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If the angles A, B, C of ΔABC are in A.P. and its sides a, b, c are in G.P., then show that a2, b2, c2 are in A.P.

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Question

If the angles A, B, C of ΔABC are in A.P. and its sides a, b, c are in G.P., then show that a2, b2, c2 are in A.P.

Sum
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Solution

A, B, C are in A.P.

∴ A + C = 2B

We know that A + B + C = 180°

∴ 2B + B = 180°

∴ 3B = 180°

∴ ∠B = 60°  .......(i)

Also, it is given that sides a, b, c are in G.P.

∴ ac = b2  .......(ii)

Consider, cos B = `("a"^2 + "c"^2 - "b"^2)/(2"ac")`  .......[By cosine rule]

∴ cos(60°) = `("a"^2 + "c"^2 - "b"^2)/(2"b"^2)`  .......[From (i) and (ii)]

∴ `1/2 = ("a"^2 + "c"^2 - "b"^2)/(2"b"^2)`

∴ b2 = a2 + c2 – b2

∴ a2 + c2 = 2b2

∴ a2, b2, c2 are in A.P.

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Chapter 1.3: Trigonometric Functions - Short Answers II

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