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# In triangle ABC,,point D is on side BC, BD=2*DC,cosDAC=,cosC=,AC+BC=?

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In triangle ABC,$${\mathtt{AB}} = {\sqrt{{\mathtt{2}}}}$$,point D is on side BC, BD=2*DC,cosDAC=$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}$$,cosC=$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}$$,AC+BC=?

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Jun 24, 2015

#1
+26357
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In triangle ABC,$${\mathtt{AB}} = {\sqrt{{\mathtt{2}}}}$$,point D is on side BC, BD=2*DC,cosDAC=$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}$$,cosC=$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}$$,AC+BC=?

$$\small{ \begin{array}{l} \angle DAC = \arccos(\frac{3\cdot\sqrt{10}}{10}) = 18.4349488229\ensurement{^{\circ}}\\\\ \angle C = \arccos(\frac{2\cdot\sqrt{5}}{5}) = 26.5650511771\ensurement{^{\circ}}\\\\ \angle DAC + \angle C = 45 \ensurement{^{\circ}} \\\\ \angle ADC = 180 \ensurement{^{\circ}} -45 \ensurement{^{\circ}} = 135\ensurement{^{\circ}}\\\\ \angle BDA = 180 \ensurement{^{\circ}} -135 \ensurement{^{\circ}} = 45\ensurement{^{\circ}}\\\\ \end{array} }$$

Now we define:

DC = x,   BD = 2x,

H = foot (of a perpendicular) from A on line BC between B and D. ($$\small{\angle BDA = 45\ensurement{^{\circ}}}$$ !!!)

h = AH

u = BH,  v = HD,  DB = u+v = 2x

$$\small{ \begin{array}{l} H = \text{foot (of a perpendicular) from }~ A \text{ to line \overline{BC} } \\ h = \overline{AH}\qquad u = \overline{BH}\qquad v = \overline{HD}\qquad x = \overline{DC}\qquad \overline{BC} = 3x \end{array} }\\\\ \small{ \begin{array}{rcl} \tan{(45\ensurement{^{\circ}} )}=1 &=& \frac{h}{v} \\ h &=& v\\\\ \tan{(C)} = 0.5 &=& \frac{h}{v+x} \\ 0.5 &=& \frac{v}{v+x} \\ v+x &=& 2v\\ v &=& x\\\\ \overline{BD} = u+v &=& 2x \\ u&=& 2x-v \\ u &=& 2x -x \\ u &=& x\\\\ u^2+h^2 &=& \sqrt{2}^2 \quad | \quad u= 2x-v\\ (2x-v)^2 + v^2 &=& 2 \quad | \quad v=x\\ (2x-x)^2 + x^2 &=& 2 \\ x^2+x^2 &=& 2\\ 2x^2 &=& 2 \\ x^2 &=& 1\\ x &=& 1\\\\ \overline{BC} = 3x = 3\cdot 1 = 3\\\\ \frac {\overline{AC}} { \sin{(135\ensurement{^{\circ}})} }&=&\frac{x}{\sin{(DAC)}}\\\\ \frac {\overline{AC}} {\sin{(135\ensurement{^{\circ}} )}} &=&\frac{1}{ \sin{(DAC)} }\\\\ \overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}} { \sin{(DAC)} } \\\\ \end{array} }\\\\$$

$$\small{ \begin{array}{rcl} \overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}} { \sin{ ( 18.4349488229 \ensurement{^{\circ}} )} } \\\\ \overline{AC} &=& 2.23606797750 \\\\ \overline{AC}+\overline{BC}& =& 2.23606797750 +3\\\\ \mathbf{ \overline{AC}+\overline{BC}} &\mathbf{ =}& \mathbf{5.23606797750} &\\ \hline \end{array} }$$

Jun 24, 2015

#1
+26357
+13

In triangle ABC,$${\mathtt{AB}} = {\sqrt{{\mathtt{2}}}}$$,point D is on side BC, BD=2*DC,cosDAC=$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}$$,cosC=$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}$$,AC+BC=?

$$\small{ \begin{array}{l} \angle DAC = \arccos(\frac{3\cdot\sqrt{10}}{10}) = 18.4349488229\ensurement{^{\circ}}\\\\ \angle C = \arccos(\frac{2\cdot\sqrt{5}}{5}) = 26.5650511771\ensurement{^{\circ}}\\\\ \angle DAC + \angle C = 45 \ensurement{^{\circ}} \\\\ \angle ADC = 180 \ensurement{^{\circ}} -45 \ensurement{^{\circ}} = 135\ensurement{^{\circ}}\\\\ \angle BDA = 180 \ensurement{^{\circ}} -135 \ensurement{^{\circ}} = 45\ensurement{^{\circ}}\\\\ \end{array} }$$

