heureka

what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24

$$\\(1) \qquad Area=\dfrac{a\cdot b}{2}\cdot\sin{(C)}\\\\
(2) \qquad b=a \cdot \dfrac{\sin{(C)}} {\sin{(A)}} }\\\\
(3) \qquad \sin{(A)}= \sin{( 180\ensurement{^{\circ}}-(B+C) )} = \sin{( B+C ) }$$

$$Area=\dfrac{a^2}{2} \cdot \dfrac{ \sin{(B)} \cdot\sin{(C)} } {\sin{(B+C)}} = \dfrac{a^2}{2} \cdot
\left( \dfrac{ 1 } { \cot{(B)} + \cot{(C)} } } \right)\\\\\\
Area = \dfrac{3^2}{2} \cdot
\left( \dfrac{ 1 } { \cot{(24\ensurement{^{\circ}})} + \cot{(24\ensurement{^{\circ}})} } } \right)\\\\
Area = \dfrac{3^2}{2} \cdot
\left( \dfrac{ 1 } { 2\cdot \cot{(24\ensurement{^{\circ}})} } } \right)\\\\
\boxed{
Area = \dfrac{3^2}{4} \cdot \tan{(24\ensurement{^{\circ}})} } } \\\\
Area = \dfrac{3^2}{4} \cdot 0.4452286853\\\\
Area = \dfrac{9}{4} \cdot 0.4452286853\\\\
Area = 1.0017645419$$

How many triangles do we have at:

1. Step 6 ?
2. Step 8 ?
3. Step 19 ?
4. Step 50 ?

I. The "empty" triangles in the middle are included:

$$\\
\begin{array}{rcrcrcr}
s_1 &=& 1 &=& 1 &=& 1\\
s_2 &=& 3s_1+1 &=& 4 &=& 1+3\\
s_3 &=& 3s_2+1 &=& 13 &=& 1+3+3^2\\
s_4 &=& 3s_3+1 &=& 40 &=& 1+3+3^2+3^3\\
s_5 &=& 3s_4+1 &=& 121 &=& 1+3+3^2+3^3+3^4\\
s_6 &=& 3s_5+1 &=& 364 &=& 1+3+3^2+3^3+3^4+3^5\\
\cdots\\
s_n &=& 3s_{n-1}+1&=& && \dfrac{3^n-1} {2}
\end{array}$$

$$\\
\rm{a.)}\qquad s_6 = \dfrac{3^6-1} {2} = 364 \\\\
\rm{b.)}\qquad s_8 = \dfrac{3^8-1} {2} = 9841\\\\
\rm{c.)}\qquad s_{19} = \dfrac{3^{19}-1} {2} = 581130733\\\\
\rm{d.)}\qquad s_{50} = \dfrac{3^{50}-1} {2} = 358~ 948~ 993~ 845~ 926~ 294 ~385~124$$

II. The "empty" triangles in the middle aren't included:

$$\\
\begin{array}{rcrcrcr}
s_1 &=& 1 &=& 1 &=& 3^0\\
s_2 &=& 3s_1 &=& 3 &=& 3^1\\
s_3 &=& 3s_2 &=& 9 &=& 3^2\\
s_4 &=& 3s_3 &=& 27 &=& 3^3\\
s_5 &=& 3s_4 &=& 81 &=& 3^4\\
s_6 &=& 3s_5 &=& 243 &=& 3^5\\
\cdots\\
s_n &=& 3s_{n-1}&=& && 3^{n-1}
\end{array}$$

$$\\\rm{a.)}\qquad s_6 = 3^5 = 243
\\\\\rm{b.)}\qquad s_8 = 3^7 = 2187
\\\\\rm{c.)}\qquad s_{19} = 3^{18} = 387420489
\\\\\rm{d.)}\qquad s_{50} = 3^{49} = 239 ~299 ~329 ~230 ~617 ~529 ~590 ~083$$

