Question

If in a triangle ABC, AB = 5 units, \(\angle B = \cos^{-1}\left(\frac{3}{5}\right)\) and radius of circumcircle of \(\Delta\)ABC is 5 units, then the area (in sq. units) of \(\Delta\)ABC is :

• A. \(10 + 6\sqrt{2}\)
• B. \(8 + 2\sqrt{2}\)
• C. \(6 + 8\sqrt{3}\)
• D. \(4 + 2\sqrt{3}\)

Hint:

Use the sine rule \(c = 2R\sin C\) to get angle C, then area \(= 2R^2 \sin A \sin B \sin C\).

Answer 1

The Correct Option is C

Here c = AB = 5, R = 5, \(\cos B = \frac{3}{5}\) so \(\sin B = \frac{4}{5}\).
Sine rule: \(\frac{c}{\sin C} = 2R\), so \(\sin C = \frac{5}{10} = \frac{1}{2}\), hence C = 30° (C = 150° is impossible because then B + C > 180°, as B is about 53°).
\(\cos C = \frac{\sqrt{3}}{2}\).
\(\sin A = \sin(B + C) = \sin B\cos C + \cos B\sin C = \frac{4}{5}\cdot\frac{\sqrt{3}}{2} + \frac{3}{5}\cdot\frac{1}{2} = \frac{4\sqrt{3} + 3}{10}\).
Area \(= 2R^2\sin A\sin B\sin C = 2\cdot 25\cdot\frac{4\sqrt{3}+3}{10}\cdot\frac{4}{5}\cdot\frac{1}{2} = 2(4\sqrt{3}+3) = 8\sqrt{3} + 6\).
Answer: (3) \(6 + 8\sqrt{3}\)