heureka

A circle centre O and radius 2cm is inscribed in an equilateral triangle ABC and touches the side BC at N. What is AN?

$$\\
\small{\text{when~ $ h = \overline{AN} $~ and one side is a}}\\\\
\qquad
\tan{(60\ensurement{^{\circ}})}
= \dfrac{h}{\frac{a}{2}} \qquad
\tan{(30\ensurement{^{\circ}})}
= \dfrac{r}{\frac{a}{2}} \\ \\
\dfrac{a}{2} = \dfrac{h}{ \tan{(60\ensurement{^{\circ}})}
} = \dfrac{r}{ \tan{(30\ensurement{^{\circ}})} }\\\\\\
h= r\cdot \dfrac{ \tan{(60\ensurement{^{\circ}})} }{ \tan{(30\ensurement{^{\circ}})} }
\small{\text{$
\qquad \tan{(60\ensurement{^{\circ}})} = \sqrt{3}
\qquad \tan{(30\ensurement{^{\circ}})} = \dfrac{\sqrt{3} }{3}
$}} \\\\\\
h= r\cdot \dfrac{ \sqrt{3}}{ \frac{\sqrt{3} }{3} }\\\\\\
h= 3\cdot r\\\\
h= 3\cdot 2~\text{cm}\\\\
h=6~\text{cm}\\\\
\mathbf{\overline{AN} = 6~\text{cm}}$$

Der Stern Alpha Centauri ist lt. Wiki 4,367 Lichtjahre von der Sonne entfernt. Das Raumschiff Helios 1 hat lt. Wiki eine Geschwindigkeit von 252792 km/h. Wie lange würde die Reise mit Helios 1 von der Sonne zum Stern Alpha Centauri dauern?

$$\small{\text{
$
\begin{array}{lll}
$Die Lichtgeschwindigkeit ist$&
c = 299 792 458 ~\dfrac{m}{ s } \\\\
$Die Flugzeit in Jahren berechnet sich zu $&
\left(c\cdot 3,6~\dfrac{km}{Lh.}\right) \cdot \dfrac{4,367~ Lj.}{252792 ~\dfrac{km}{h}}= 299 792 458 ~\cdot 3,6 \cdot \dfrac{4,367}{252792 } = 18644,17 ~$ Jahre $\\
\end{array}
$}}\\\\$$

cos(300) is equal to sin(?)

$$\small{
\cos{(300\ensurement{^{\circ}})} \text{ is equal to } \sin{(90\ensurement{^{\circ}}-300\ensurement{^{\circ}} )}
= \sin{(-210\ensurement{^{\circ}} )}
= \sin{(-210\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} )}
=\sin{(150 \ensurement{^{\circ}} } )}
}$$

(gcd = greatest common divisor)

1. There is only a solution when gcd( 1027, 712 ) = 1 or in other words 1027 and 712 are relatively prime.

2. Using the Euclidian algorithm to calcutate gcd(1027,712)

$$\small{
\begin{array}{|r|r|r|r|}
\hline
&&&\\
a & b & q& r \\
\hline
&&&\\
1027 & 712 & 1 & 315 \\
712 & 315 & 2 & 82 \\
315 & 82 & 3 & 69 \\
82 & 69 & 1 & 13 \\
69 & 13 & 5 &4 \\
13 & 4 &3 & \textcolor[rgb]{1,0,0}{1} \\
4 & 1 & 4 & 0 \\
&&&\\
\hline
\end{array}
}$$

