Question

2sin(A+B)=root3 and tanB= 1 then find the value of sin2B​

Answer 1

Answer:

Step-by-step explanation:

Given that  2sin(A + B) = $\sqrt{3}$    and     tanB = 1    to find    sin2B

first we solve  2sin(A + B) =$\sqrt{3}$

by dividing both side by 2

⇒ sin(A + B) =√3/2

by taking the sin inverse

⇒ A + B = arcsin(√3/2)

⇒A + B = 60   ································(1)

therefore tanB = 1

⇒ B = arctan(1)

⇒B = 45 ..................................(2)

then we solve equation (1) and (2)

A + B = 60

A = 60 - 45

A = 15

the value of sin2B

sin2B = sin(2×45)

sin2B = sin90

⇒sin2B = 1

in summary

A = 15,    B = 60,    and  sin2B = 1