Question

Amit standing on a horizontal plane finds a bird flying at a distance of 200 m from him at an elevation of 30 degree Depak standing on the roof of a 50 M high building find the angle of elevation of the same but to be 45 and Deepak iron opposite sides of a bird find the distance of the bird from Deepak

Answer 1

Answer:

Distance between girl and bird is  50√2 m  = 70.71 m.

Step-by-step explanation:

Given information are

Distance between Amit and Bird = 200 m

 Height of building = 50 m

Angle of elevation Amit = 30°

Angle of elevation Deepak = 45°

To find: Distance between Deepak and Bird

In Δ XYZ

$sin\,30^{\circ}=\frac{XZ}{XY}$

$\frac{1}{2}=\frac{XZ}{200}$

$XZ=\frac{200}{2}$

XZ = 100 m

⇒ XZ = XM + MZ ( from figure )

   100 =  XM + 50    ( ∵ MZ = LN = 50 m )

    XM = 100 - 50

    XM = 50 m

Now in Δ XLM

$sin\,45^{\circ}=\frac{XM}{XL}$

$\frac{1}{\sqrt{2}}=\frac{50}{XL}$

XL = 50√2 = 70.71 m

Therefore, Distance between girl and bird is  50√2 m = 70.71 m.

Answer 2

Solution:-

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

∠A = 30°

∠D = 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/√2 = 50/BD

=> BD = 50√2 = 70.7 m

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

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@GauravSaxena01