Question

The heights of two towers are 180 meters and 60 meters respectively. If the angle of elevation of the top of the first tower from the base of the second tower is 60°, what is the angle of elevation of the top of the second tower from the base of the first tower.

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Answer 1

Step-by-step explanation:

in longer tower of 180 m

we have angle of elevation as 60

tan 60 =root3 = p/b

root3 = 180 / b

b = 180root3

now second tower height 60 m

tan x = p /b

tan x = 60 /180root3

tan x = 1 /3root 3

x = 30°

Answer 2

❥${\red{\underline{\underline{\large{\mathtt{Question:-}}}}}}$

The heights of two towers are 180 meters and 60 meters respectively. If the angle of elevation of the top of the first tower from the base of the second tower is 60°, what is the angle of elevation of the top of the second tower from the base of the first tower.

❥${\red{\underline{\underline{\large{\mathtt{Solution:-}}}}}}$

• Height of the 1st tower(AB)=180 m

• Height of the 2nd tower(CD)=60m

• Angle of elevation of the top of the 1st tower from the base of 2nd tower= 60°

★ In case of ∆ ABC,

$\frac{AB}{BC}$= tan60°

→$\frac{180}{BC}$=√3

→BC = 60√3

${\green{\boxed{\boxed{\large{\bold{BC=60\sqrt3}}}}}}}$

★In case of ∆ DCB,

$\frac{DC}{BC}$= tan∅

→$\frac{60}{60\sqrt3}$= tan∅

→$\frac{1}{\sqrt3}$= tan∅

→${\purple{\boxed{\bold{\theta=30}}}}$

❥${\red{\underline{\underline{\large{\mathtt{Answer:-}}}}}}$

The angle of elevation of the top of the 2nd tower from the base of the first is 30°.