Question

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

Answer 1

The total sum of the angles in a triangle is 180
So
$\angle A+\angle B+\angle C={180}^{\circ }$
given
$\angle A=x^{\circ },\angle B=(3x-2{)}^{\circ },\angle C=y^{\circ }$

$\left(x\right)+(3x-2)+\left(y\right)=180$

$4x+y=180+2$

$4x+y=182$

$y=182-4x$....(1)

given

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

$(y-3x+2)=9^{\circ }$

Put the value of y from equation (1)

$182-4x-3x+2=9$

$-7x=9-2-182$

$-7x=-175$

$x=25$

put in equation (1)

$y=182-4\times 25$

$y=182-100$

$y=82$

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

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

$,\angle C=y^{\circ }={82}^{\circ }$