Question:

If one angle of a parallelogram is 24° less than twice the smallest angle then the largest angle of the parallelogram is

(a) 68°

(b) 102°

(c) 112°

(d) 176°

Solution:

(c) $112^{\circ}$

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

$T$ he sum of adjacent angles of a parallelogram is $180^{\circ}$.

$\therefore x+2 x-24=180$

$\Rightarrow 3 x-24=180$

$\Rightarrow 3 x=180+24$

$\Rightarrow 3 x=204$

$\Rightarrow x=\frac{204}{3}$

$\Rightarrow x=68$

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

$L$ argest angle $=(180-68)^{\circ}=112^{\circ}$