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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 $\angle C=82^o$.  
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