نویسنده : امیر انصاری

پاسخ

1. $$
   j^2 = h^2+l^2 -2hl \cos J\\
   2hl \cos J = h^2+l^2 -j^2\\
   \cos J = \frac{h^2+l^2 -j^2}{2hl}\\
   \angle{J} = \cos^{-1} \biggl( \frac{h^2+l^2 -j^2}{2hl} \biggr)\\
   \angle{J} = \cos^{-1} \biggl( \frac{11^2+17^2-10^2}{2(11)(17)} \biggr)\\
   \angle{J} = 34.016...^{\circ}\\
   \angle{J} \approx 34^{\circ}
   $$
   
2. $$
   \cos L = \frac{n^2+m^2-l^2}{2nm}\\
   \angle{L} = \cos^{-1} \biggl( \frac{n^2+m^2-l^2}{2nm} \biggr)\\
   \angle{L} = \cos^{-1} \biggl( \frac{21.9^2+10.4^2-18^2}{2(21.9)(10.4)} \biggr)\\
   \angle{L} =54.616...\\
   \angle{L} \approx 55^{\circ}
   $$
   
3. $$
   \cos P=\frac{q^2+r^2-p^2}{2qr}\\
   \angle{P} = \cos^{-1} \biggl( \frac{q^2+r^2-p^2}{2qr} \biggr)\\
   \angle{P} = \cos^{-1} \biggl( \frac{6^2+9^2-14^2}{2(6)(9)} \biggr)\\
   \angle{P} = 137.010...^{\circ}\\
   \angle{P} \approx 137^{\circ}
   $$
   
4. $$
   \cos C = \frac{a^2+b^2-c^2}{2ab}\\
   \angle{C} = cos^{-1} \biggl( \frac{a^2+b^2-c^2}{2ab} \biggr)\\
   \angle{C} = cos^{-1} \biggl( \frac{20^2+13^2-31^2}{2(20)(13)} \biggr)\\
   \angle{C} =138.924...^{\circ}\\
   \angle{C} \approx 139^{\circ}
   $$