Question

The angles of depression of two boats on a river from the top of a pole 30 m high on the bank of the river are 60° and 75°. If the boats are in line with the pole, find the distance between the boats to the nearest metre.
Ans=9 m
Please solve quickly its urgent.

Answer 1

Given:

The angles of depression of two boats on a river from the top of a pole 30 m high on the bank of the river are 60° and 75°.

The boats are in line with the pole

To find:

The distance between the boats to the nearest metre.

Solution:

To solve the given problem we will use the following formula of trigonometric ratios:

$\boxed{\bold{tan \: \theta = \frac{Perpendicular}{Base} }}$

 

Referring to the figure attached below, we have

"AB" → height of the pole = 30 m

"CD" → the distance between the two boats

"∠EAC" = "∠ACB" = 75° → angle of depression of the first boat from the top of the pole

"∠EAD" = "∠ADB" = 60° → angle of depression of the second boat from the top of the pole

Considering Δ ABC, we have

θ = 75°

Perpendicular = AB

Base = BC

∴  $tan \:\theta = \frac{AB}{BC}$

substituting the given values

$\implies tan \: 75\° = \frac{30}{BC}$

$\implies 3.732 = \frac{30}{BC}$

$\implies BC = \frac{30}{3.732}$  

$\implies \bold{BC = 8.038}$ ....... (i)

Considering Δ ABD, we have

θ = 60°

Perpendicular = AB

Base = BD = BC + CD

∴  $tan \:\theta = \frac{AB}{BC + CD}$

substituting the given values and the value of BC from (i)

$\implies tan \: 60\° = \frac{30}{8.038 + CD}$

$\implies 1.732 = \frac{30}{8.038 + CD}$

$\implies 13.921 + 1.732CD = 30$

$\implies 1.732CD = 30 - 13.921$

$\implies 1.732CD = 16.079$

$\implies CD = \frac{16.079}{1.732}$

$\implies \bold{CD = 9.28\:m}$ ≈ $\bold{9 \:m}$

Thus, the distance between the two boats to the nearest metre → 9 m

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