If one of the angles of a triangle is 130°,

(d) Let angles of a triangle  be ∠A, ∠B and ∠C.

In $\triangle A B C$,

$\angle A+\angle B+\angle C=180^{\circ}$    [sum of all interior angles of a triangle is $180^{\circ}$ ]

$\Rightarrow \quad \frac{1}{2} \angle A+\frac{1}{2} \angle B+\frac{1}{2} \angle C=\frac{180^{\circ}}{2}=90^{\circ}$

[dividing both sides by 2]

$\Rightarrow \quad \frac{1}{2} \angle B+\frac{1}{2} \angle C=90^{\circ}-\frac{1}{2} \angle A$

$\Rightarrow \quad\left[\because\right.$ in $\left.\triangle O B C, \angle O B C+\angle B C O+\angle C O B=180^{\circ}\right]$

$\left[\right.$ since, $\frac{\angle B}{2}+\frac{\angle C}{2}+\angle B O C=180^{\circ}$ as $B O$ and $O C$ are the angle

bisectors of $\angle A B C$ and $\angle B C A$, respectively]

$\Rightarrow \quad 180^{\circ}-\angle B O C=90^{\circ}-\frac{1}{2} \angle A$

$\therefore \quad \angle B O C=180^{\circ}-90^{\circ}+\frac{1}{2} \angle A=90^{\circ}+\frac{1}{2} \angle A$

$=90^{\circ}+\frac{1}{2} \times 130^{\circ}=90^{\circ}+65^{\circ}$ $\left[\therefore \angle A=130^{\circ}\right.$ (given) $]$

$=155^{\circ}$

Hence, the required angle is $155^{\circ} .$