In triangle ABC< D is on AC such that AD=BC and BD=DC, `/_DBC=2x` and `/_BAD=3x` where each angle is in degree. Then find x

Explanation

In `Delta ABC, (AC)/(sin 5x) = (BC)/(sin 3x)`
`rArr (a + p)/(sin 5x) = (a)/(sin 3 x)`....(i)

In `Delta BDN, cos 2x = (a)/(2p)`
or `a = 2p cos 2x`
From Eq. (i) `(2p cos 2 x + p)/(sin 5 x) = (2p cos 2x)/(sin 3x)`
or `2 sin 3x cos 2x + sin 3x = 2 sin 5x cos 2x`
or `sin 5x + sin x + sin 3x = sin 7 x + sin 3x`
or `sin 7 x - sin 5 x = sin x`
or `2 cos 6x sin x = sin x`
or `cos 6x = (1)/(2)`
`rArr x = 10^(@)`