Question

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.

NCERT Class X
Mathematics - Mathematics

Chapter _SOME APPLICATIONS OF
TRIGONOMETRY

Answer 1

Let the distance between the foot of the tower and the foot of the building be d.

height of the building = H
height of the tower = h = 50 m

Tan 30⁰ = 1/√3 = H/d
Tan 60⁰ = √3 = h/d
               =>  (1/√3)/(√3) = H / h
             =>   H  =  h/3  =>   50 m / 3  = 16.667 meters

Answer 2

Answer:

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:

$\sf In \: \Delta BDC$

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

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

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

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

___________________

$\sf In \: \Delta ABD,$

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

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

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

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

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

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