Saved Bookmarks
| 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^(@)`
|
|