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; ?

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