Question

In the given figure, $AOB$ is a straight line. If $\angle AOC=(3x+10{)}^{\circ }$ and $\angle BOC=(4x-26{)}^{\circ }$, then $\angle BOC=?$

(a) ${96}^{\circ }$

(b) ${86}^{\circ }$

(c) ${76}^{\circ }$

(d) ${106}^{\circ }$.

Answer 1

Since, the sum of the linear pair of angles is always equals to ${180}^{\circ }$.
Given angles $\angle AOC=(3x+10{)}^{\circ }$ and $\angle BOC=(4x-26{)}^{\circ }$ forms a linear pair.
So,$\angle AOC+\angle BOC={180}^{\circ }$
$\Rightarrow 3x+10+4x-26=180$
$\Rightarrow 7x-16=180$
$\Rightarrow 7x=180+16$
$\Rightarrow 7x=196$
$\therefore x={28}^{\circ }$
Hence, $\angle BOC=(4\times 28)-26$,
$\angle BOC=112-26$
$\therefore \angle BOC={86}^{\circ }$.