In a cyclic quadrilateral with one angle being $60°$, find the area given $\cos 60° = \frac{4 + 25 - c^2}{2 × 5}$.
Step-by-Step Solution
Key Concept: In a cyclic quadrilateral, opposite angles are supplementary; use the cosine rule and area formulas for triangles.
From $\cos 60° = \frac{1}{2}$, we have $\frac{1}{2} = \frac{4 + 25 - c^2}{2 × 5}$, so $10 = 29 - c^2$, giving $c^2 = 19$. For the supplementary angle $120°$, $\cos 120° = -\frac{1}{2} = \frac{a^2 + b^2 - 19}{ab}$. This gives $a^2 + b^2 + ab = 19$. The area of the quadrilateral is $\frac{1}{2} × 2 × 5 \sin 60° + \frac{1}{2} ab \sin 120° = 5\sqrt{3} + 4\sqrt{3} = 9\sqrt{3}$, which approximates to $12$ units.
Correct Answer: 12