Question

${\huge{\tt{\red{❦︎Q}{\blue{ᴜᴇ}{\green{ᴇs}{\purple{ᴛɪ}{\orange{ᴏɴ}{\pink{࿐}}}}}}}}}$



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Answer 1

${\huge{\tt{\red{❦︎A}{\blue{n}{\green{s}{\purple{w}{\orange{e}{\pink{r}}}}}}}}}$

42.3..

for rest..refer the attachment pls..

Answer 2

⠀⠀⠀⠀⠀⠀${\rm{\purple{\underline{\underline{★Required \:Answer★}}}}}$

42.3 m

⠀⠀⠀⠀⠀${\rm{\pink{\underline{\underline{★Solution★}}}}}$

${\rm{\red{\underline{\underline{Given : }}}}}$

Angle of elevation of the bird flying at a distance of 100m from boy = 30°

Angle of elevation from a girl standing on a 20m high building = 45°

Both (boy and girl) are in opposite sides of the bird.

${\rm{\blue{\underline{\underline{To \: Find: }}}}}$

Distance of the girl from the bird.

${\rm{\purple{\underline{\underline{Formula \: Used: }}}}}$

$\longmapsto\sin \: 30° \: = \frac{1}{2}$

$\longmapsto\sin \: 45° \: = \frac{1}{ \sqrt{2} }$

${\rm{\orange{\underline{\underline{Calculations : }}}}}$

Let A be the position of the bird and E and C be the positions of the girl and the boy, respectively.

Then,

∠ACB = 30°,

∠AED = 45°,

AC = 100 m,

EF = 20 m

In right triangle ACB,

we have,

$\longmapsto\sin \: 30° \: = \frac{AB}{AC}$

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

$\longmapsto \: AB = 50 m$

$\rm{AD = AB-BD}$

$\rm{AD = 50 - EF}$

$\rm{AD = 50 - 20 = 30 m}$

In right triangle ADE,

we have,

$\longmapsto\sin \: 45° \: = \frac{AD}{AE}$

$\longmapsto \frac{1}{ \sqrt{2} } \: = \frac{30}{AE}$

$\rm{AE = 30√2 = 42.3 m}$

Therefore, the distance of bird from the girl is 42.3m