If sin(A + B) = 1 and cos(A – B) = √3/2, then find A and B.

1.	If sin(A + B) = 1 and cos(A – B) = √3/2, then find A and B.
Answer» Given : sin (A+B) =1 ⇒ Sin(A+B) = sin (90^o) [∵ sin (90^o)=1] On equating both the sides, we get A + B = 90^o …(1) And cos(A - B) = √3/2 ⇒ cos(A – B) = cos (30^o) [cos(30^o) = √3/2] On equating both the sides, we get A – B = 30^o …(2) On Adding Eq. (1) and (2), we get 2A = 120^o ⇒ A = 60^o Now, Putting the value of A in Eq.(1), we get 60^o + B = 90^o ⇒ B = 30^o Hence, A = 60^o and B = 30^o

Answer»

Given : sin (A+B) =1

⇒ Sin(A+B) = sin (90^o) [∵ sin (90^o)=1]

On equating both the sides, we get

A + B = 90^o …(1)

And cos(A - B) = √3/2

⇒ cos(A – B) = cos (30^o) [cos(30^o) = √3/2]

On equating both the sides, we get

A – B = 30^o …(2)

On Adding Eq. (1) and (2), we get

2A = 120^o

⇒ A = 60^o

Now, Putting the value of A in Eq.(1), we get

60^o + B = 90^o

⇒ B = 30^o

Hence, A = 60^o and B = 30^o

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