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Question
In any ΔABC, with usual notations, prove that b2 = c2 + a2 – 2ca cos B.
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Solution
Let us take the angle B of ΔABC in standard position,
i.e. B as origin, X-axis along the line BC and the Y-axis perpendicular to the line BC.
In the two figures, ∠B is shown as acute in one and obtuse in the other.
∵ l(BC) = a
∴ C ≡ (a, 0)
Let A ≡ (x, y)
Since l (BA) = c, we have
cos B = `x/c` and sin B = `y/c`
∴ x = c cos B and y = c sin B
∴ A ≡ (c cos B, c sin B)
∴ By the distance formula
b2 = AC2 = (a – c cos B)2 + (0 – c sin B)2
= a2 – 2ca cos B + c2 cos2B + c2 sin2B
= c2(cos2B + sin2B) + a2 – 2ca cos B
∴ b2 = c2 + a2 – 2ca cos B.
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