cos(A-B = 3
Sin (A+B) = 1
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Answer:
Given sin(A+B)=1
⇒sin(A+B)=sin90
o
(∵sin90
o
=1)
⇒A+B=90
o
…………(1)
Again, cos(A−B)=
3
/2
⇒cos(A−B)=cos30
o
(∵cos30
o
=
3
/2)
⇒A−B=30
o
…………(2)
Adding (1)+(2)
A+B+A−B=90
o
+30
o
⇒2A=120
o
⇒A=120/2
⇒A=60
o
Putting A=60
o
in equation-(2) we get
A−B=30
o
⇒60
o
−B=30
o
⇒60
o
−30
o
=B
⇒B=30
o
∴A=60
o
;B=30
o
.
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