Question

Find sin(a+B)

Answer 1

Answer:

Cos A=5/13

we have

Cos² A=$1-Sin ²A$

25/169=1-Sin²A

sin²A=1-25/169

sin²A=144/169

Sin A=$\sqrt{144/169}=12/13$

again

Tan B=4/3

P/b=4/3

p=4

b=3

h=$\sqrt{3²+4²}=5$

Now

Sin B=p/h=4/5

in IV quadrant sin angle is negative so

Sin B=-4/5

CosB=b/h=3/5

Now

Sin(A+B):sinAcosB+CosAsinB

now

substitute value

Sin(A+B):12/13*3/5+5/13*(-4/5)=36/65-4/13

=16/65 is a required answer