Given that $\sin x = \frac{5}{13}$ and $\sin y = \frac{8}{17}$ , where x and y are acute, find $\cos(x+y)$ .

$\frac{130}{221}$

$\frac{140}{221}$

$\frac{140}{204}$

$\frac{220}{23}$

Correct answer is B

$\cos(x+y) = \cos x\cos y - \sin x\sin y$

Given $\sin$ of an angle implies we have the value of the opposite and hypotenuse of the right-angled triangle. We find the adjacent side using Pythagoras' theorem.

$Adj^{2} = Hyp^{2} - Opp^{2}$

For triangle with angle x, $adj = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12$

For triangle with angle y, $adj = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15$

$\therefore \cos x = \frac{12}{13}; \cos y = \frac{15}{17}$

$\cos(x+y) = (\frac{12}{13}\times\frac{15}{17}) - (\frac{5}{13}\times\frac{8}{17}) = \frac{180}{221} - \frac{40}{221}$

= $\frac{140}{221}$

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