Question

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Answer 1

Odpowiedź:

|AD|=x

[tex]\displaystyle \frac{x}{20-x} =\frac{2}{3} \\3x=2(20-x)\\3x=40-2x\\5x=40/:5\\|AD|= x=8\\P_{\bigtriangleup ABD}=\frac{1}{2} |AD|\cdot|AB|\sin\angle(AD,AB)\\P_{\bigtriangleup ABD}=\frac{1}{2} \cdot8\cdot20\cdot \frac{\sqrt{3} }{2} =40\sqrt{3} \\P_{\bigtriangleup ABC}=\frac{a^{2} \sqrt{3} }{4} =\frac{20^{2} \sqrt{3} }{4} =100\sqrt{3} \\\frac{P_{\bigtriangleup ABC}}{P_{\bigtriangleup ABD}} =\frac{100\sqrt{3} }{40\sqrt{3} } =\frac{5}{2} =2,5\\\underline{C}\\23.1.2\\[/tex]

[tex]|BD|^2=|AD|^2+|AB|^2-2|AD||AB|cos60^{\circ}\\|BD|^2=8^2+20^2-2\cdot8\cdot10\cdot\frac{1}{2} \\|BD|^2=64+400-160=304\\|BD|=\sqrt{304} =4\sqrt{19} \\\underline{G}\\23.2\\[/tex]

[tex]|AD|^2=|AB|^2+|BD|^2-2|AB||BD|cos\angle ABD\\8^{2} =20^{2} +(4\sqrt{19} )^{2} -2\cdot20\cdot4\sqrt{19}cos\angle ABD\\160\sqrt{19} cos\angle ABD=400+304-64\\160\sqrt{19} cos\angle ABD=640/:160\sqrt{19} \\\displaystyle cos\angle ABD=\frac{640}{160\sqrt{19} } =\frac{4}{\sqrt{19} } =\frac{4\sqrt{19} }{19} \\\underline{A}[/tex]