[tex]d_1 = 13 \ cm\\d_2 = 8 \ cm\\\phi = 60^{o}\\sin\phi = sin60^{o} = \frac{\sqrt{3}}{2}\\P = ?\\\\\\Korzystamy \ ze \ wzoru \ na \ pole \ rownolegloboku:\\\\P = \frac{1}{2}\cdot d_1\cdot d_2\cdot sin\phi\\\\P = \frac{1}{2}\cdot13 \ cm\cdot8 \ cm\cdot sin60^{o}\\\\P = 13 \ cm\cdot4 \ cm \cdot\frac{\sqrt{3}}{2}\\\\\underline{P = 26\sqrt{3} \ cm^{2}}}[/tex]
