Question

2. 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.​

Answer 1

Given:

DC = height of the tower = 50m

Solution: let AB be x = height of the building

Let y be the distance between the building and the tower.

tan60 =50/y

root3 = 50/y

y = 50/root3 metres

Now, tan30 = x/y

That is , 1/root3 = x/50/root3

1/root3 = root3x/50

3x=50

x = 50/3 metres

Which is = 16.66 metres

Therefore the height of the building is 16.66 metres. HOPE THIS ANSWER WAS HELPFUL

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:

$\sf In \: \Delta BDC$

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

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

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

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

___________________

$\sf In \: \Delta ABD,$

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

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

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

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

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

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