# 1. Find angle x in each figure:

P Pankaj Sanodiya

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

and

the sum of angles of the triangle is equal to $180^0$. So,

i) $x = 40^0$

ii) $x+45^0+45^0=180^0$

$x=180^0-90^0$

$x=90^0$

iii) $x=50^0$

iv) $100^0+x+x=180^0$

$2x=180^0-100^0$

$2x=80^0$

$x=40^0$

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

$2x=180^0-90^0$

$2x=90^0$

$x=45^0$

vi)$x+x+40^0=180^0$

$2x=180^0-40^0$

$2x=140^0$

$x=70^0$

vii)$x=180^0-120^0$

$x=60^0$

viii) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle. So,

$x+x=110^0$

$2x=110^0$

$x=55^0$

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

$x=30^0$

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