If angles A, B, C to a ∆ABC from an increasing AP, then sin B =


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) $\frac{1}{\sqrt{2}}$


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,


$\mathrm{So}, a=60^{\circ}$

Therefore, $\angle B=60^{\circ}$

Hence, $\sin \angle B=\frac{\sqrt{3}}{2}$

So answer is (b)

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