Question

(ii) The heights of two towers are 180 metres and 60 metres respectively. If the angle of elevation of the top of the first tower from the foot of the second tower is 60°, then find the angle of elevation of the top of the second tower from the foot of the first.

No spam___!!!​

Answer 1

★${\purple{\huge{\underline{\mathbb{Given:-}}}}}$

ᴥHeight of first tower(AB)= 180 m

ᴥHeight of second tower(DC)= 60 m

ᴥThe angle of elevation of the top of the first tower from the foot of the second tower[angle ACB=60°]

★${\purple{\huge{\underline{\mathbb{To\:find:-}}}}}$

The angle of elevation of the top of the second tower from the foot of the first.

★${\purple{\huge{\underline{\mathbb{Solution:-}}}}}$

Suppose,

The angle of elevation of the top of the second tower from the foot of the first [angle CBD]= ∅

∆ ABC is a right triangle.

tan60° = AB/BC

→ √3 = 180/BC

→ BC = 60√3

∆BCD is a right triangle.

tan∅ = DC/BC

→ tan∅ = 60/60√3

→ tan∅ = 1/√3

→ tan∅ = tan30°

→ ∅ = 30°

★${\green{\huge{\underline{\mathbb{Answer:-}}}}}$

The angle of elevation of the top of the second tower from the foot of the first is 30°.

Answer 2

$\huge\underline\mathfrak\black {To find }$

The heights of two towers are 180 metres and 60 metres respectively. If the angle of elevation of the top of the first tower from the foot of the second tower is 60°, then find the angle of elevation of the top of the second tower from the foot of the first.

$\huge\underline\mathfrak\red{Solution }$

let AB the first tower and CD be the second tower.

then , AB = 180 m and CD = 60 m

and angle ADB = 60° and angle CBD = ß

$\tan(60) = \frac{AB}{BD}$

$\sqrt{3} = \frac{180}{BD}$

$BD = \frac{180}{ \sqrt{3} }$

BD = 60√3

↪again from the right triangle ∆BCD

$\tan( \beta ) = \frac{CD}{BD}$

$\tan( \beta ) = \frac{60}{60 \sqrt{3} }$

$\tan( \beta ) = \frac{1}{ \sqrt{3} }$

this implies that

ß = 30°

$\huge\underline\mathfrak\purple{Answer}$

Hence ,the angle of elevation of the top of the second tower from the foot of the first tower is 30°.