Question

In ∆ABC, if `(cos "A")/"a" = (cos "B")/"b"`, then show that it is an isosceles triangle

Answer 1

In ∆ABC by sine rule, we have

`"a"/"sin A" = "b"/"sin B" = "k"`

∴ a = k sin A, b = k sin B 

Now, `(cos "A")/"a" = (cos "B")/"b"`    .......[Given]

∴ `"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

Hence, ∆ABC is an isosceles triangle.