نویسنده : امیر انصاری

پاسخ

1. $$
   p^2=q^2+r^2-2qr \cos P\\
   p^2 = 28^2+29^2-2(28)(29) \cos 52^{\circ}\\
   p = \sqrt{28^2+29^2-2(28)(29) \cos 52^{\circ}}\\
   p =25.003...\\
   p \approx 25.0 \text{ km}\\
   \text{ }\\[2ex]
   \cos Q=\frac{r^2 + p^2 - q^2}{2rp}\\
   \angle{Q} = \cos^{-1} \biggl( \frac{r^2 + p^2 - q^2}{2rp} \biggr)\\
   \angle{Q} = \cos^{-1} \biggl( \frac{29^2 + 25^2 - 28^2}{2(29)(25)} \biggr)\\
   \angle{Q} = 61.943...^{\circ}\\
   \angle{Q} \approx 62^{\circ}\\
   \text{ }\\[2ex]
   \angle{R} = 180^{\circ} - 62^{\circ} - 52^{\circ} = 66^{\circ}
   $$
   
2. $$
   \cos R = \frac{s^2 + t^2 -r^2}{2st}\\
   \angle{R} = \cos^{-1} \biggl( \frac{s^2 + t^2 -r^2}{2st} \biggr)\\
   \angle{R} = \cos^{-1} \biggl( \frac{9.1^2 + 6.8^2 - 5^2}{2(9.1)(6.8)} \biggr)\\
   \angle{R} = 32.781...^{\circ}\\
   \angle{R} \approx 33^{\circ}\\
   \text{ }\\[2ex]
   \cos S = \frac{r^2 + t^2 -s^2}{2rt}\\
   \angle{R} = \cos^{-1} \biggl( \frac{r^2 + t^2 -s^2}{2rt} \biggr)\\
   \angle{R} = \cos^{-1} \biggl( \frac{5^2 + 6.8^2 - 9.1^2}{2(5)(6.8)} \biggr)\\
   \angle{R} = 99.796...^{\circ}\\
   \angle{R} \approx 100^{\circ}\\
   \text{ }\\[2ex]
   \angle{T} = 180^{\circ} - 100^{\circ} - 33^{\circ} = 47^{\circ}
   $$