The angles of a triangle are $ x, y $ and $ 40^{\circ} $. The difference between the two angles $ x $ and $ y $ is $ 30^{\circ} $. Find $ x $ and $ y $.

Given:

The angles of a triangle are \( x, y \) and \( 40^{\circ} \). The difference between the two angles \( x \) and \( y \) is \( 30^{\circ} \).

To do:

We have to find \( x \) and \( y \).

Solution:

We know that,

The sum of the angles of a triangle is $180^o$

Therefore,

$x+y+40^o=180^o$

$x+y=180^o-40^o$

$x+y=140^o$..........(i)

The difference between the two angles \( x \) and \( y \) is \( 30^{\circ} \).

This implies,

$x-y=30^o$........(ii)

Adding (i) and (ii), we get,

$2x=140^o+30^o$

$x=\frac{170^o}{2}$

$x=85^o$

$\Rightarrow y=85^o-30^o$

$y=55^o$

Hence, $x=85^o$ and $y=55^o$.