What is the value of a+b+\sqrt{2}c equal to?

Consider the following for the next two (02) items that follow : The angles A, B and C of a triangle ABC are in the ratio 3:5:4.

  1. A. 3a
  2. B. 2b
  3. C. 3b
  4. D. 2c

Correct Answer: C. 3b

Explanation

The angles are in the ratio 3:5:4. Adding gives 12 parts = 180^\circ, so A = 45^\circ, B = 75^\circ, C = 60^\circ. By the Sine Rule, a = 2R\sin 45^\circ = \sqrt{2}R, b = 2R\sin 75^\circ = R\frac{\sqrt{6}+\sqrt{2}}{2}, and c = 2R\sin 60^\circ = \sqrt{3}R. Dropping R for proportional length, a+b+\sqrt{2}c = \sqrt{2} + \frac{\sqrt{6}+\sqrt{2}}{2} + \sqrt{6} = \frac{3\sqrt{6}+3\sqrt{2}}{2}, which clearly equals 3b.

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