Question 24

∆PQR is right angled at Q. If m$$\angle$$P = $$60^\circ$$, then find the value of $$(sec R + \frac{1}{2})$$

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 : $$(sec R + \frac{1}{2})$$

= $$sec(30^\circ)+\frac{1}{2}$$

= $$\frac{2}{\sqrt3}+\frac{1}{2}$$

= $$\frac{(\sqrt{3}+4)}{2\sqrt{3}}$$

=> Ans - (B)

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