In the given figure, AOB is a straight line. If ∠AOC = (3x + 10)° and ∠BOC (4x − 26)°, then ∠BOC = ?

(b) 86°

We have

$\angle A O C+\angle B O C=180^{\circ} \quad$ [Since $A O B$ is a straight line ]

$\Rightarrow 3 x+10+4 x-26=180^{\circ}$

$\Rightarrow 7 x=196^{\circ}$

$\Rightarrow x=28^{\circ}$

$\therefore \angle B O C=[4 \times 28-26]^{\circ}$

Hence, $\angle B O C=86^{\circ}$.