Question

A boy observes that the angle of elevation of the bird flying at a distance of 100 m is 30 degree. at same distance of a boy,a girl finds the angle of elevation of the same bird from a building 20 m high is 45 degree. find the distance of the bird from the girl.

Answer 1

This is ur ans hope it will help you in case of any doubt comment below

Answer 2

Distance of bird from girl is 30√2 meters.

Step-by-step explanation:

Given:

• A boy observes that the angle of elevation of the bird flying at a distance of 100 m is 30°.
• At same distance of a boy,a girl finds the angle of elevation of the same bird from a building 20 m high is 45°.

To find:

• Find the distance of the bird from the girl.

Solution:

Concept to be used:

• Draw the figure to represent the suitation mark label.
• Apply application of trigonometry to find the required result.
• $sin \: {30}^{ \circ} = \frac{1}{2}$
• $sin \: {45}^{ \circ} = \frac{1}{\sqrt{2}}$

Step 1:

Draw the suitation in figure.

*see the attachment.

Position of bird: Point E

Position of boy: Point F

Position of girl: Point B

Building: BC of 20m

Distance of bird from boy: EF (100 m)

Step 2:

Calculate ED.

Apply trigonometric ratio in ∆FED.

$sin \: {30}^{ \circ} = \frac{ED}{EF}$

or

$\frac{1}{2} = \frac{ED}{100}$

or

$\bf ED = 50 \: m$

Step 3:

Calculate EA.

$ED=EA+AB$

or

$EA=ED-AB$

or

$EA=50 - 20$

or

$\bf EA = 30 \: m$

Step 4:

Calculate BE.

Apply trigonometry ratio in ∆ EAB.

$sin \: {45}^{ \circ} = \frac{EA}{EB}$

or

$\frac{1}{ \sqrt{2} } = \frac{30}{EB}$

or

$\bf EB = 30 \sqrt{2} \: m$

Thus,

Distance of bird from girl is 30√2 meters.

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