Question

Find the radius of the circle escribed to the triangle ABC (Shown in the figure below) on the side BC if $\angle NAB =30^{\circ} ; \angle BAC =30^{\circ} ; AB = AC =5$.

• A. $\frac{(10 \sqrt{2}+5 \sqrt{3}-5)(2-\sqrt{3})}{2 \sqrt{2}}$
• B. $\frac{(10 \sqrt{2}+5 \sqrt{3}-5)}{2 \sqrt{2}}(2-\sqrt{3})$
• C. $\frac{(10 \sqrt{2}+5 \sqrt{3}-5)}{2 \sqrt{2}}(2+\sqrt{3})$
• D. $\frac{(10 \sqrt{2}+5 \sqrt{2}+1)}{2 \sqrt{3}}(\sqrt{3}-1)$

Answer 1

Answer: (A)

Correct answer is (a) $\frac{(10 \sqrt{2}+5 \sqrt{3}-5)(2-\sqrt{3})}{2 \sqrt{2}}$