Question

$\sin(a + b) = \frac{ \sqrt{3} }{2}$
$\cos(a - b ) = \frac{1}{2}$
Find 'A' and 'B'​

Answer 1

Answer:

A=B=30degree

Step-by-step explanation:

Step-by-step explanation:let (a+b)=A

Step-by-step explanation:let (a+b)=A (a-b)=B

Step-by-step explanation:let (a+b)=A (a-b)=Bsin(A+B)=sinA cosB+cosA sinB

Step-by-step explanation:let (a+b)=A (a-b)=Bsin(A+B)=sinA cosB+cosA sinBsin(a-b)=√[1-cos^2(a-b)]=√[1-(3/4)]=1/2

Step-by-step explanation:let (a+b)=A (a-b)=Bsin(A+B)=sinA cosB+cosA sinBsin(a-b)=√[1-cos^2(a-b)]=√[1-(3/4)]=1/2following the above procedure

Step-by-step explanation:let (a+b)=A (a-b)=Bsin(A+B)=sinA cosB+cosA sinBsin(a-b)=√[1-cos^2(a-b)]=√[1-(3/4)]=1/2following the above procedure cos(a+b)=√3/2

Step-by-step explanation:let (a+b)=A (a-b)=Bsin(A+B)=sinA cosB+cosA sinBsin(a-b)=√[1-cos^2(a-b)]=√[1-(3/4)]=1/2following the above procedure cos(a+b)=√3/2putting the value of A&B we get,

Step-by-step explanation:let (a+b)=A (a-b)=Bsin(A+B)=sinA cosB+cosA sinBsin(a-b)=√[1-cos^2(a-b)]=√[1-(3/4)]=1/2following the above procedure cos(a+b)=√3/2putting the value of A&B we get,sin(a+b+a-b)=sin(a+b)cos(a-b)+cos(a+b)sin(a-b)

Step-by-step explanation:let (a+b)=A (a-b)=Bsin(A+B)=sinA cosB+cosA sinBsin(a-b)=√[1-cos^2(a-b)]=√[1-(3/4)]=1/2following the above procedure cos(a+b)=√3/2putting the value of A&B we get,sin(a+b+a-b)=sin(a+b)cos(a-b)+cos(a+b)sin(a-b)putting the values mentioned in the question we get,

sin2a=(√3/2*1/2)+(√3/2*1/2)=√3/4+√3/4=√3/2

sin2a=√3/2=sin60,so,

a=30

sin (a+b)=√3/2=sin60

a+b=60,a=30,so b=30