The supplementary angle of an angle is$\dfrac{9}{4}$times its complementary angle. Find the measure of the supplementary angle (in degree).
$(a)144$
$(b)126$
$(c)162$
$(d)108$

Hint: Take a variable angle and write down its supplementary and complementary angle , now make equations using given information to proceed.

Let the angle be$x$, then its complementary angle will be$\left( {{{90}^0} - x} \right)$
and supplementary angle will be$\left( {{{180}^0} - x} \right)$
now, it is given in the question that the supplementary angle of an angle is$\dfrac{9}{4}$times its complementary angle, that is
\[({180^0} - x) = \left( {\dfrac{9}{4}} \right)({90^0} - x)\]
\[4 \times ({180^0} - x) = 9 \times ({90^0} - x)\]
\[4 \times ({180^0} - x) = {810^0} - 9x\]
\[{720^0} - 4x = {810^0} - 9x\]
\[ - 4x + 9x = {810^0} - {720^0}\]
\[5x = {90^0}\]
$x = \dfrac{{{{90}^0}}}{5}$
\[\therefore x = {18^0}\]
Therefore, the supplementary angle\[ = {180^0} - {18^0} = {162^0}\]
Hence, the required solution is$(c)162$.

Note: Assume the angles of the triangle according to the conditions given in the solution and then further evaluate to obtain the required solution.