Question

Amit, standing in a horizontal plane, finds a bird flying at a distance of 200m from him at an elevation of 30º. Deepak standing on the roof of 50m high building, finds the angle of elevation of the same bird to be 45º.Amot and Deepak are no opposite sidea of the bird. Find the distance of the bird form Deepak.

Answer 1

Answer:

Let Amit is standing at point A,

bird at point B, deepak at point d

height of the bird = BC,

let E be a point on BC corresponding to point D at 50 m height

given,

AB = 200

ZA = 30 °

2D = 45 °

In triangle ABC

Sin 30 = BC / AB

=> 1/2 = BC / 200

=> BC = 100 m

=> BE = BC - EC = 100 - 50 = 50 m

In triangle,

BED, sin 45 = BE / BD

=> 1/12 = 50 / BD

=> BD = 50V2 = 70.7 m

Hence the distance of the bird from deepak is 70.7 m or approx 71

hope it helped u thank you

Answer 2

❥ Let Amit is standing at bottom of the building at B and observing the bird at C with angle of elevation 30°

❥ It is assumed that given distance 200m is BC, the distance along line of sight

We have ,

$⇒sin \: {30}^{</p><p>o} = \frac{CD}{BC}$

$⇒ \frac{1}{2} = \frac{CD}{200}$

$⇒ CD = \frac{200}{2}$

$⇒CD = 100m$

❥ If Deepika is seeing the same bird at top of the building With the Angle of elevation 45°,

We have

$⇒sin \: {45}^{</p><p>o} = \frac{CE}{AC}$

$⇒ \frac{1}{ \sqrt{2} } = \frac{CE}{AC}$

$⇒AC = \sqrt{2CE}$

$⇒AC = (CD−DE) \sqrt{2}$

$⇒AC = (100−50) \sqrt{2}$

$⇒AC = 50 \sqrt{2}$

$⇒AC \: =50×1.414$

$⇒AC = 70.7 \: ≈ \: 71m$

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