The greatest common divisor gcd(1027,712) $$\textcolor[rgb]{1,0,0}{=1}$$

and we can go on

3. Using Extended Euclidean algorithm to calculate the first solution

$$\small{
\begin{array}{|r|r|r|r||r|r|}
\hline
&&&&&\\
a & b & q& r &x&y\\
\hline
&&&&&\\
1027 & 712 & 1 & 315 & \textcolor[rgb]{1,0,0}{-165} & 73 - 1 (-165) = \textcolor[rgb]{1,0,0}{238}\\
712 & 315 & 2 & 82 & 73 & -19 - 2\cdot 73 = -165\\
315 & 82 & 3 & 69 & -19 & 16 -3(-19) = 73\\
82 & 69 & 1 & 13 & 16 & -3 -1\cdot 16 = -19\\
69 & 13 & 5 &4 & - 3 & 1 -5(-3) = 16\\
13 & 4 &3 & 1 & 1 & 0 - 3\cdot 1 = -3 \\
4 & 1 & 4 & 0 & 0 & 0\cdot 4 + 1 = 1\\
&&&&&\\
\hline
\end{array}
}$$

$$\small{\text{
The first solution is $ (-165,238) \qquad 1027 \cdot (\textcolor[rgb]{1,0,0}{ -165} ) + 712 \cdot \textcolor[rgb]{1,0,0}{238} = 1
$}}\\\\$$

4. All Solutions:

$$\small{\text{$
\begin{array}{lcl}
\boxed{
a\cdot x + b \cdot y = 1 }\\\\
\mathrm{First~ solution~~} (x_0,~ y_0)\\
\mathrm{All~ solutions~~} \left(x_0+\dfrac{z\cdot b}{ gcd(a,b) } ,~ y_0 - \dfrac{z\cdot a}{ gcd(a,b) }\right)\\
\end{array}
$}}\\\\$$

$$\small{\text{$
\begin{array}{lcl}
\boxed{
1027\cdot x + 712 \cdot y = 1 }\\\\
\mathrm{First~ solution~~} (x_0=-165,~ y_0=238)\\
\mathrm{All~ solutions~~} \left(-165+\dfrac{z\cdot 712}{ 1 } ,~ 238 - \dfrac{z\cdot 1027}{ 1 }\right)\\
\end{array}
$}}\\\\$$

we have:

$$\small{\text{$
\begin{array}{lcl}
\left\{~\left(~ -165 + 712\cdot z,~ 238 - 1027\cdot z\right)~|~z \in Z ~
\right\} \\
\end{array}
$}}\\\\$$

what is b in

  3            2

'''''''''''''=''''''''''

3b+4      b - 4

$$\small{
\begin{array}{rll}
\dfrac{3}{3b+4} &=& \dfrac{2}{b-4} \\\\
\dfrac{3b+4}{3} &=& \dfrac{b-4}{2} \\\\
b + \dfrac43 &=& \dfrac{b}{2} - 2 \\\\
\dfrac{b}{2} &=& - 2 - \dfrac43 \\\\
\dfrac{b}{2} &=& -\dfrac{10}3 \\\\
b &=& -\dfrac{20}{3}\\\\
b &=& -6\dfrac{2}{3}\\\\
\end{array}
}$$

show me the process of solving (1-cot200)(1-cot25)

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]\\\\
&=& \left[ 1 -
\dfrac { 1 }
{ \tan{ ( 200\ensurement{^{\circ}} ) } }
\right]
\cdot
\left[ 1 -
\dfrac { 1 }
{ \tan{ ( 25\ensurement{^{\circ}} ) } }
\right]\\\\
&=& \left[
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) } - 1 }
{ \tan{ ( 200\ensurement{^{\circ}} ) } }
\right]
\cdot
\left[
\dfrac { \tan{ ( 25\ensurement{^{\circ}} ) } - 1 }
{ \tan{ ( 25\ensurement{^{\circ}} ) } }
\right]\\\\
&=&
\dfrac { \left[\tan{ ( 200\ensurement{^{\circ}} ) } - 1\right]\cdot \left[ \tan{ ( 25\ensurement{^{\circ}} ) } - 1 \right] }
{ \tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \tan{ ( 200\ensurement{^{\circ}} ) } - \tan{ ( 25\ensurement{^{\circ}} ) }
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ \tan{ ( 200\ensurement{^{\circ}} ) } + \tan{ ( 25\ensurement{^{\circ}} ) } \right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\end{array}
$}}$$

