Question

from the top of a cliff 120 metre high the angle of depression to the top and bottom of a tower are 30° and 60° respectively. find the height of a tower.

Answer 1

Answer:

Step-by-step explanation:

consider the angle of dep. from the cliff to the bottom of the tower.

tan60= √3 = 120m/a(m)

⇒ a = (120/√3)m

keep it aside for the next step.

take up the angle of depression to the top of the tower from the cliff.

tan30 = 1/√3 = b(m)/(120/√3)m

⇒ b = 40m

h = (120-b)m

⇒ h = 80m

Answer 2

The height of a tower is 80 m.

The figure is in the attachment below.

Given:

Let AB = 120 m be the height of the cliff.

The angle of depression of the top and bottom of the tower is 30° and 60°.

Let DC = h m be the height of the tower.

To find: The height of a tower.

Solution:

Step 1: Find AC :

In ∆ABC,

tan 60° = $\frac{AB}{BC} = \frac{P}{B}$

$\sqrt3 = \frac{120}{BC} \\\\BC = \frac{120}{ \sqrt3} \ m \\\\BC = \frac{120 \times \sqrt3 }{\sqrt3 \times \sqrt3 } \ m$

[ Rationalising the denominator]

$BC = \frac{120\ \times \ \sqrt3}{3} \\\\BC = 40 \sqrt3$

Step 2: Find BC:

In ∆BDF,

$tan \ 30\° = \frac{ BF}{DF} = \frac{P}{B} \\\\\frac{1}{\sqrt3} = \frac{ BF}{40 \sqrt3} \\\\BF = \frac{1 \ \times \ 40 \sqrt3 }{\sqrt3}$

[DF = BC]

BF = 40 m

Step 3: Find the height of the tower DC:

DC = AB - BF

DC = (120 - 40) m

DC = 80 m

Hence, the height of the tower is 80 m.

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