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Question
Prove that in any triangle ABC, cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`, where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.
Solution
Here, in the given figure, the components of c are c cos A and c sin A.
∴ `vec"CD"` = b – c cos A
In ΔBDC,
a2 = CD2 + BD2
⇒ a2 = (b – c cos A)2 + (c sin A)2
⇒ a2 = b2 + c2 cos2A – 2bc cos A + c2 sin2A
⇒ a2 = b2 + c2 (cos2A + sin2A) – 2bc cos A
⇒ a2 = b2 + c2 – 2bc cos A
⇒ 2bc cos A = b2 + c2 – a2
∴ cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`
Hence Proved.
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