Question

Find the measure of all the angles of a parallelogram , if one angle is ${24}^{\circ }$ less than twice the smallest angle.

Answer 1

Let smallest angle of the ||gm = x

Then second angle = $2x-{24}^{\circ }$

But these are consecutive angles

$\therefore x+(2x-{24}^{\circ })={180}^{\circ }$

$\Rightarrow x+2x-{24}^{\circ }={180}^{\circ }$

$\Rightarrow 3x={180}^{\circ }+{24}^{\circ }={204}^{\circ }$

$\Rightarrow x=\frac{{204}^{\circ }}{3}={68}^{\circ }$

$\therefore$ Smallest angle = ${68}^{\circ }$

and second angle = $2\times {68}^{\circ }-{24}^{\circ }$

$={136}^{\circ }-{24}^{\circ }={112}^{\circ }$

$\because$ The opposite angles of a || gm are equal

$\therefore$ Other two angles will be ${68}^{\circ }and{112}^{\circ }$

Hence angles are ${68}^{\circ },{112}^{\circ },{68}^{\circ },{112}^{\circ }$