نویسنده : امیر انصاری

1. در \(\triangle{ABC}\) ، \(\angle{C}=31^{\circ}\)، \(a=5.6 \text{ cm}\)، و \(c=3.9 \text{ cm}\) .
   
2. در \(\triangle{PQR}\) ، \(\angle{Q}=43^{\circ}\)، \(p=20 \text{ cm}\)، و \(q=15 \text{ cm}\) .
   
3. در \(\triangle{XYZ}\) ، \(\angle{X}= 53^{\circ}\)، \(x=8.5 \text{ cm}\)، و \(z=12.3 \text{ cm}\) .
   

پاسخ

1. $$
   \frac{\sin A}{a} = \frac{\sin C}{c}\\
   \frac{\sin A}{5.6} = \frac{\sin 31^{\circ}}{3.9}\\
   \sin A = \frac{5.6 \sin 31^{\circ}}{3.9}\\
   \angle{A} = \sin^{-1} \biggl( \frac{5.6 \sin 31^{\circ}}{3.9} \biggr)\\
   \angle{A} \approx 48^{\circ}\\
   \angle{A} = 180^{\circ} - 48^{\circ} = 132^{\circ}\\
   $$
   اگر \(\angle{A} = 48^{\circ}\) :
   $$
   \angle{B} = 180^{\circ} - 48^{\circ} - 31^{\circ} \\
   \angle{B} = 101^{\circ}\\
   \frac{b}{\sin B} = \frac{c}{\sin C}\\
   \frac{b}{\sin 101^{\circ}} = \frac{3.9}{\sin 31^{\circ}}\\
   b \approx 7.4 \text{ cm}
   $$
   اگر \(\angle{A} = 132^{\circ}\) :
   $$
   \angle{B} = 180^{\circ} - 132^{\circ} - 31^{\circ} \\
   \angle{B} = 17^{\circ}\\
   \frac{b}{\sin B} = \frac{c}{\sin C}\\
   \frac{b}{\sin 17^{\circ}} = \frac{3.9}{\sin 31^{\circ}}\\
   b \approx 2.2 \text{ cm}
   $$
   
2. $$
   \frac{\sin P}{p} = \frac{\sin Q}{q}\\
   \frac{\sin P}{20} =\frac{\sin 43^{\circ}}{15}\\
   \sin P = \frac{20 \sin 43^{\circ}}{15}\\
   \angle{P} = \sin^{-1} \biggl( \frac{20 \sin 43^{\circ}}{15} \biggr)\\
   \angle{P} \approx 65^{\circ}\\
   \angle{P} = 180^{\circ} - 65^{\circ} = 115^{\circ}\\
   \text{ }\\[2ex]
   \text{if } \angle{P} = 65^{\circ}\\
   \angle{R} = 180^{\circ} - 65^{\circ} - 43^{\circ} = 72^{\circ}\\
   \frac{r}{\sin R} = \frac{q}{\sin Q}\\
   \frac{r}{\sin 72^{\circ}} = \frac{15}{\sin 43^{\circ}}\\
   r \approx 20.9 \text{ cm}\\
   \text{ }\\[2ex]
   \text{if } \angle{P} = 115^{\circ}\\
   \angle{R} = 180^{\circ} - 115^{\circ} - 43^{\circ} = 22^{\circ}\\
   \frac{r}{\sin R} = \frac{q}{\sin Q}\\
   \frac{r}{\sin 22^{\circ}} = \frac{15}{\sin 43^{\circ}}\\
   r \approx 8.2 \text{ cm}
   $$
   
3. $$
   \frac{\sin Z}{z} = \frac{\sin X}{x}\\
   \frac{\sin Z}{12.3} = \frac{\sin 53^{\circ}}{8.5}\\
   \sin Z = \frac{12.3 \sin 53^{\circ}}{8.5}\\
   \sin Z \approx 1.156\\
   \angle{Z} = \sin^{-1} (1.156)\\
   \angle{Z} = \text{undefined} (\text{ تعریف نشده })
   $$
   از آنجا که معکوس سینوس در \(\angle{Z}\) تعریف نشده است، این مسأله هیچ پاسخی ندارد.