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МатематикаЗадание 45. Найдите x – длину стороны ΔABC. 1) По теореме синусов: Ответ: 2) Ответ: 3) Ответ: 4) Ответ: 5) Ответ: 6) Ответ: 7) sinA=0,3, sinC=0,2 Ответ: 8) sinC=1/4, sinB=5/6 Ответ:
Ответ:
Задание 45. Найдите x – длину стороны ΔABC.
1)
Дано: треугольник ABC, AB = 7, ∠B = 45°, ∠C = 30°.
Найти: BC = x.
По теореме синусов: \[\frac{AB}{\sin C} = \frac{BC}{\sin A}\]
\[\frac{7}{\sin 30^\circ} = \frac{x}{\sin (180^\circ - 45^\circ - 30^\circ)}\]
\[\frac{7}{\frac{1}{2}} = \frac{x}{\sin 105^\circ}\]
\[105^\circ = 60^\circ + 45^\circ\]
\[\sin (60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\]
\[\frac{7}{\frac{1}{2}} = \frac{x}{\frac{\sqrt{6} + \sqrt{2}}{4}}\]
\[14 = \frac{4x}{\sqrt{6} + \sqrt{2}}\]
\[4x = 14(\sqrt{6} + \sqrt{2})\]
\[x = \frac{7}{2}(\sqrt{6} + \sqrt{2})\]
Ответ: \[\frac{7}{2}(\sqrt{6} + \sqrt{2})\]
2)
Дано: треугольник ABC, AC = √2, ∠B = 45°, ∠C = 60°.
Найти: AB = x.
По теореме синусов: \[\frac{AB}{\sin C} = \frac{AC}{\sin B}\]
\[\frac{x}{\sin 60^\circ} = \frac{\sqrt{2}}{\sin 45^\circ}\]
\[\frac{x}{\frac{\sqrt{3}}{2}} = \frac{\sqrt{2}}{\frac{\sqrt{2}}{2}}\]
\[\frac{2x}{\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{2}}\]
\[\frac{2x}{\sqrt{3}} = 2\]
\[2x = 2\sqrt{3}\]
\[x = \sqrt{3}\]
Ответ: \[\sqrt{3}\]
3)
Дано: треугольник ABC, AB = 3√2, ∠B = 105°, ∠C = 30°.
Найти: AC = x.
По теореме синусов: \[\frac{AC}{\sin B} = \frac{AB}{\sin C}\]
\[\frac{x}{\sin 105^\circ} = \frac{3\sqrt{2}}{\sin 30^\circ}\]
\[\frac{x}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{3\sqrt{2}}{\frac{1}{2}}\]
\[\frac{4x}{\sqrt{6} + \sqrt{2}} = 6\sqrt{2}\]
\[4x = 6\sqrt{2}(\sqrt{6} + \sqrt{2})\]
\[4x = 6(\sqrt{12} + 2)\]
\[4x = 6(2\sqrt{3} + 2)\]
\[x = \frac{6(2\sqrt{3} + 2)}{4}\]
\[x = 3(\sqrt{3} + 1)\]
Ответ: \[3(\sqrt{3} + 1)\]
4)
Дано: треугольник ABC, BC = 5√2, ∠B = 105°, ∠A = 45°.
Найти: AC = x.
По теореме синусов: \[\frac{AC}{\sin B} = \frac{BC}{\sin A}\]
\[\frac{x}{\sin 105^\circ} = \frac{5\sqrt{2}}{\sin 45^\circ}\]
\[\frac{x}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{5\sqrt{2}}{\frac{\sqrt{2}}{2}}\]
\[\frac{4x}{\sqrt{6} + \sqrt{2}} = 10\]
\[4x = 10(\sqrt{6} + \sqrt{2})\]
\[x = \frac{5}{2}(\sqrt{6} + \sqrt{2})\]
Ответ: \[\frac{5}{2}(\sqrt{6} + \sqrt{2})\]
5)
Дано: треугольник ABC, AB = √6, ∠B = 60°, ∠A = 75°.
Найти: AC = x.
∠C = 180° - 60° - 75° = 45°
По теореме синусов: \[\frac{AC}{\sin B} = \frac{AB}{\sin C}\]
\[\frac{x}{\sin 60^\circ} = \frac{\sqrt{6}}{\sin 45^\circ}\]
\[\frac{x}{\frac{\sqrt{3}}{2}} = \frac{\sqrt{6}}{\frac{\sqrt{2}}{2}}\]
\[\frac{2x}{\sqrt{3}} = \frac{2\sqrt{6}}{\sqrt{2}}\]
\[\frac{2x}{\sqrt{3}} = 2\sqrt{3}\]
\[2x = 6\]
\[x = 3\]
Ответ: 3
6)
Дано: треугольник ABC, BC = 10√2, ∠B = 45°, ∠C = 75°.
Найти: AB = x.
∠A = 180° - 45° - 75° = 60°
По теореме синусов: \[\frac{AB}{\sin C} = \frac{BC}{\sin A}\]
\[\frac{x}{\sin 75^\circ} = \frac{10\sqrt{2}}{\sin 60^\circ}\]
\[\frac{x}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{10\sqrt{2}}{\frac{\sqrt{3}}{2}}\]
\[\frac{4x}{\sqrt{6} + \sqrt{2}} = \frac{20\sqrt{2}}{\sqrt{3}}\]
\[4x = \frac{20\sqrt{2}}{\sqrt{3}}(\sqrt{6} + \sqrt{2})\]
\[4x = \frac{20}{\sqrt{3}}(\sqrt{12} + 2)\]
\[4x = \frac{20}{\sqrt{3}}(2\sqrt{3} + 2)\]
\[x = \frac{5}{\sqrt{3}}(2\sqrt{3} + 2)\]
\[x = 10 + \frac{10}{\sqrt{3}}\]
\[x = 10 + \frac{10\sqrt{3}}{3}\]
Ответ: \[10 + \frac{10\sqrt{3}}{3}\]
7)
Дано: треугольник ABC, ∠A, sin A = 0.3, ∠C, sin C = 0.2, AC = 4.
Найти: AB = x.
По теореме синусов: \[\frac{AB}{\sin C} = \frac{AC}{\sin B}\]
\[\frac{x}{\sin C} = \frac{4}{\sin A}\]
\[\frac{x}{0.2} = \frac{4}{0.3}\]
\[0.3x = 0.8\]
\[x = \frac{8}{3}\]
Ответ: \[\frac{8}{3}\]
8)
Дано: треугольник ABC, BC = 9, ∠C, sin C = 1/4, ∠B, sin B = 5/6.
Найти: AC = x.
По теореме синусов: \[\frac{AC}{\sin B} = \frac{BC}{\sin C}\]
\[\frac{x}{\frac{5}{6}} = \frac{9}{\frac{1}{4}}\]
\[\frac{6x}{5} = 36\]
\[6x = 180\]
\[x = 30\]
Ответ: 30
