Question

if the angle of depression from the top of a building to a point situated 10√3m away from the bottom of the building is 60°, then the height of the building will be​

Answer 1

(Diagram given in attachment)

★ Given :-

• Angle of depression = ∠DAC = 60°
• Distance = BC = 10√3 m

★ To find :-

➟ Height of the building = AB = ?

★ Solution :-

We know that,

∠DAC = ∠ACB = 60°          [∵ Alternate Interior Angles]

Now,

In ΔABC

$\bf{tan\: 60^{\circ} = \dfrac{AB}{BC}}\\\\\\\Rightarrow \sf{\dfrac{1}{\sqrt{3}}= \dfrac{AB}{BC}}\\\\\\\\\Rightarrow \sf{\dfrac{1}{\sqrt{3}}= \dfrac{AB}{10\sqrt{3}}}\\\\\\\\\Rightarrow \sf{AB\sqrt{3} = 10\sqrt{3} }$

$\\\rm{Cancelling \: \sqrt{3} \:on \: both\: sides,}$

$\\\boxed{\boxed{\bf{AB = 10 \: m}}}$

∴ Height of the building = 10 m

Know more:

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