# 1.  Find the value of the unknown x in the following diagrams

As we know that the sum of the internal angles of the triangle is equal to $180^0$. So,

i) $x+50^0+60^0=180^0$

$x=180^0-50^0-60^0$

$x=70^0$

ii) $x+30^0+90^0=180^0$

$x=180^0-30^0-90^0$

$x=60^0$

iii) $x+30^0+110^0=180^0$

$x=180^0-30^0-110^0$

$x=40^0$

iv) $x+x+50^0=180^0$

$2x=180^0-50^0$

$2x=130^0$

$x=65^0$

v) $x+x+x=180^0$

$3x=180^0$

$x=60^0$

vi) $x+2x^0+90^0=180^0$

$3x=180^0-90^0$

$3x=90^0$

$x=30^0$

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