In a triangle ABC, \angle A=75^{\circ} and \angle B=45^{\circ}. What is 2a-b equal to?
- A. c
- B. \sqrt{2}c ✓
- C. 2c
- D. 2\sqrt{2}c
Correct Answer: B. \sqrt{2}c
Explanation
The third angle \angle C = 180^{\circ} - (75^{\circ} + 45^{\circ}) = 60^{\circ}. Using the Sine Rule: \frac{a}{\sin 75^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 60^{\circ}} = k. This gives a = k\frac{\sqrt{3}+1}{2\sqrt{2}}, b = k\frac{1}{\sqrt{2}}, and c = k\frac{\sqrt{3}}{2}. Then 2a - b = 2(k\frac{\sqrt{3}+1}{2\sqrt{2}}) - k\frac{1}{\sqrt{2}} = k\frac{\sqrt{3}+1}{\sqrt{2}} - k\frac{1}{\sqrt{2}} = k\frac{\sqrt{3}}{\sqrt{2}}. Notice that \sqrt{2}c = \sqrt{2}(k\frac{\sqrt{3}}{2}) = k\frac{\sqrt{3}}{\sqrt{2}}. Hence, 2a - b = \sqrt{2}c.