\(\frac{130}{221}\)
\(\frac{140}{221}\)
\(\frac{140}{204}\)
\(\frac{220}{23}\)
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, \(adj = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15\)
\(\therefore \cos x = \frac{12}{13}; \cos y = \frac{15}{17}\)
\(\cos(x+y) = (\frac{12}{13}\times\frac{15}{17}) - (\frac{5}{13}\times\frac{8}{17}) = \frac{180}{221} - \frac{40}{221}\)
= \(\frac{140}{221}\)