The angle between the external bisectors of two angles of a triangle is 60°. Then the third angle of the triangle is

∠ABC + ∠CBP = 180°

⇒ ∠B + 2 ∠1 = 180°

⇒ 2∠1 = 180° – ∠B

$\Rightarrow \angle 1=90^{\circ}-\frac{1}{2} \angle B$

Again, ∠ACB + ∠QCB = 180°

$\Rightarrow \angle 2=90^{\circ}-\frac{1}{2} \angle C$

In Δ BOC, ∠1 + ∠2 + ∠BOC = 180°

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

$\angle \mathrm{C}+\angle \mathrm{BOC}=180^{\circ}$

$\Rightarrow \angle \mathrm{BOC}=\frac{1}{2}(\angle \mathrm{B}+\angle \mathrm{C})$

$=\frac{1}{2}\left(180^{\circ}-\angle \mathrm{A}\right)$

$\Rightarrow \angle \mathrm{BOC}=90^{\circ}-\frac{1}{2} \angle \mathrm{A}$

$\Rightarrow 60^{\circ}=90^{\circ}-\frac{1}{2} \angle \mathrm{A}$

$\Rightarrow \angle \mathrm{A}=60^{\circ}$