Question

Find the measures of an angle if $6$ times its complement is $12$ less than twice its supplement.

Answer 1

$\Rightarrow$ Let the angle be $x$

$\Rightarrow$ Complement of an angle $={90}^o-x$

$\Rightarrow$ Supplement of an angle $={180}^o-x$

According to the question,

$\Rightarrow$ $6({90}^o-x)+12=2({180}^o-x)$

$\Rightarrow$ ${540}^o-6x+12={360}^o-2x$

$\Rightarrow$ ${552}^o-{360}^o=4x$

$\Rightarrow$ ${192}^o=4x$

$\therefore$ $x=\frac{{192}^o}{4}={48}^o$