If sin (A + B) = 1 and cos (A -B) = √3÷2, then find the value of A and B.
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Answer:
sin(A + B) = 1
sin(A + B) =sin 90
A +B= 90
cos(A-B)= √3/2
cos(A-B)= cos 30
A-B=30
from 1 and 2 equation
A+B=90/A-B=30
by elimination method
2A =120
A =120/2
A=60.
by putting the value of A in equation 1st
A+B=90
60+B=90
B=90-60
B=30
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