Question 24

ΔPQR is right angled at Q. If m$$\angle$$P = 60°, then ﬁnd the value of $$(cot R + \frac{\sqrt{3}}{2})$$.

Solution

Sum of angles of $$\triangle$$ PQR = $$\angle P+\angle Q+\angle R=180^\circ$$

=> $$60^\circ+90^\circ+\angle R=180^\circ$$

=> $$\angle R=180^\circ-150^\circ=30^\circ$$

To find : $$(cot R + \frac{\sqrt{3}}{2})$$

= $$cot(30^\circ)+\frac{\sqrt3}{2}$$

= $$\sqrt3+\frac{\sqrt3}{2}$$

= $$\frac{3\sqrt3}{2}$$

=> Ans - (A)

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: