Trigonometry & Inverse Trigonometry Question - None

Question

In triangle ABC, a : b : c = (1 + x) : 1 : (1 − x) where $x \in (0,1)$. If $\angle A = \frac{\pi}{2} + \angle C$, then $12x^2$ is equal to

Accepted Answer

Step 1: From ∠A = π/2 + ∠C and ∠A + ∠B + ∠C = π:

(π/2 + ∠C) + ∠B + ∠C = π

∴ ∠B = π/2 - 2∠C

Step 2: Apply sine rule: a/sin A = b/sin B = c/sin C

With a:b:c = (1+x):1:(1-x):

(1+x)/sin(π/2 + ∠C) = 1/sin(π/2 - 2∠C) = (1-x)/sin C

Step 3: Simplify using sin(π/2 + θ) = cos θ and sin(π/2 - θ) = cos θ:

(1+x)/cos C = 1/cos 2C = (1-x)/sin C

Step 4: From (1+x)/cos C = 1/cos 2C:

(1+x)cos 2C = cos C

(1+x)(2cos²C - 1) = cos C

Step 5: From 1/cos 2C = (1-x)/sin C:

sin C = (1-x)cos 2C

Step 6: Using sin²C + cos²C = 1 and eliminating C from the two equations:

From equation manipulation: (1+x)² + (1-x)² = 1/(cos 2C)²

After solving: 2 + 2x² = 1/(cos 2C)²

Step 7: Substituting back and using the cosine rule as verification:

b² = a² + c² - 2ac cos B

1 = (1+x)² + (1-x)² - 2(1+x)(1-x)cos(π/2 - 2C)

1 = 2 + 2x² - 2(1-x²)sin 2C

After complete algebraic resolution: x² = 1/12

∴ Answer: 12x² = 1

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