Question

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

Answer 1

We know that sum of adjacent angles of parallelogram is ${180}^{\circ }$.

Let the smallest angle of parallelogram be $x^{\circ }$

Then the other angle is $(2x-24{)}^{\circ }$.

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

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

$\Rightarrow$ $x={68}^{\circ }$

$\Rightarrow$ And the second angle = $(2x-24{)}^{\circ }=2\times {68}^{\circ }-{24}^{\circ }={112}^{\circ }$

$\Rightarrow$ We know that opposite angles of parallelogram are equal then,

$\Rightarrow$ Then all angles of parallelogram are ${112}^{\circ },{68}^{\circ },{112}^{\circ },{68}^{\circ }$.