1.

Let triangle ABC be an isosceles triangle with AB=AC. Suppose that the angle bisector of its angle B meets the side AC at a point D and that BC=BD+AD. Find angle A (in degrees).

Answer»

Solution :`BC=BD+AD`
`a=p+q`
or `(a)/(p)=1+(q)/(p)`
USING SINE law in `DELTAABD`
`(sin3x)/(sin2x)=1+(SINX)/(sin4x)implies(sin3x.sin4x-sinx.sin2x)/(sin2x.sin4x)=1`
`2sin3x.sin4x-2sinx.sin2x=2sin2x.sin4x`
`(cosx-cos7x)-(cosx-cos3x)=cos2x-cos6x`
`cos3x-cos7x=cos2x-cos6x`
`2sin5x.sin2x=2sin4x.sin2x`
as `sin5x=sin4x`
`5x+4x=180^(@)""implies""x=20^(@),"""HENCE "angleA=180^(@)-80^(@)=100^(@)`


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