جواب تمرین صفحه 66 درس 3 هندسه یازدهم (روابط طولی در مثلث)

الف

\(\begin{array}{l}{a^2} = {20^2} + {15^2} - 2 \times 20 \times 15 \times \cos {60^ \circ } = 400 + 225 - 300\\ \Rightarrow {a^2} = 325 \Rightarrow a = 5\sqrt {13} \end{array}\)

ب

\(\frac{{5\sqrt {13} }}{{\sin {{60}^ \circ }}} = \frac{{15}}{{\sin C}} \Rightarrow \sin C = \frac{{15 \times \frac{{\sqrt 3 }}{2}}}{{5\sqrt {13} }} \simeq 0/72 \Rightarrow \widehat C \simeq {46^ \circ }\)

پ

\(\sin {60^ \circ } = \frac{{BH}}{{AB}} \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{BH}}{{15}} \Rightarrow BH = \frac{{15\sqrt 3 }}{2}\)

5DC=CE= در نتیجه مثلث DCE متساوی الساقین است و چون یک زاویه 60 درجه دارد، پس متساوی الاضلاع است؛ یعنی 5DE= . در مثلث DCE زاویه α یک زاویه خارجی است؛ پس:

\(\alpha = {60^ \circ } + {60^ \circ } \Rightarrow \alpha = {120^ \circ }\)

\(\begin{array}{l}AB = 60 \times 1 = 60km\quad ,\quad BC = 40 \times 0/5 = 20km\\\\\\A{C^2} = {60^2} + {20^2} - 2 \times 60 \times 20 \times \cos {150^ \circ } = 3600 + 400 + 1200\sqrt 3 \\\\ \Rightarrow A{C^2} = 400\left( {10 + 3\sqrt 3 } \right)\\\\ \Rightarrow AC = 20\sqrt {10 + 3\sqrt 3 } \end{array}\)

\(\begin{array}{l}A\mathop B\limits^\Delta M:\quad {c^2} = {(\frac{a}{2})^2} + A{M^2} - 2 \times \frac{a}{2} \times AM \times \cos \alpha \\\\{c^2} = \frac{{{a^2}}}{4} + A{M^2} - a \times AM \times \cos \alpha \;\;\;\;\left( 1 \right)\\\\A\mathop C\limits^\Delta M:\quad {b^2} = {(\frac{a}{2})^2} + A{M^2} - 2 \times \frac{a}{2} \times AM \times \cos ({180^ \circ } - \alpha )\\\\{\kern 1pt} {b^2} = \frac{{{a^2}}}{4} + A{M^2} + a \times AM \times \cos \alpha \;\;\;\left( 2 \right)\\\\\mathop \Rightarrow \limits^{\left( 1 \right)\;,\;\left( 2 \right)} \;\;{b^2} + {c^2} = \frac{{{a^2}}}{2} + 2A{M^2}\end{array}\)

\(\begin{array}{l}AB = c = 4\quad ,\quad AC = b = 6\quad ,\quad BC = a = 8\\\\{b^2} + {c^2} = \frac{{{a^2}}}{2} + 2A{M^2}\\\\ \Rightarrow A{M^2} = \frac{{2({b^2} + {c^2}) - {a^2}}}{2} = \frac{{2(36 + 16) - 64}}{2} = 20\\\\ \Rightarrow AM = 2\sqrt 5 \end{array}\)

\(\begin{array}{l}A\mathop B\limits^\Delta D:\quad \\\\A{B^2} = A{D^2} + D{B^2} - 2AD \cdot DB \cdot \cos \alpha \end{array}\)

\({\mathop \Rightarrow \limits^{ \times DC} \;\;\;A{B^2} \cdot DC = A{D^2} \cdot DC + D{B^2} \cdot DC - 2DC \cdot AD \cdot DB \cdot \cos \alpha \quad \left( 1 \right)}\)

\(\begin{array}{l}A\mathop C\limits^\Delta D:\quad \\\\A{C^2} = A{D^2} + C{D^2} - 2AD \cdot DC \cdot \cos \left( {{{180}^\circ } - \alpha } \right)\end{array}\)

\({\mathop \Rightarrow \limits^{ \times DB} \;\;A{B^2} \cdot DB = A{D^2} \cdot DB + DB \cdot D{C^2} + 2DB \cdot AD \cdot DC \cdot \cos \alpha \quad \left( 2 \right)}\)

\(\mathop \Rightarrow \limits^{\left( 1 \right) + \left( 2 \right)} \;\;A{B^2} \cdot DC + A{B^2} \cdot DB = A{D^2} \cdot DC + D{B^2} \cdot DC - 2DC \cdot AD \cdot DB \cdot \cos \alpha \)

\({ + A{D^2} \cdot DB + DB \cdot D{C^2} + 2DB \cdot AD \cdot DC \cdot \cos \alpha }\)

\({ \Rightarrow A{B^2} \cdot DC + A{C^2} \cdot DB = A{D^2}\left( {DC + DB} \right) + BD.DC\left( {DC + DB} \right)}\)

\({ \Rightarrow A{B^2} \cdot DC + A{C^2} \cdot DB = A{D^2} \cdot BC + DB \cdot DC \cdot BC}\)

\({DC = DB = \frac{a}{2}\quad ,\quad AC = b\quad ,\quad AB = c}\)

\({\frac{a}{2} \times {c^2} + \frac{a}{2} \times {b^2} = A{D^2} \times a + \frac{a}{2} \times \frac{a}{2} \times a}\)

\({ \Rightarrow \frac{a}{2}({c^2} + {b^2}) = \frac{a}{2}(2A{D^2} + \frac{{{a^2}}}{2})}\)

\({ \Rightarrow {c^2} + {b^2} = 2A{D^2} + \frac{{{a^2}}}{2}}\)

\(\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{AB = AC = BC = 8\;,\;AD = 7\;,\;DB = x}\\\;\\{DC = 8 - x\;,\;DB < DC}\end{array}} \right.}\\{}\\{A{B^2}.DC + A{C^2}.DB = A{D^2}.BC + DB.DC.BC}\\{}\\{ \Rightarrow 64\left( {8 - x} \right) + 64x = 49 \times 8 + 8x\left( {8 - x} \right)}\\{}\\{ \Rightarrow 64 \times 8 - 64x + 64x = 49 \times 8 + 8x\left( {8 - x} \right)}\\{}\\{\mathop \Rightarrow \limits^{ \div 8} \;\;64 = 49 + 8x - {x^2} \Rightarrow {x^2} - 8x + 15 = 0}\\{}\\{ \Rightarrow \left( {x - 3} \right)\left( {x - 5} \right) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x = 3}\\{x = 5}\end{array}} \right.}\\{}\\{\mathop \Rightarrow \limits^{DB < DC} \;\;x = DB = 3\quad ,\quad DC = 5}\end{array}\)