Question

If one angle of a parallelogram is ${24}^{\circ }$ less than twice the smallest angle then the largest angle of the parallelogram is

(a) ${68}^{\circ }$ (b) ${102}^{\circ }$

(c) ${112}^{\circ }$ (d) ${176}^{\circ }$

Answer 1

Let smallest angle = x

then largest angle = $2x-{24}^{\circ }$

But$x+2x-{24}^{\circ }={180}^{\circ }$

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

$\Rightarrow 3x={180}^{\circ }+24={204}^{\circ }$

$\Rightarrow x=\frac{204}{3}={68}^{\circ }$

$\therefore$ largest angle = ${180}^{\circ }-{68}^{\circ }={112}^{\circ }$