Trapez ABCD

Strecke AD e mal Wurzel 6

Strecke CD e mal wurzel 2

Winkel bei C 120° Winkel bei D 135°

Umfang und Flächeninhalt nur mit e und Wurzeln berechnen

$$\\
\small{\text{$
\begin{array}{rcl}
a=\overline{AD}=e\sqrt{6}\quad
b=\overline{CD}=e\sqrt{2}\quad
c=\overline{CB}= \?\quad
d=\overline{AB}=p+b+q\quad
\alpha=ADC = 135 \ensurement{^{\circ}}\quad
\beta=DCB = 120 \ensurement{^{\circ}}
\end{array}
$}}\\
\small{\text{$
\begin{array}{rcl}
\cos{(\alpha-90\ensurement{^{\circ}})}&=&\dfrac{h}{a}\\
h &=& a\cdot \cos{(\alpha-90\ensurement{^{\circ}})}\\
h &=& e\sqrt{6} \cdot \cos{(135-90\ensurement{^{\circ}})}\\
h &=& e\sqrt{6} \cdot \cos{(45\ensurement{^{\circ}})} \quad | \quad \cos{(45\ensurement{^{\circ}})} = \dfrac{ \sqrt{2} } {2}\\
h &=& e\sqrt{6} \cdot \dfrac{ \sqrt{2} } {2}\\
h &=& e \cdot \dfrac{ \sqrt{4\cdot 3} } {2}\\
h &=& e \cdot 2 \cdot \dfrac{ \sqrt{3} } {2}\\
h &=& e \sqrt{3} \\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\sin{(\alpha-90\ensurement{^{\circ}})}&=&\dfrac{p}{a}\\
p &=& a\cdot \sin{(\alpha-90\ensurement{^{\circ}})}\\
p &=& e\sqrt{6} \cdot \sin{(135-90\ensurement{^{\circ}})}\\
p &=& e\sqrt{6} \cdot \sin{(45\ensurement{^{\circ}})} \quad | \quad \sin{(45\ensurement{^{\circ}})} = \dfrac{ \sqrt{2} } {2}\\
p &=& e\sqrt{6} \cdot \dfrac{ \sqrt{2} } {2}\\
p &=& e \cdot \dfrac{ \sqrt{4\cdot 3} } {2}\\
p &=& e \cdot 2 \cdot \dfrac{ \sqrt{3} } {2}\\
p &=& e \sqrt{3} \\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\tan{( 180\ensurement{^{\circ}} - \beta)}&=&\dfrac{h}{q}\\
q &=& \dfrac{ h } { \tan{( 180\ensurement{^{\circ}} - \beta)} } \\\\
q &=& \dfrac{ e\sqrt{3} } { \tan{( 180\ensurement{^{\circ}} - 120\ensurement{^{\circ}} )} } \\\\
q &=& \dfrac{ e\sqrt{3} } { \tan{( 60\ensurement{^{\circ}} )} }
\quad | \quad \tan{( 60\ensurement{^{\circ}}) } = \sqrt{3} \\\\
q &=&\dfrac{ e\sqrt{3} } { \sqrt{3} }\\\\
q &=& e
\end{array}
$}}$$

$$\begin{array}{rcl}
\sin{( 180\ensurement{^{\circ}} - \beta)}&=&\dfrac{h}{c}\\
c &=& \dfrac{ h } { \sin{( 180\ensurement{^{\circ}} - \beta)} } \\\\
c &=& \dfrac{ e\sqrt{3} } { \sin{( 180\ensurement{^{\circ}} - 120\ensurement{^{\circ}} )} } \\\\
c &=& \dfrac{ e\sqrt{3} } { \sin{( 60\ensurement{^{\circ}} )} }
\quad | \quad \sin{( 60\ensurement{^{\circ}}) } = \dfrac{1}{2}\sqrt{3} \\\\
c &=&\dfrac{ e\sqrt{3} } {\dfrac{1}{2}\sqrt{3} }\\\\
c &=& 2e
\end{array}
$}}$$

$$\\d = p + b + q \qquad d = e\sqrt{3}+ e\sqrt{2} + e\\\\
\begin{array}{rcl}
U&=&a+b+c+d \\
U&=& e\sqrt{6} +e\sqrt{2} + 2e+e\sqrt{3}+ e\sqrt{2} + e\\
U&=& 3e +2e\sqrt{2}+ e\sqrt{3} + e\sqrt{6}\\
U&=& e \left( 3 +2\sqrt{2}+ \sqrt{3} + \sqrt{6} \right)
\end{array}$$

$$\boxed{
U&=& e \left( 3 +2\sqrt{2}+ \sqrt{3} + \sqrt{6} \right)
}$$

$$\begin{array}{rcl}
A&=& \left( \dfrac{d+b}{2}\ \right) \cdot h\\\\
A&=& \left( \dfrac{ e\sqrt{3}+ e\sqrt{2} + e+ e\sqrt{2} }{2}\ \right) \cdot e\sqrt{3}\\\\
A&=& \left( \dfrac{ e + 2e\sqrt{2} + e\sqrt{3} }{2}\ \right) \cdot e\sqrt{3}\\\\
A&=& \left( e + 2e\sqrt{2} + e\sqrt{3} \right)
\cdot \dfrac{ e\sqrt{3} } {2}\\\\
A&=& \left( 1 + 2\sqrt{2} + \sqrt{3} \right)
\cdot \dfrac{ e^2\sqrt{3} } {2}
\end{array}$$

$$\boxed{
A= \left( 1 + 2\sqrt{2} + \sqrt{3} \right)
\cdot \dfrac{ e^2\sqrt{3} } {2}
}$$

i+i²+i³+i⁴ = ?

