In an acute triangle ABC, if sin2(A+B-C) = 1 and tan(B+C-A)=√3, than the value of angle 
B is
সমাধান
সঠিক উত্তর: Option B
দেয়া আছে, sin2(A+B-C) = 1
=> sin2(A+B-C) = sin900 [ ∴sin900 = 1 ]
=> 2(A+B-C) = 900
A+B-C = 450 ...............(i)
আবার, tan(B+C-A)=√3
=> tan(B+C-A)= tan600 [ ∴tan600 = √3 ]
=> B+C-A= 600 ...........(ii)
এখন, (i) এবং (ii) নং সমীকরণ যোগ করে পাই
A+B-C = 450
B+C-A= 600
2B = 1050
B = 1050/2 = 52Λ1/2
=> sin2(A+B-C) = sin900 [ ∴sin900 = 1 ]
=> 2(A+B-C) = 900
A+B-C = 450 ...............(i)
আবার, tan(B+C-A)=√3
=> tan(B+C-A)= tan600 [ ∴tan600 = √3 ]
=> B+C-A= 600 ...........(ii)
এখন, (i) এবং (ii) নং সমীকরণ যোগ করে পাই
A+B-C = 450
B+C-A= 600
2B = 1050
B = 1050/2 = 52Λ1/2
