Question

( sin [n+B]=1 )
( sin [A B]=sin 40 )
( A+B=90 )
( cos [n-B]=frac{sqrt{3}}{2} )
( cos {A-B]=cos 30 )
( A-B=30 )
trom
( A+B=90 )
( frac{A-B=30}{2 n=120} )
( vec{n}=60 Rightarrow overrightarrow{B=30} )

# COSO here A > B. Find 19. If sin(A + B) = 1 and COS(A - B) = 13, where A > B. Find the value of B? (1) 30Ⓡ (2) 45° (3) 60° (4) 90°

Solution