Now we define:

DC = x,   BD = 2x,

H = foot (of a perpendicular) from A on line BC between B and D. ($$\small{\angle BDA = 45\ensurement{^{\circ}}}$$ !!!)

h = AH

u = BH,  v = HD,  DB = u+v = 2x

$$\small{ \begin{array}{l} H = \text{foot (of a perpendicular) from }~ A \text{ to line \overline{BC} } \\ h = \overline{AH}\qquad u = \overline{BH}\qquad v = \overline{HD}\qquad x = \overline{DC}\qquad \overline{BC} = 3x \end{array} }\\\\ \small{ \begin{array}{rcl} \tan{(45\ensurement{^{\circ}} )}=1 &=& \frac{h}{v} \\ h &=& v\\\\ \tan{(C)} = 0.5 &=& \frac{h}{v+x} \\ 0.5 &=& \frac{v}{v+x} \\ v+x &=& 2v\\ v &=& x\\\\ \overline{BD} = u+v &=& 2x \\ u&=& 2x-v \\ u &=& 2x -x \\ u &=& x\\\\ u^2+h^2 &=& \sqrt{2}^2 \quad | \quad u= 2x-v\\ (2x-v)^2 + v^2 &=& 2 \quad | \quad v=x\\ (2x-x)^2 + x^2 &=& 2 \\ x^2+x^2 &=& 2\\ 2x^2 &=& 2 \\ x^2 &=& 1\\ x &=& 1\\\\ \overline{BC} = 3x = 3\cdot 1 = 3\\\\ \frac {\overline{AC}} { \sin{(135\ensurement{^{\circ}})} }&=&\frac{x}{\sin{(DAC)}}\\\\ \frac {\overline{AC}} {\sin{(135\ensurement{^{\circ}} )}} &=&\frac{1}{ \sin{(DAC)} }\\\\ \overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}} { \sin{(DAC)} } \\\\ \end{array} }\\\\$$

$$\small{ \begin{array}{rcl} \overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}} { \sin{ ( 18.4349488229 \ensurement{^{\circ}} )} } \\\\ \overline{AC} &=& 2.23606797750 \\\\ \overline{AC}+\overline{BC}& =& 2.23606797750 +3\\\\ \mathbf{ \overline{AC}+\overline{BC}} &\mathbf{ =}& \mathbf{5.23606797750} &\\ \hline \end{array} }$$

heureka Jun 24, 2015
#2
+583
+5

Thank you,heureka!

I noticed that I could find the measure of angle ADC in this way.

angle ADC +angle DAC +angle ACD =180 degrees

cosADC=$$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}$$

$$\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)} = {\mathtt{0}}$$

therefore,cosADC=$$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}$$=$${\mathtt{\,-\,}}\left({\mathtt{1}}{\mathtt{\,\times\,}}\left({\mathtt{cosDAC}}{\mathtt{\,\times\,}}{\mathtt{cosACD}}{\mathtt{\,-\,}}{\mathtt{sinDAC}}{\mathtt{\,\times\,}}{\mathtt{sinACD}}\right)\right) = {\mathtt{sinDAC}}{\mathtt{\,\times\,}}{\mathtt{sinACD}}{\mathtt{\,-\,}}{\mathtt{cosDAC}}{\mathtt{\,\times\,}}{\mathtt{cosACD}}$$

given that $$\underset{\,\,\,\,{{\rightarrow {\mathtt{cosdac, cosacd}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{cosDAC}}={\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\\ {\mathtt{cosACD}}={\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}\end{array}\right)}$$

so $$\underset{\,\,\,\,{{\rightarrow {\mathtt{sindac, sinacd}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{sinDAC}}={\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\\ {\mathtt{sinACD}}={\frac{{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}\end{array}\right)}$$

cosADC=sinDAC*sinACD-cosDAC*cosACD=$${\frac{{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}{\mathtt{\,-\,}}{\frac{{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{50}}}}}{{\mathtt{50}}}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{50}}}}}{{\mathtt{10}}}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}$$

angle ACD=arccos -sqrt(2)/2=135 (the measure of angle ACD is less than 180 degrees,ut more than 90 degrees)

AC=$${\mathtt{AC}} = {\frac{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{135}}^\circ\right)}}{\left[{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\right]}} = {\frac{\left[{\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right]}{\left[{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\right]}} = {\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}} = {\sqrt{{\mathtt{5}}}}$$

AC+BC=$${\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}$$

Jun 25, 2015