An exterior angle of a triangle is `108^0` and its interior opposite angles are in the ratio 4:5. The angles of the triangle are `48^0,\ 60^0,\ 72^0` (b) `50^0,\ 60^0,\ 70^0` `52^0,\ 56^0,\ 72^0` (d) `42^0,\ 60^0,\ 76^0`

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°`.