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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