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Question
In ΔABC, if `"cos A"/"a" = "cos B"/"b"`, then show that it is an isosceles triangle.
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Solution
Given: `"cos A"/"a" = "cos B"/"b"` ....(1)
By sine rule,
`"a"/"sin A" = "b"/"sin B" = "k"`
∴ a = k sin A, b = k sin B
∴ (1) gives,
`"cos A"/"k sin A" = "cos B"/"k sin B"`
∴ `"cos A"/"sin A" = "cos B"/"sin B"`
∴ sin A cos B = cos A sin B
∴ sin A cos B – cos A sin B = 0
∴ sin (A – B) = 0 = sin 0
∴ A – B = 0
∴ A = B
∴ The triangle is an isosceles triangle.
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