 \degree \\ \\ \sf : \implies \: \: height \: of \: tower \: = 50 \: m$

$\sf : \implies \: \: in \: right \: angled \: triangle \: = \triangle \: DBC \\ \\ \: \sf : \implies \: \tan( \theta) = \frac{perpendicular}{base} \\ \\ \sf : \implies \: \tan(60 \degree) = \frac{dc}{bc} \\ \\ \sf : \implies \: \sqrt{3} = \frac{50}{bc} \\ \\ \sf : \implies \: bc \: = \frac{50}{ \sqrt{3} }$

$\sf : \implies \: in \: right \: angled \: triangle \: = \triangle \: ABC \\ \\ \: \sf : \implies \: \tan( \theta) = \frac{perpendicular}{base} \\ \\ \sf : \implies \: \tan(30 \degree) = \frac{ab}{bc} \\ \\ \sf : \implies \: \frac{1}{ \sqrt{3} } = \frac{ab}{ \frac{50}{ \sqrt{3} } }$

$\sf : \implies \: \dfrac{1}{ \sqrt{3} } = \dfrac{ab}{50} \times \sqrt{3} \\ \\ \sf : \implies \: ab \: = \frac{50}{3} m$

$\sf : \implies \: \green{{ \underline{height \: of \: building \: = \frac{50}{3}m }}}$

Answer 2

Given :

• The angle of elevation of the top of a building from the foot of the tower is 30°.

• The angle of elevation of the top of tower from the foot of the building is 60°.

• Height of the tower is 50 m.

• Let the height of the building be h.

Step-by-step explanation:

$\sf In \: \Delta BDC$

$:\implies \sf tan \: \theta = \dfrac{Perpendicular}{Base}$

$:\implies \sf tan \: 60^{\circ} = \dfrac{CD}{BD}$

$:\implies \sf \sqrt{3} = \dfrac{50}{BD}$

$:\implies \sf BD = \dfrac{50}{ \sqrt{3} } \: m$

___________________

$\sf In \: \Delta ABD,$

$\dashrightarrow\:\: \sf tan \: 30 ^{\circ} = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{h}{ \dfrac{50}{ \sqrt{3} } }$

$\dashrightarrow\:\: \sf h = \dfrac{ 50 }{ \sqrt{3} \times \sqrt{3} }$

$\dashrightarrow\:\: \underline{ \boxed{\sf Height= \dfrac{50}{3} \: m }}$

$\therefore\:\underline{\textsf{Height of the building is \textbf{50/3 meter}}}.$

__________________

$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$