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

Answer

It is given that,
$\angle\text{A}=\text{x}^\circ ....(\text{i})$
$\angle\text{B}=(3\text{x}-2)^\circ\ ...(\text{ii})$
$\angle\text{C}=\text{y}^\circ\ ....(\text{iii})$
And, $\angle\text{C}-\angle\text{B}=9^\circ\ ....(\text{iv})$
Putting $\angle\text{C}=\text{y}^ \circ$ and $\angle\text{B}=(3\text{x}-2)^\circ$ in equation (iv), we get
y - (3x - 2) = 9
⇒ y - 3x + 2 = 9
⇒ y - 3x = 9 - 2
⇒ -3x + y = 7 .....(v)
We know that, the sum of angles of a triangle is 180°
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
⇒ x + 3x - 2 + y = 180°
⇒ 4x - 2 + y = 180°
⇒ 4x + y = 180 + 2
⇒ 4x + y = 182 .......(vi)
Subtracting equation (v) from equatoin (vi) we get
4x + 3x = 182 - 7
⇒ 7x = 175
$\Rightarrow\text{x}=\frac{175}{7}$
⇒ x = 25
Putting x = 25 in equation (v) we get
-3 × 25 + y = 7
⇒ -75 + y = 7
⇒ y = 7 + 75
⇒ y = 82
$\therefore\angle\text{A}=\text{x}^\circ=25^\circ$
$\angle\text{B}=(3\text{x}-2)^\circ$
(3 × 25 - 2)°
= (75 - 2)
= 73°
And, $\angle\text{C}=\text{y}^\circ=82^\circ$

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