Answer:
[tex] \frac{63}{65} [/tex]
Step-by-step explanation:
We have to find the value of sinB & cosA . So to find the value of sinB , let's use the identity
[tex] { \sin }^{2} b + { \cos }^{2} b = 1[/tex]
By using the identity above gives
[tex] { \sin}^{2} b + ( { \frac{5 }{13} })^{2} = 1[/tex]
[tex] = > { \sin}^{2} b = 1 - \frac{25}{169} = \frac{144}{169} [/tex]
[tex] = > \sin(b) = \sqrt{ \frac{144}{169} } = \frac{12}{13} [/tex]
Now to find the value of cosA , we'll use the same identity.
[tex] {( \frac{3}{5} })^{2} + { \cos}^{2} a = 1[/tex]
[tex] = > { \cos }^{2} a = 1 - \frac{9}{25} = \frac{16}{25} [/tex]
[tex] = > \cos(a) = \sqrt{ \frac{16}{25} } = \frac{4}{5} [/tex]
Now we know that
[tex] \sin(a + b)=\sin(a)\cos(b)+\sin(b) \cos(a) [/tex]
So value of sin(a+b) =
[tex] \frac{3}{5} \times \frac{5}{13} + \frac{12}{13} \times \frac{4}{5} [/tex]
[tex] = \frac{3}{13} + \frac{48}{65} [/tex]
[tex] = \frac{63}{65} [/tex]
