Question

if in a triangle ABC right angled at B, AB= 6 units and BC= 8 units then find the value of sinAcosC+cosAsinC

Answer 1

Answer:

AB=6units

BC=8units

AC*2=6*2+8*2

ac*2=36+64

ac=√100

ac=10

sinA=8/10

sinC=6/10

cosA=6/10

cosC=8/10

(8/10)(8/10)+(6/10)(6/10)=1