Question

In a triangle, the length of the two larger sides are 24 and 22, respectively. If the angles are in AP, then the third side is

• A. \(12 + 2\sqrt{3}\)
• B. \(12 - 2\sqrt{3}\)
• C. \(2\sqrt{3} + 2\)
• D. \(2\sqrt{3} - 2\)

Hint:

Use Law of Sines and angle sum properties.

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \text{Angles in AP} \Rightarrow B = 60^\circ \]
Step 2: Calculation / Simplification}
Larger sides: \(a = 24, b = 22\). By Law of Sines:
\(\frac{\sin A}{24} = \frac{\sin 60^\circ}{22} = \frac{\sqrt{3}/2}{22} = \frac{\sqrt{3}}{44}\)
\(\sin A = 24 \cdot \frac{\sqrt{3}}{44} = \frac{6\sqrt{3}}{11}\)
\(\cos A = \sqrt{1 - \frac{108}{121}} = \frac{\sqrt{13}}{11}\)
\(C = 180^\circ - (A + 60^\circ)\)
\(\sin C = \sin(A + 60^\circ) = \sin A \cos 60^\circ + \cos A \sin 60^\circ\)
\(= \frac{6\sqrt{3}}{11} \cdot \frac{1}{2} + \frac{\sqrt{13}}{11} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3} + \sqrt{39}}{22}\)
By Law of Sines: \(\frac{c}{\sin C} = \frac{22}{\sin 60^\circ} = \frac{44}{\sqrt{3}}\)
\(c = \frac{44}{\sqrt{3}} \cdot \frac{3\sqrt{3} + \sqrt{39}}{22} = 2(3\sqrt{3} + \sqrt{39})/\sqrt{3} = 12 + 2\sqrt{3}\)
Step 3: Final Answer
\[ 12 + 2\sqrt{3} \]