# Please solve 8c(i,ii) Dear Student,

8 ( c ) Given : Sin B  = $\frac{4}{5}$  , We know :  Sin   ,So

Sin B = $\frac{\mathrm{AD}}{\mathrm{AB}}$
We assume ratio coefficient of sides =  x , So  AD  = 4 x  and AB = 5 x

Now we apply Pythagoras theorem we get :

AB2 = AD2 + BD2 , Substitute given values we get :

(5x)2 = (4x)2 + BD2 ,

25x2 =  16x2 + BD2 ,

BD2 = 9x2,

BD = 3 x

But we know BD  = 15 cm ( Given ) , So

3 x = 15

= 5 , So

AD = 4x  = 4 ( 5 ) = 20  cm

And

AB = 5 x = 5 ( 5 ) = 25 cm

And

Given : tan C  = 1  , We know : tan   ,So

tan C = $\frac{\mathrm{AD}}{\mathrm{DC}}$ , So

1 = $\frac{20}{\mathrm{DC}}$ ,

DC =  20 cm

Now we apply Pythagoras theorem in triangle ACD we get :

AC2 = AD2 + DC2 , Substitute values we get :

AC2 = 202 + 202 ,

AC2 = 400 + 400 ,

AC2 = 800 ,

AC = 20

Therefore,

AD  = 20 cm , AB = 25 cm , DC = 20 cm and AC = 28.28 cm                                               ( Ans )

ii ) tan B  =    And We know :  Cos   ,So Cos B =

To Show : , We take L.H.S. and substitute values , As :

Therefore,

L.H.S. = R.H.S.                                            ( Hence proved )