Question

Find the measure of all the angles of a parallelogram, if one angle is ${12}^{\circ }$ more than twice the smallest angle. [3 MARKS]

Answer 1

Concept : 1 Mark
Equation : 1 Mark
Calculation : 1 Mark




Let the smallest angle be $x.$

The other angle is $12+2x$

These angles are adjacent and therefore, supplementary.

$x+(12+2x)$= ${180}^{\circ }$

$3x+12$ = ${180}^{\circ }$

$x$ = ${56}^{\circ }$
$\Rightarrow$ $12+2x$ = $124$
$\therefore$ angles are ${56}^{\circ },{124}^{\circ },{56}^{\circ },{124}^{\circ }$