7. Zadatak

Prvo rješenje.

Neka su $a$, $b$ i $c$ stranice trokuta, redom nasuprot kutova $\alpha$, $\beta$ i $\gamma$. Primijenimo poučak o sinusima.

$$\frac a{\sin\alpha}=2R\Rightarrow\sin\alpha=\frac a{2R},\qquad \frac b{\sin\beta}=2R\Rightarrow\sin\beta=\frac b{2R},\qquad \frac c{\sin\gamma}=2R\Rightarrow\sin\gamma=\frac c{2R}.$$

Tada početna jednakost prelazi u

$$\frac{\frac{c^2}{4R^2}+\frac{b^2}{4R^2}-\frac{a^2}{4R^2}}{\frac b{2R}\cdot\frac c{2R}}=\sqrt3$$

odnosno

$$\frac{c^2+b^2-a^2}{bc}=\sqrt3.$$

Slijedi $c^2+b^2-a^2=\sqrt3bc$.

Prema poučku o kosinusu

$$\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$$

pa je

$$\cos\alpha=\frac{\sqrt3bc}{2bc}=\frac{\sqrt3}{2}\Rightarrow\alpha=30^\circ.$$

Drugo rješenje.

Zapišimo $\alpha=\pi-(\beta+\gamma)$.

$$\frac{\sin^2\gamma+\sin^2\beta-\sin^2\alpha}{\sin\beta\sin\gamma} =\frac{\sin^2\gamma+\sin^2\beta-\sin^2(\beta+\gamma)}{\sin\beta\sin\gamma}.$$ $$\frac{\sin^2\gamma+\sin^2\beta-\sin^2\beta\cos^2\gamma-\sin^2\gamma\cos^2\beta-2\sin\beta\sin\gamma\cos\beta\cos\gamma}{\sin\beta\sin\gamma}.$$

Ako u brojniku grupiramo prvi i četvrti pribrojnik, te drugi i treći pribrojnik, dobivamo

$$\frac{\sin^2\gamma\sin^2\beta+\sin^2\gamma\sin^2\beta-2\sin\beta\sin\gamma\cos\beta\cos\gamma}{\sin\beta\sin\gamma}$$

odnosno

$$\frac{2\sin\beta\sin\gamma(\sin\beta\sin\gamma-\cos\beta\cos\gamma)}{\sin\beta\sin\gamma}=-2\cos(\beta+\gamma).$$

Iz uvjeta zadatka slijedi

$$-2\cos(\beta+\gamma)=\sqrt3\Rightarrow\beta+\gamma=\frac{5\pi}{6}.$$

Mjera traženog kuta $\alpha$ je $\frac\pi6=30^\circ$.