Объяснение:
∠C=60∘; AC = \sqrt{2} + \sqrt{6}AC=2+6 ; BC = 2\sqrt{2}BC=22 .
1) Найдём \angle C∠C :
Сумма внутренних углов треугольника равна {180}^{\circ}180∘ .
\Rightarrow \angle C = {180}^{\circ} - (\angle A + \angle B) = {180}^{\circ} - ({45}^{\circ} + {75}^{\circ}) = {60}^{\circ}⇒∠C=180∘−(∠A+∠B)=180∘−(45∘+75∘)=60∘
2) Найдём BCBC :
По теореме синусов: \dfrac{AB}{\sin(C)} = \dfrac{BC}{\sin(A)}sin(C)AB=sin(A)BC
\Rightarrow BC = \dfrac{AB \cdot \sin(A)}{\sin(C)} = \dfrac{2\sqrt{3} \cdot \sin({45}^{\circ})}{\sin({60}^{\circ})} = \dfrac{2\sqrt{3}\cdot \dfrac{\sqrt{2} }{2} }{\dfrac{\sqrt{3} }{2} }=\sqrt{3} \cdot\sqrt{2}\cdot \dfrac{2}{\sqrt{3} } = 2\sqrt{2}⇒BC=sin(C)AB⋅sin(A)=sin(60∘)23⋅sin(45∘)=2323⋅22=3⋅2⋅32=22
3) Найдём ACAC :
Пусть xx - AC.AC.
По теореме косинусов:
AB = \sqrt{{BC}^{2} + {AC}^{2} - 2 \cdot BC \cdot AC \cdot \cos(C)}AB=BC2+AC2−2⋅BC⋅AC⋅cos(C)
(2\sqrt{3})^{2} = (2\sqrt{2})^{2} + x^{2} - 2 \cdot (2\sqrt{2}) \cdot x \cdot cos(60^{\circ})(23)2=(22)2+x2−2⋅(22)⋅x⋅cos(60∘)
\begin{gathered}12 = 8 + x^{2} - (2\sqrt{2})x\\12 - 8 - x^{2} +( 2\sqrt{2}) x = 0\\-x^{2} + (2\sqrt{2}) x +4 = 0\\x^{2} - (2\sqrt{2}) x - 4 = 0\end{gathered}12=8+x2−(22)x12−8−x2+(22)x=0−x2+(22)x+4=0x2−(22)x−4=0
\begin{gathered}\\x = \dfrac{-(-2\sqrt{2})\pm\sqrt{(-2\sqrt{2})^{2}-4\cdot1\cdot(-4) } }{2\cdot1}\end{gathered}x=2⋅1−(−22)±(−22)2−4⋅1⋅(−4)
\begin{gathered}x = \dfrac{2\sqrt{2}\pm\sqrt{8 + 16} }{2} \\x = \dfrac{2\sqrt{2} \pm2\sqrt{6} }{2} \\ x_{1} = \sqrt{2} + \sqrt{6} \\x_{2} = \sqrt{2 } - \sqrt{6}\end{gathered}x=222
Объяснение:
∠C=60∘; AC = \sqrt{2} + \sqrt{6}AC=2+6 ; BC = 2\sqrt{2}BC=22 .
Объяснение:
1) Найдём \angle C∠C :
Сумма внутренних углов треугольника равна {180}^{\circ}180∘ .
\Rightarrow \angle C = {180}^{\circ} - (\angle A + \angle B) = {180}^{\circ} - ({45}^{\circ} + {75}^{\circ}) = {60}^{\circ}⇒∠C=180∘−(∠A+∠B)=180∘−(45∘+75∘)=60∘
2) Найдём BCBC :
По теореме синусов: \dfrac{AB}{\sin(C)} = \dfrac{BC}{\sin(A)}sin(C)AB=sin(A)BC
\Rightarrow BC = \dfrac{AB \cdot \sin(A)}{\sin(C)} = \dfrac{2\sqrt{3} \cdot \sin({45}^{\circ})}{\sin({60}^{\circ})} = \dfrac{2\sqrt{3}\cdot \dfrac{\sqrt{2} }{2} }{\dfrac{\sqrt{3} }{2} }=\sqrt{3} \cdot\sqrt{2}\cdot \dfrac{2}{\sqrt{3} } = 2\sqrt{2}⇒BC=sin(C)AB⋅sin(A)=sin(60∘)23⋅sin(45∘)=2323⋅22=3⋅2⋅32=22
3) Найдём ACAC :
Пусть xx - AC.AC.
По теореме косинусов:
AB = \sqrt{{BC}^{2} + {AC}^{2} - 2 \cdot BC \cdot AC \cdot \cos(C)}AB=BC2+AC2−2⋅BC⋅AC⋅cos(C)
(2\sqrt{3})^{2} = (2\sqrt{2})^{2} + x^{2} - 2 \cdot (2\sqrt{2}) \cdot x \cdot cos(60^{\circ})(23)2=(22)2+x2−2⋅(22)⋅x⋅cos(60∘)
\begin{gathered}12 = 8 + x^{2} - (2\sqrt{2})x\\12 - 8 - x^{2} +( 2\sqrt{2}) x = 0\\-x^{2} + (2\sqrt{2}) x +4 = 0\\x^{2} - (2\sqrt{2}) x - 4 = 0\end{gathered}12=8+x2−(22)x12−8−x2+(22)x=0−x2+(22)x+4=0x2−(22)x−4=0
\begin{gathered}\\x = \dfrac{-(-2\sqrt{2})\pm\sqrt{(-2\sqrt{2})^{2}-4\cdot1\cdot(-4) } }{2\cdot1}\end{gathered}x=2⋅1−(−22)±(−22)2−4⋅1⋅(−4)
\begin{gathered}x = \dfrac{2\sqrt{2}\pm\sqrt{8 + 16} }{2} \\x = \dfrac{2\sqrt{2} \pm2\sqrt{6} }{2} \\ x_{1} = \sqrt{2} + \sqrt{6} \\x_{2} = \sqrt{2 } - \sqrt{6}\end{gathered}x=222