Three angles of ∆ABC are in the ratio 3 : 4 : 5, taken in order. If side AB is extended to X, what is the measure of exterior angle ∠CBX?

Correct option is C

Given:

Angles of $$\triangle$$​ ABC are in ratio =3:4:5

Formula Used:

Exterior angle = Sum of interior opposite angles

Sum of angles of a triangle = $$180^o$$​​

Solution: 

Let the angles A, B and C are in ratio 3x,4x and 5x respectively

Then 3x + 4x + 5x = 180^o

12x = $$180^o$$​​

x = $$\frac{180}{12} = 15^o$$​​

​$$\angle A = 3x = 45^o$$​​

​$$\angle B = 4x = 60^o$$​​

​$$\angle C = 5x = 75^o$$​​

For $$\angle CBX$$​ the interior opposite angles are$$\angle A \text{ and }\angle C$$​​

​$$\angle CBX = \angle A + \angle C = 45^ o + 75^o =120^o$$​​