+0  
 
+5
420
2
avatar+519 

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=?

(picture might not draw to scale)

fiora  Jun 24, 2015

Best Answer 

 #1
avatar+18715 
+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
Sort: 

2+0 Answers

 #1
avatar+18715 
+13
Best Answer

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
avatar+519 
+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

angle ADC=180-(angle DAC+angle ACD)

cosADC=cos[180-(angle DAC+angle ACD)]

cosADC=$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)} = {\mathtt{0}}$$

therefore,cosADC=$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}$$=$${\mathtt{\,-\,}}\left({\mathtt{1}}{\mathtt{\,\times\,}}\left({\mathtt{cosDAC}}{\mathtt{\,\times\,}}{\mathtt{cosACD}}{\mathtt{\,-\,}}{\mathtt{sinDAC}}{\mathtt{\,\times\,}}{\mathtt{sinACD}}\right)\right) = {\mathtt{sinDAC}}{\mathtt{\,\times\,}}{\mathtt{sinACD}}{\mathtt{\,-\,}}{\mathtt{cosDAC}}{\mathtt{\,\times\,}}{\mathtt{cosACD}}$$

given that $$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{cosdac, cosacd}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{cosDAC}}={\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\\
{\mathtt{cosACD}}={\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}\end{array}\right)}$$

so $$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{sindac, sinacd}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{sinDAC}}={\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\\
{\mathtt{sinACD}}={\frac{{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}\end{array}\right)}$$

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=$${\mathtt{AC}} = {\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{135}}^\circ\right)}}{\left[{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\right]}} = {\frac{\left[{\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right]}{\left[{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\right]}} = {\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}} = {\sqrt{{\mathtt{5}}}}$$ 

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

fiora  Jun 25, 2015

1 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details