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Clearly $\gamma$ has to be the angle oposite to the side of lenght 7. The law of cosines gives me $$7^2 = 5^2+3^2 - 2\cdot 5\cdot3\cos{\gamma}\Longleftrightarrow\cos{\gamma}=-\frac{7^2-5^2-3^2}{2\cdot5\cdot 3}=-\frac{1}{2}.$$

This value of $\cos{\gamma}$ means there can be two values of $\gamma$ satisfying it in $0\leq\gamma \leq 2\pi$. So $\sin{\gamma}$ can be either negative or positive. Using the trigonometric identity $\sin^2{\theta}+\cos^2{\theta}=1$, I obtain $$\sin{\gamma}=\pm\sqrt{1-\cos{\gamma}}=\pm\sqrt{1-\left(-\frac{1}{2}\right)^2}=\pm\frac{\sqrt{3}}{2}.$$

Using the positive value I get the correct answer of $$\cos{\gamma} + \sin{\gamma}=-\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}-1}{2}.$$

Why is it incorrect to use the negative value of $\sin{\gamma}$ ?

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As $0<\gamma<\pi$ (actually $\frac{\pi}{2}<\gamma<\pi$), we have $\sin\gamma>0$

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  • $\begingroup$ How did you conclude that $\pi/2 < \gamma < \pi$? $\endgroup$ – Parseval May 7 '17 at 17:59
  • $\begingroup$ $\cos\gamma$ is negative. So $\gamma$ must be obtuse. $\endgroup$ – CY Aries May 7 '17 at 18:00
  • $\begingroup$ True, damn I should think of it myself. $\endgroup$ – Parseval May 7 '17 at 18:01
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For $0^\circ<\gamma<180^\circ$ – the range of an angle in a non-degenerate triangle – $\sin\gamma$ must be positive.

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  • $\begingroup$ Do you agree with the OP that $\cos\gamma=-\frac12$ ? $\endgroup$ – John Bentin May 7 '17 at 17:54
  • $\begingroup$ @JohnBentin Yes, the answer is correct. $\endgroup$ – Parcly Taxel May 7 '17 at 17:56
  • $\begingroup$ Yes I understand. Thanks! $\endgroup$ – Parseval May 7 '17 at 18:00
  • $\begingroup$ OK—then why do you say that the triangle is acute and $\gamma<90^\circ$ ? $\endgroup$ – John Bentin May 7 '17 at 18:12
  • $\begingroup$ @JohnBentin Please remove your downvote. Must have been an oversight on my part. $\endgroup$ – Parcly Taxel May 7 '17 at 18:13

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