$$\\\small{ \text{Formula: $
\begin{array}{rcl}
&&\\
&&\\
&&\\
&&\\
\boxed{
\tan{ (200\ensurement{^{\circ}}+25\ensurement{^{\circ}} ) }
=
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) }
+ \tan{ ( 25\ensurement{^{\circ}} ) } }
{ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }
\qquad \tan{ (200\ensurement{^{\circ}}+25\ensurement{^{\circ}} ) }
= \tan{ (225\ensurement{^{\circ}} ) }
= \tan{ (180\ensurement{^{\circ}}+45\ensurement{^{\circ}} ) }
= \tan{ (45\ensurement{^{\circ}} ) }
= 1
}
\end{array}
$}} \\\\
\small{ \text{$
\begin{array}{rcl}
1 &=&
\dfrac { \tan{ ( 200\ensurement{^{\circ}} ) }
+ \tan{ ( 25\ensurement{^{\circ}} ) } }
{ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) } }\\\\
1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
&=&
\tan{ ( 200\ensurement{^{\circ}} ) + \tan{ ( 25\ensurement{^{\circ}} ) } } \\\\
\end{array}
$}}$$

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ \tan{ ( 200\ensurement{^{\circ}} ) } + \tan{ ( 25\ensurement{^{\circ}} ) } \right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
- \left[ 1 - \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
\right]
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
-1 + \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
+1
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\end{array}
$}}$$

$$\small{ \text{$
\begin{array}{rcl}
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
&=&
\dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
+ \tan{ ( 200\ensurement{^{\circ}} ) }
\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
&=&
2\cdot \dfrac
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}
{
\tan{ ( 200\ensurement{^{\circ}} ) }\cdot \tan{ ( 25\ensurement{^{\circ}} ) }
}\\\\
\mathbf{
\left[ 1 - \cot{(200\ensurement{^{\circ}})} ] \cdot [ 1 - \cot{(25\ensurement{^{\circ}})} \right]
}&\mathbf{
=}& \mathbf{2 }
\end{array}
$}}$$

Factor 

$$\\4(6x+8)+(3x^2+4x)\\\\
=4(6x+8)+x(3x+4)\\\\
=4\cdot 2(3x+4) + x(3x+4)\\\\
=8(3x+4) + x(3x+4)\\\\
=(3x+4)(8 + x)$$

Simplify  Express your answer in simplest radical form in terms of .

$$\boxed{~~ \text{Formula: } \sqrt[n]{a\cdot b}
= \sqrt[n]{a}\cdot \sqrt[n]{b} ~~}$$

$$\root 3 \of {x \root 3 \of {x \root 3 \of {x \sqrt{x}}}} \\\\
= \root 3 \of {x} \cdot \root 3 \of {\root 3 \of {x \root 3 \of {x \sqrt{x}}}}\\\\
=
\root 3 \of {x} \cdot
\root 9 \of {x \root 3 \of {x \sqrt{x}}}\\\\
=
\root 3 \of {x} \cdot
\root 9 \of {x} \cdot
\root 9 \of {\root 3 \of {x \sqrt{x}}}\\\\
=
\root 3 \of {x} \cdot
\root 9 \of {x} \cdot
\root 27 \of {x \sqrt{x}}\\\\
=
\root 3 \of {x} \cdot
\root 9 \of {x} \cdot
\root 27 \of {x} \cdot
\root 27 \of {\sqrt{x}}\\\\
=
\root 3 \of {x} \cdot
\root 9 \of {x} \cdot
\root 27 \of {x} \cdot
\root 54 \of {x}\\\\
=
x^{\frac13}
\cdot x^{\frac19}
\cdot x^{\frac1{27}}
\cdot x^{\frac1{54}}\\\\
=
x^{\frac13+\frac19+\frac1{27}+ \frac1{54}}\\\\
=
x^{\frac13 \cdot \frac{18}{18}
+\frac19 \cdot \frac{6}{6}
+\frac1{27} \cdot \frac{2}{2} + \frac1{54}}\\\\
=x^{\frac{18+6+2+1}{54} } \\\\$$

$$\\=x^{\frac{27}{54}}\\\\
=x^{\frac{1}{2} } \\\\
\mathbf{= \sqrt{x}}$$