Решение треугольников с использованием теорем синусов и косинусов.
a) ∠A = 60°, ∠B = 40°, c = 14
- Найдем угол C: ∠C = 180° - ∠A - ∠B = 180° - 60° - 40° = 80°
- По теореме синусов: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
- $$a = \frac{c \cdot \sin A}{\sin C} = \frac{14 \cdot \sin 60°}{\sin 80°} \approx \frac{14 \cdot 0.866}{0.985} \approx 12.3$$
- $$b = \frac{c \cdot \sin B}{\sin C} = \frac{14 \cdot \sin 40°}{\sin 80°} \approx \frac{14 \cdot 0.643}{0.985} \approx 9.1$$
Ответ: ∠C = 80°, a ≈ 12.3, b ≈ 9.1
б) ∠A = 30°, ∠C = 75°, b = 4,5
- Найдем угол B: ∠B = 180° - ∠A - ∠C = 180° - 30° - 75° = 75°
- Так как ∠B = ∠C, треугольник равнобедренный, значит a = b = 4.5
- По теореме синусов: $$\frac{b}{\sin B} = \frac{c}{\sin C}$$
$$\frac{4.5}{\sin 75°} = \frac{c}{\sin 75°}$$
$$c = \frac{4.5 \cdot \sin 75°}{\sin 75°} = 4.5$$
Ответ: ∠B = 75°, a = 4.5, c = 4.5
в) ∠A = 80°, a = 16, b = 10
- По теореме синусов: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\frac{16}{\sin 80°} = \frac{10}{\sin B}$$
$$\sin B = \frac{10 \cdot \sin 80°}{16} \approx \frac{10 \cdot 0.985}{16} \approx 0.616$$
- ∠B = arcsin(0.616) ≈ 38.0°
- Найдем угол C: ∠C = 180° - ∠A - ∠B = 180° - 80° - 38° = 62°
- $$c = \frac{a \cdot \sin C}{\sin A} = \frac{16 \cdot \sin 62°}{\sin 80°} \approx \frac{16 \cdot 0.883}{0.985} \approx 14.3$$
Ответ: ∠B ≈ 38.0°, ∠C ≈ 62°, c ≈ 14.3
г) ∠B = 45°, ∠C = 70°, a = 24,6
- Найдем угол A: ∠A = 180° - ∠B - ∠C = 180° - 45° - 70° = 65°
- $$b = \frac{a \cdot \sin B}{\sin A} = \frac{24.6 \cdot \sin 45°}{\sin 65°} \approx \frac{24.6 \cdot 0.707}{0.906} \approx 19.2$$
- $$c = \frac{a \cdot \sin C}{\sin A} = \frac{24.6 \cdot \sin 70°}{\sin 65°} \approx \frac{24.6 \cdot 0.940}{0.906} \approx 25.5$$
Ответ: ∠A = 65°, b ≈ 19.2, c ≈ 25.5
д) ∠A = 60°, a = 10, b = 7
- $$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\frac{10}{\sin 60°} = \frac{7}{\sin B}$$
$$\sin B = \frac{7 \cdot \sin 60°}{10} \approx \frac{7 \cdot 0.866}{10} \approx 0.606$$
- ∠B = arcsin(0.606) ≈ 37.3°
- Найдем угол C: ∠C = 180° - ∠A - ∠B = 180° - 60° - 37.3° = 82.7°
- $$c = \frac{a \cdot \sin C}{\sin A} = \frac{10 \cdot \sin 82.7°}{\sin 60°} \approx \frac{10 \cdot 0.992}{0.866} \approx 11.5$$
Ответ: ∠B ≈ 37.3°, ∠C ≈ 82.7°, c ≈ 11.5
e) a = 6,3, b = 6,3, ∠C = 54°
- Так как a = b, треугольник равнобедренный, ∠A = ∠B.
- ∠A = ∠B = (180° - ∠C) / 2 = (180° - 54°) / 2 = 126° / 2 = 63°
- По теореме синусов: $$\frac{c}{\sin C} = \frac{a}{\sin A}$$
$$c = \frac{a \cdot \sin C}{\sin A} = \frac{6.3 \cdot \sin 54°}{\sin 63°} \approx \frac{6.3 \cdot 0.809}{0.891} \approx 5.7$$
Ответ: ∠A = 63°, ∠B = 63°, c ≈ 5.7
ж) b = 32, с = 45, ∠A = 87°
- По теореме косинусов: $$a^2 = b^2 + c^2 - 2bc \cdot \cos A$$
$$a^2 = 32^2 + 45^2 - 2 \cdot 32 \cdot 45 \cdot \cos 87°$$
$$a^2 = 1024 + 2025 - 2880 \cdot 0.052 \approx 3049 - 149.8 \approx 2899.2$$
- a = √2899.2 ≈ 53.8
- $$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{2899.2 + 2025 - 1024}{2 \cdot 53.8 \cdot 45} \approx \frac{3900.2}{4842} \approx 0.805$$
- ∠B = arccos(0.805) ≈ 36.4°
- ∠C = 180° - ∠A - ∠B = 180° - 87° - 36.4° = 56.6°
Ответ: a ≈ 53.8, ∠B ≈ 36.4°, ∠C ≈ 56.6°
з) а = 14, b = 18, c = 20
- По теореме косинусов:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{18^2 + 20^2 - 14^2}{2 \cdot 18 \cdot 20} = \frac{324 + 400 - 196}{720} = \frac{528}{720} \approx 0.733$$
- ∠A = arccos(0.733) ≈ 42.8°
- $$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{14^2 + 20^2 - 18^2}{2 \cdot 14 \cdot 20} = \frac{196 + 400 - 324}{560} = \frac{272}{560} \approx 0.486$$
- ∠B = arccos(0.486) ≈ 60.9°
- ∠C = 180° - ∠A - ∠B = 180° - 42.8° - 60.9° = 76.3°
Ответ: ∠A ≈ 42.8°, ∠B ≈ 60.9°, ∠C ≈ 76.3°
и) а = 6, b = 7,3, c = 4,8
- По теореме косинусов:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{7.3^2 + 4.8^2 - 6^2}{2 \cdot 7.3 \cdot 4.8} = \frac{53.29 + 23.04 - 36}{70.08} = \frac{40.33}{70.08} \approx 0.576$$
- ∠A = arccos(0.576) ≈ 54.8°
- $$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{6^2 + 4.8^2 - 7.3^2}{2 \cdot 6 \cdot 4.8} = \frac{36 + 23.04 - 53.29}{57.6} = \frac{5.75}{57.6} \approx 0.1$$
- ∠B = arccos(0.1) ≈ 84.3°
- ∠C = 180° - ∠A - ∠B = 180° - 54.8° - 84.3° = 40.9°
Ответ: ∠A ≈ 54.8°, ∠B ≈ 84.3°, ∠C ≈ 40.9°