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1. Find angle x in each figure:

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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

$x=30^0$

Posted by

Pankaj Sanodiya

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