Question

if sin(A+B)=1 and cos(A-B)=√3/2;o<A+B lessthanequal 90,A>B then find A and B​

Answer 1

Step-by-step explanation:

sin(A+B)=1,but sin 90°=1

therefore,A+B=90°

cos(A-B)=(√3/2),but cos30°=(√3/2)

therefore,A-B=30°

adding

A+B+A-B=120°

2A=120°

A=60°

therefore,A+B=90°

60°+B=90°

B=30°

Answer 2

Answer:

$a + b = 90 \\ 60 + b = 90 \\ b = 90 - 60 \\ \: \: \: = 30$