Question

in a triangle ABC right angled at B,AB=6units and BC=8units,then find the value of sin A.cos C+cos A.sin C.

Answer 1

by pythagorous theorum ac=10
therefore,
sinA=BC/AB=8/10
cosC=BC/AC=8/10
sinC=AB/AC=6/10
cosA=AB/AC=6/10
sinA.cosC+sinC.cosA=[8×8/10×10]+[6×6/10×10]
                                       =64/100+36/100
                                        =100/100
                                      =1