If ∠ A and ∠ B are acute angles such that sinA = sinB then prove that ∠ A = ∠ B.
Consider ΔABC to be a right - angled triangle.
∴ sinA = BC/AB
sinB = AC/AB
Given that sinA = sinB
BC/AB = AC/AB
→ BC = AC
→ ∠ A = ∠ B (In a triangle, angles opposite to equal angles are equal)
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