# 2.  Find angles x and y in each figure.

i) As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal. So,

$y+120^0=180^0$

$y=180^0-120^0$

$y=60^0$

Now,  As we know the sum of internal angles of a triangle is 180. so,

$x+60^0+60^0=180^0$

$x=180^0-60^0-60^0$

$x=60^0$

Hence, $x=60^0\:\:and\:\:y=60^0$.

ii)  As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal.

AND

the sum of internal angles of a triangle is 180. so,

$x+x+90^0=180^0$

$2x=90^0$

$x=45^0$

Also,

$y=180^0-x$

$y=180^0-45^0$

$y=135^0$.

Hence $x=45^0\:\:and\:\:y=135^0$.

iii) As we know when two lines are intersecting, the opposite angles are equal.

And

the sum of internal angles of a triangle is 180. so,

$x+x+92^0=180^0$

$2x=180^0-92^0$

$2x=88^0$

$x=44^0$

Now, As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle

$y=x+92^0$

$y=44^0+92^0$

$y=136^0$

Hence $x=44^0\:\:and\:\:y=136^0$.

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