In a $\triangle ABC$, $\angle A = x^o, \angle B = (3x– 2)^o, \angle C = y^o$. Also, $\angle C - \angle B = 9^o$. Find the three angles.

Given:

In a $\triangle ABC$, $\angle A = x^o, \angle B = (3x– 2)^o, \angle C = y^o$. Also, $\angle C - \angle B = 9^o$.

To do:

We have to find the three angles.

Solution:

We know that,

Sum of the angles in a triangle is $180^o$. 

Therefore,

$\angle A+\angle B+\angle C=180^o$

$x^o+(3x-2)^o+y^o=180^o$

$4x^o+y^o=180^o+2^o$

$y^o=182^o-4x^o$....(i)

$\angle C - \angle B = 9^o$

$y^o-(3x-2)^o=9^o$

$y^o-3x^o+2^o=9^o$

$182^o-4x^o-3x^o=9^o-2^o$   (From (i))

$7x^o=182^o-7^o$

$7x^o=175^o$

$x^o=\frac{175^o}{7}$

$x^o=25^o$

$y^o=182^o-4(25^o)$     (From (i))

$y^o=182^o-100^o$

$y^o=82^o$

This implies,

$\angle A = x^o=25^o$

$\angle B = (3x– 2)^o=3(25^o)-2^o=75^o-2^o=73^o$

$\angle C = y^o=82^o$$

The three angles are $\angle A=25^o$, $\angle B=73^o$ and $\angle C=82^o$.