Answer
$\frac{1}{15}(4+6\sqrt 2)$
Work Step by Step
sin x = $\frac{1}{3}$
Pythagoras theorem:
$1^{2}+{a}^{2}=3^{2}$
a=$\sqrt 8$
cos x = $\frac{\sqrt 8}{3}$
sec y = $\frac{5}{4}$
cos y = $\frac{1}{sec y}$=$\frac{4}{5}$
Pythagoras theorem:
$b^{2}+{4}^{2}=5^{2}$
b=3
sin y = $\frac{3}{5}$
Expand:
sin ( x + y )
= sin x cos y + sin y cos x
= $\frac{1}{3}\times\frac{4}{5} + \frac{3}{5}\times\frac{\sqrt 8}{3}$
=$\frac{4+6\sqrt 2}{15}$
=$\frac{1}{15}(4+6\sqrt 2)$