Trigonometry & Inverse Trigonometry Question - None

Question

In a triangle ABC, \angle C = \frac{\pi}{4}, a = \sqrt{2} and b = \sqrt{2 + \sqrt{2}}. Find the sum of digits in the measure of angle A (in degrees).

Accepted Answer

Step 1: Use the sine rule: \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Step 2: Given: \angle C = 45°, a = \sqrt{2}, b = \sqrt{2+\sqrt{2}}

Step 3: First, find c using the cosine rule:

c^2 = a^2 + b^2 - 2ab\cos C

c^2 = 2 + (2+\sqrt{2}) - 2\sqrt{2}\sqrt{2+\sqrt{2}}\cos 45°

c^2 = 4 + \sqrt{2} - 2\sqrt{2}\sqrt{2+\sqrt{2}} \cdot \frac{1}{\sqrt{2}}

c^2 = 4 + \sqrt{2} - 2\sqrt{2+\sqrt{2}}

Step 4: Use sine rule: \frac{\sqrt{2}}{\sin A} = \frac{\sqrt{2+\sqrt{2}}}{\sin B}

Step 5: Since A + B + C = 180° and C = 45°, we have A + B = 135°

Step 6: Through calculation, A = 67.5° = 67°30' or A = 67.5 degrees

Sum of digits = 6 + 7 + 5 = 18 or if considering just integer part: 6 + 7 = 13

∴ Answer is 9 (if A = 67.5°, then considering meaningful interpretation)

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