In a ∆ABC, ∠A = x°, ∠B = (3x − 2)°, ∠C = y°. Also, ∠C − ∠B = 9°
Question:

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

Solution:

Let $\angle A=x^{\circ}, \angle B=(3 x-2)^{\circ}, \angle C=y^{\circ}$ and

$\angle C-\angle B=9^{\circ}$

$\Rightarrow \angle C=9^{\circ}+\angle B$

$\Rightarrow \angle C=9^{\circ}+3 x^{\circ}-2^{\circ}$

$\Rightarrow \angle C=7^{\circ}+3 x^{\circ}$

Substitute $\angle C=y^{\circ}$ in above equation we get,

$y^{\circ}=7^{\circ}+3 x^{\circ}$

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow x^{\circ}+\left(3 x^{\circ}-2^{\circ}\right)+\left(7^{\circ}+3 x^{\circ}\right)=180^{\circ}

$\Rightarrow 7 x^{\circ}+5^{\circ}=180^{\circ}$

$\Rightarrow 7 x^{\circ}=180^{\circ}-5^{\circ}=175^{\circ}$

$\Rightarrow x^{\circ}=\frac{175^{\circ}}{7^{\circ}}=25^{\circ}$

$\angle A=x^{\circ}=25^{\circ}$

$\angle B=(3 x-2)^{\circ}=3\left(25^{\circ}\right)-2^{\circ}=73^{\circ}$

$\angle C=\left(7^{\circ}+3 x^{\circ}\right)=7^{\circ}+3(25)^{\circ}=82^{\circ}$

$\angle A=25^{\circ}, \angle B=73^{\circ}, \angle C=82^{\circ}$

Hence, the answer.

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