Question

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

Answer 1

see this solution if it is wrong then plz tell me

Answer 2

Answer:

$30\sqrt{2}$

Step-by-step explanation:

Refer the attached figure

AC = 100

∠ACE=30°

DE=BF=20 m

Let AE be h

AD is the distance of bird from girl.

In ΔABC

$Sin \theta = \frac{Perpendicular}{Hypotenuse}$

$Sin30^{\circ} = \frac{AB}{AC}$

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

$50=AB$

AF = AB -FB

AF = 50 - 20

AF= 30 m

Now In ΔADF

$Sin \theta = \frac{Perpendicular}{Hypotenuse}$

$Sin45^{\circ} = \frac{AF}{AD}$

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

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

$AD=30\sqrt{2}$

Hence the distance of the bird from the girl  is $30\sqrt{2}$