In ` A B C ,` if `/_A=100^0,\ A D` bisects `/_A\ a n d\ /_A D\ _|_B C` . Then, `/_B=` `50^0` (b) `90^0` (c) `40^0` (d) `100^0`

Explanation

Let /_1, /_2 and ∠/_3 be the angles of the triangle and /_4 be its exterior angle.
∠4 = 108°(Given)
∠1: ∠2 = 4: 5 (Given)
Let, ∠1 = 4k ,∠2 = 5k
Now, ∠1 + ∠2 = 108°(Exterior angle theorem)
4k + 5k = 108°
9k = 108° k = 12°
Thus, `∠1 = 4*12 = 48°,∠2 = 5*12 = 60°`
We know that,` ∠1 + ∠2 + ∠3 = 180° `
`48° + 60° + ∠3 = 180°`
`108° + ∠3 = 180°`
`∠3 = 180° – 108° = 72°`
.Thus, angles of triangle are `48°, 60°, 72°`.