Question

The angle of elevation of the top of a hill at the foot of the tower is 600
and the angle of
elevation of the top of the tower from the foot of the hill is 300
. If the tower is 50 m high,
What is the height of the hill?

Answer 1

EXPLANATION.

→ Let AB be the Tower and CD be the hill.

→ Let CD = X m.

→ < ACB = 30° and < CAD = 60°.

→ in right triangle ∆BCA,

→ tan ø = perpendicular/base = p/b

→ tan (30°) = AB/AC

→ tan (30°) = 50/AC

→ 1/√3 = 50/AC

→ AC = 50√3 M.

→ in right triangle ∆DAC,.

→ tan ø = p/b

→ tan (60°) = DC/AC

→ √3 = x / AC

→ √3 = x / 50√3

→ x = 150 M.

→Heights of hill is = 150 M.

Answer 2

Answer:

★ Given :-

• The angle of elevation of the top of a hill at the foot of the tower is 60°.
• The angle of elevation of the top of the tower from the foot of the hill is 30°.
• The height of the tower is 50 m.

★ To Find :-

• What is the height of the hill.

★ Solution :-

Let, AB be the tower and CD be the hill.

And,

• ∠ACB = 30°
• ∠CAD = 60°
• AB = 50 m

➣ Let, CD = xm

❖ In ∆BAC, we have,

$\implies$ cot30° = $\dfrac{AC}{AB}$

$\implies$ $\sqrt{3}$ = $\dfrac{AC}{50}$

➠ AC = 50$\sqrt{3}$

❖ In ∆ACD, we have,

$\implies$ tan60° = $\dfrac{CD}{AC}$

$\implies$ $\sqrt{3}$ = $\dfrac{x}{50\sqrt{3}}$

$\implies$ $\sf{x = 50 × 3}$

➠ x =《 150 m 》

$\therefore$ The height of the hill is $\boxed{\bold{\large{150\: m}}}$

_______________________________