$$\begin{array}{rcl}
i+i^2+i^3+i^4&\\
&=& {( \underbrace{ 1+\underbrace{i^2}_{=-1} }_{=0} ) \cdot(i+i^2)\\
i+i^2+i^3+i^4&=& 0
\end{array}$$

1 - cos(2x) + cos(6x) - cos(8x) = 0       x =?

$$\small{\text{$
\begin{array}{rcl}
1-\cos{(2x)}+\cos{(6x)}-\cos{(8x)}&=&0\\
1-\cos{(2x)} &=&\cos{(8x)}-\cos{(6x)}\\
&&$Formula:\\
&&
$~\boxed{\mathbf{ \cos{2x}=1-2\sin^2{(x)} \quad or \quad 1-\cos{2x}=2\sin^2{(x)} }}\\
2\sin^2{(x)} &=& \cos{(8x)}-\cos{(6x)}\\
&&$Formula:\\
&&
$~\boxed{\mathbf{
\cos{(a)}-\cos{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\sin{ \left( \dfrac{b-a}{2} \right)}
}}\\
&&
\cos{(8x)}-\cos{(6x)}=2\sin{ \left( \dfrac{14x}{2} \right)}\sin{ \left( \dfrac{-2x}{2} \right)}\\
&&
\cos{(8x)}-\cos{(6x)}= -2\sin{ (7x) }\sin{(x)}\\
2\sin^2{(x)} &=& -2\sin{ (7x) }\sin{(x)}\\
\sin^2{(x)} &=& -\sin{ (7x) }\sin{(x)}\\
\sin^2{(x)} +\sin{ (7x) }\sin{(x)} &=& 0\\
\sin{(x)} \left[ \sin{(x)} + \sin{ (7x) } \right] &=& 0\\
&&$Formula:\\
&&
$~\boxed{\mathbf{
\sin{(a)}+\sin{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\cos{ \left( \dfrac{b-a}{2} \right)}
}}\\
&&
\sin{(x)}+\sin{(7x)}=2\sin{ \left( \dfrac{8x}{2} \right)}\cos{ \left( \dfrac{6x}{2} \right)}\\
&&
\sin{(x)}+\sin{(7x)}=2\sin{ (4x) }\cos{ (3x) }\\
\sin{(x)}\cdot 2 \cdot\sin{ (4x) }\cos{ (3x) } &=& 0\\
\end{array}
$}}$$

The solution set of the given equation is:

$$\\\boxed{\mathbf{
\sin{(x)}=0 \qquad \textcolor[rgb]{150,0,0}{x = k\cdot\pi \qquad k \in \mathrm{Z} }
}}\\
\boxed{\mathbf{
\sin{(4x)}=0 \qquad 4x = k\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = k\cdot \dfrac{\pi}{4} \qquad k \in \mathrm{Z} }
}}\\
\boxed{\mathbf{
\cos{(3x)}=0 \qquad 3x = \pm \dfrac{\pi}{2}+ k\cdot 2\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = \pm\dfrac{\pi}{6}
+ k\cdot \dfrac{2}{3}~\pi \qquad k \in \mathrm{Z} }
}}$$

Find the exact values of tan (cos^-1(2/3)) and cos (sin^-1(-3/5)).

tan (cos^-1(2/3))

$$\small{\text{$ \boxed{\tan{ \left(~ \arccos{(x)} ~\right) }=\pm\dfrac {\sqrt{1-x^2}} {x}
=\pm\dfrac {\sqrt{1-\left(\dfrac{2}{3}\right)^2}} { \dfrac{2}{3} } =\pm \dfrac{3}{2} \cdot \sqrt{ 1-\dfrac{4}{9} }
=\pm \dfrac{3}{2} \cdot \dfrac{ \sqrt{5}}{3}
=\pm \dfrac{ \sqrt{5} }{2}
}
$}}$$

cos (sin^-1(-3/5))

$$\small{\text{$ \boxed{\cos{ \left(~ \arcsin{(x)} ~\right) }=\pm\sqrt{1-x^2}=\pm \sqrt{1+ \left( \dfrac{-3}{5} \right)^2}=\pm \dfrac{\sqrt{34} }{ 5 } }$}}$$

Which is the equivalent of 6 14' 48" written in decimal form ?

