Question

the angle of elevation of the top of a building from the foot of the to Tower is 30 degree and the angle of elevation of the top of the tower from the foot of the building is 60 degree if the tower is 50 metre high find the height of the building​

Answer 1

Step-by-step explanation:

Height of Building is 20 m

Step-by-step explanation:

Given: Height of Tower, CD = 60 m

Angle of elevation foot of tower to top of building = 30°

Angle of elevation foot of building to top of tower = 60°

To find: Height of Building , AB

In ΔCDB,

tan\,60^{\circ}=\frac{CD}{CB}tan60

∘

=

CB

CD

\sqrt{3}=\frac{60}{CB}

3

=

CB

60

CB=\frac{60}{\sqrt{3}}CB=

3

60

In ΔABC,

tan\,30^{\circ}=\frac{AB}{CB}tan30

∘

=

CB

AB

\frac{1}{\sqrt{3}}=\frac{AB}{CB}

3

1

=

CB

AB

CB=AB\sqrt{3}CB=AB

3

from above,

AB\sqrt{3}=\frac{60}{\sqrt{3}}AB

3

=

3

60

AB=20AB=20

Therefore, Height of Building is 20 m

Answer 2

AB is the height of tower = 50m

DC is the height of building = h m

Angle of elevation to the top of base of building = 60°

$\large\tt\red{In∆ABC,}$

$\longrightarrow$$\large\tt\purple{\angle{60°}}$

$\longrightarrow$$\large\tt\purple{tan\:60°=\frac{AB}{BC}=\frac{50}{√3}=BC}$

$\large\tt\red{In\:∆DCB,}$

$\longrightarrow$$\large\tt\purple{tan\:30°=\frac{DC}{BC}}$

$\longrightarrow$$\large\tt\purple{\frac{1}{√3}=\frac{ \frac{h}{50} }{ \sqrt{3} }}$

$\longrightarrow$$\large\tt\purple{\frac{h√3}{50}=\frac{1}{√3}}$

$\longrightarrow$$\large\tt\purple{3h=50}$

$\longrightarrow$$\large\tt\purple{h=\frac{50}{3}}$

$\longrightarrow$$\large\tt\purple{16.33m}$