Answer:

∠B=90

0

=π/2

So, (I) sinAcosC+cosAsinC=sin(A+C)=sin(π−B)

=sinB=sinπ/2=1

(II) cosAcosC−sinAsinC=cos(A+C)=cos(π−B)

=−cosB=−cos(π/2)=0

Verified answer

Answer:

angle B =90 degree

Hope it helps you blink