$$\boxed{~6\ensurement{^{\circ}} ~14' ~48" ~}\\\\
\small{\text{$
\dfrac{ \dfrac{48"}{60}+14' }{60} + 6\ensurement{^{\circ}}
=\dfrac{0.8+14'}{60} + 6\ensurement{^{\circ}}
=\dfrac{14.8}{60} + 6\ensurement{^{\circ}}
=0.24666666667+ 6\ensurement{^{\circ}}
=6.24666666667\ensurement{^{\circ}}
=6.24\overline{6}\ensurement{^{\circ}}
$}}$$

cos( atan( sqrt(3) )+asin(1/3) ) ?

$$\boxed{
\mathbf{
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)}
~\right)
}
} =~ ?
}$$

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\cos{ \left(\alpha_{rad} + \beta_{rad} \right) }
= \cos{ \left(\alpha_{rad} \right) } \cdot
\cos{ \left(\beta_{rad} \right) }
- \sin{ \left(\alpha_{rad} \right) } \cdot
\sin{ \left(\beta_{rad} \right) }
} & \quad \mathbf{
\alpha_{rad} = \arctan{(a)} \quad a= \sqrt{3} } \\
& \mathbf{
\quad \beta_{rad} = \arcsin{(b)} \quad b= \dfrac{1}{3} }
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot
\sin{ \left(\arcsin{(b)} \right) } \\
&=&
\cos{ \left( \arctan{(a)} \right) } \cdot
\cos{ \left(\arcsin{(b)} \right) }
- \sin{ \left( \arctan{(a)} \right) } \cdot b
\end{array}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{1}{\sqrt{1+a^2}}
=\pm\dfrac{1}{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{1}{2}
}
$}}$$

$$\small{\text{
$
\boxed{
\cos{ \left(~ \arcsin{(b)} ~\right) }
=\pm\sqrt{1-b^2}
=\pm \sqrt{1+ \left( \dfrac{1}{3} \right)^2}
=\pm \dfrac{\sqrt{8} }{ 3 }
}
$}}$$

$$\small{\text{
$
\boxed{
\sin{ \left(~ \arctan{(a)} ~\right) }
=\pm\dfrac{a}{\sqrt{1+a^2}}
=\pm\dfrac{ \sqrt{3} }{\sqrt{1+(\sqrt{3})^2}}
=\pm \dfrac{ \sqrt{3} }{2}
}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left( \alpha_{rad}+ \beta_{rad} \right) }
&=&
\left( \pm \dfrac{1}{2} \right) \cdot
\left( \pm \dfrac{ \sqrt{8} }{ 3 } \right)
-\left( \pm \dfrac{ \sqrt{3} }{ 2 } \right) \cdot
\left( \dfrac{1}{3} \right) \\
&=& \dfrac{1}{6} \left( \pm \sqrt{8} \pm\sqrt{3} \right)
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{ \left(~ \arctan{(~\sqrt{3}~)}
+ \arcsin{\left(\dfrac{1}{3}\right)} ~\right)
}
&=& \dfrac{1}{6} \left( + \sqrt{8} + \sqrt{3} \right)
= 0.76007965539 \\\\
&=& \dfrac{1}{6} \left( + \sqrt{8} - \sqrt{3} \right)
= 0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} + \sqrt{3} \right)
= -0.18272938620 \\\\
&=& \dfrac{1}{6} \left( - \sqrt{8} - \sqrt{3} \right)
= -0.76007965539
\end{array}
$}}$$

Der Scheinwerfer eines Leuchtturms benötigt für eine volle Umdrehung 30 Sekunden. Wie groß ist der Winkel, der von den Lichtbündel des Scheinwerfers überstrichen wird in 1;3s;17s;26s; ?

$$\small{\text{$
\begin{array}{lcr}
\dfrac {360~ \ensurement{^{\circ}}} {30~s} \cdot \mathbf{
1~s} &=& \mathbf{
12 ~ \ensurement{^{\circ}} } \\\\
\dfrac {360~ \ensurement{^{\circ}}} {30~s} \cdot \mathbf{
3~s} &=& \mathbf{
36 ~ \ensurement{^{\circ}} } \\\\
\dfrac {360~ \ensurement{^{\circ}}} {30~s} \cdot \mathbf{
17~s} &=& \mathbf{
204 ~ \ensurement{^{\circ}} } \\\\
\dfrac {360~ \ensurement{^{\circ}}} {30~s} \cdot \mathbf{
26~s} &=& \mathbf{
312 ~ \ensurement{^{\circ}} } \\\\
\end{array}
$}}$$