Question:
If angles $A, B, C$ to a $\triangle A B C$ from an increasing $A P$, then $\sin B=$
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) 1
(d) $\frac{1}{\sqrt{2}}$
Solution:
Let the angles of a triangle $\triangle A B C$ be $(a-d),(a),(a+d)$ respectively which constitute an A.P.As we know that sum of all the three angles of a triangle is $180^{\circ} .$ So,
$(a-d)+a+(a+d)=180^{\circ}$
$\mathrm{So}, a=60^{\circ}$
Therefore, $\angle B=60^{\circ}$
Hence, $\sin \angle B=\frac{\sqrt{3}}{2}$
So answer is (b)