Question

If cosA=5/13 and SinB=4/5 then find Sin(A+B) where A,B and A+b are positive acute angles..

Answer 1

cosA=5/13
∴, sinA
=√(1-cos²A)
=√(1-25/169)
=√(169-25)/169
=√(144/169)
=12/13 [neglecting the negative sign since A is an acute angle]
sinB=4/5
∴. cosB
=√(1-sin²B)
=√(1-16/25)
=√(25-16)/25
=√(9/25)
=3/5 [neglecting the negative sign since A is an acute angle]
∴, sin(A+B)
=sinAcosB+cosAsinB
=12/13×3/5+5/13×4/5
=36/65+20/65
=(36+20)/65
=56/65

Answer 2

Step-by-step explanation:

Yeah. Its correct. I have clarified my doubt