asinus

What is UX?

Im Dreieck STU sei M der Mittelpunkt von \overline{ST} und N sei auf \overline{TU},

so dass \overline{SN} eine Höhe des Dreiecks STU ist.

Wenn ST = 13, SU = 14, TU = 4 und \overline{SN} und \overline{UM} sich bei X schneiden,

was ist dann UX?

\(S(0;0)\\ T(13,0)\\ f_S(x)=\pm \sqrt{14^2-x^2}\\ f_T(x)=\pm \sqrt{4^2-(x-13)^2}\\ x^2-14^2=(x-13)^2-4^2\\ x^2-14^2=x^2-26x+13^2-4^2\\ 26x=169-16+196\\ x_U=\frac{349}{26}\\ x_U=13.4\overline{230769}\\ y_U=3.9776\)

\(U(13.423,\ 3.978)\)

to be continued

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Find the coordinates of the center of the circle.

A(22,15)

B(-25,0)

C(22,-29)

\(M_{AB}(-1.5,7.5)\\ M_{AC}(22,-7)\\ f_{AB}=-\frac{1}{\frac{15}{47}}(x-(-1.5))+7.5\\ f_{AB}=-\frac{47}{15}(x+1.5)+7.5\\ f_{AB}=-\frac{47}{15}x-4.7+7.4\\ f_{AB}=-\frac{47}{15}x+2.7\)

\(f_{AC}=-7\\ f_{AB}=f_{AC}\\ -\frac{47}{15}x+2.7=-7\\ x=\frac{15(7+2.7)}{47}\\ x=3.0957\\ y=-7\\\)

The coordinates of the center of the circle are (3.0957, -7).

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\((1+r)^2=(\overline{AB}-r)^2+1^2\\(1+r)^2=(2-r)^2+1\\ 1+2r+r^2=4-4r+r^2+1\\ 6r=4\\\color{blue}r=\frac{2}{3}\)

\(\color{blue}The\ radius\ of\ the\ circle\ is\ \frac{2}{3}.\)

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In triangle ABC,  the angle bisector of  BAC meets  BC at D.

If   BAC=60°, CAD=45°, and  AD=24, then find the area of triangle ABC.

\(\color{blue}If\ D\ lies\ on\ the\ bisector\ of\ \angle BAC=60°, \angle CAD \ cannot\ be\ equal\ to\ 45°.\)

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Find the area of the strip.

Let a be the distance of the base 10 touched by the strip.

Let alpha be the slope angle of the strip.

\(tan(\alpha)=\frac{8}{10-a}\\ sin(\alpha )=\frac{1}{a}\\ \alpha =atan(\frac{8}{10-a})=asin(\frac{1}{a})\\ atan(\frac{8}{10-a})-asin(\frac{1}{a})=0\)

\(\color{green}WolframAlpha\\ \color{green}a=\frac{8\sqrt{163}}{63}-\frac{10}{63}\\\)

\(\color{blue}a=1.4625\\ A=8\cdot 1.4625\\ \color{blue}A=11.7\)

\(\color{blue}The\ area\ of\ the\ strip\ is\ 11.7\ .\)

\(A\in \mathbb R\ |0< A< \infty\)

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What is the radius of the circle?

\(\dfrac{83^\circ \cdot \pi r^2}{360^\circ}=60 \pi\\ r=\sqrt{\frac{60\pi \cdot 360}{83\pi}}=16.132\\ \color{blue}The\ radius\ of\ the\ circle\ is\ 16.132.\)

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The length of AP is 4, and the circle has a radius of 5.  Find the length AB.

\(f_{circle}(x)=\sqrt{5^2-x^2}\\ \color{blue}A\ (0,5)\\ \color{blue}P\ (4,5)\\ f_{PB}(x)=-\sqrt{4^2-(x-4)^2}+5=-\sqrt{16-x^2+8x-16}+5\\ f_{PB}(x)=-\sqrt{8x-x^2}+5 \)

\(\sqrt{25-x^2}=-\sqrt{8x-x^2}+5\\ \sqrt{5-x}\sqrt{x+5}=5-\sqrt{8-x}\sqrt{x}\\ (WolframAlpha)\\ x_{B}\in \{0,\frac{200}{41}\}\\ y_b=\sqrt{25-4.\overline{87804}\ ^2}=1.0976\\ B\ (\frac{200}{41},\sqrt{20.\overline{12195}}) \\ \color{blue}B\ (4.\overline{87804},1.0976)\)

 \(\overline{AB}= \sqrt{(y_A-y_B)^2+(x_B-x_A)^2}=\sqrt{(5-1.0976)^2+(4.\overline{87804}-0)^2}\\ \color{blue}\overline{AB}=6.2469\)

\(\color{blue}The\ length\ \overline{AB}\ is\ 6.2469.\)  

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\(The\ length\ of\ \overline{AB}\ is\ 4,\ find\ the\ length\ AB.\\ What's\ this\ nonsense?\)

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 How many miles did I drive?

\(s=40mph\cdot t=55mph\cdot (t-\frac{25}{60}h)\\ 40mph\cdot t=55mph\cdot t-55mph\cdot \frac{25}{60}h\\ (55-40)mph\cdot t=55mph\cdot \frac{25}{60}h\\ t=\dfrac{55\cdot \frac{25}{60}h}{15}\\ t=1.52\overline 7h\)

\(s=40mph\cdot1.52\overline 7h\\ s=61\frac{1}{9}\ miles\\ \color{blue}61\frac{1}{9}\ miles\ dit\ i\ drive.\\ \)

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 Find the area of triangle AMN.

\(\overline{AB} = c=5, \overline{AC} = b=5\ and\ \overline{BC}=a = 8\)

\( A = s ( s − a ) ( s − b ) ( s − c )\ Heron's\ formula\\ s=\dfrac{a+b+c}{2}=\dfrac{8+5+5}{2}=9\\ A_{ABC} = 9 ( 9 − 8 ) ( 9 − 5) ( 9 − 5 )\\ A_{ABC}=144\\ Any\ line\ through\ the\ center\ of\ gravity\ of\ a\ triangle\ divides\ it\ into\ equal\ parts.\\A_{AMN=}\frac{1}{2}\cdot A_{ABC}=\frac{1}{2}\cdot 144\\ \color{blue}A_{AMN}=72\)

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 PQ = 8, QR = 5, and PR = 10.

\(f_{PR}(x)=0\\ 5^2=8^2+10^2-2\cdot 8\cdot 10\cdot cos\ P\\ 160\cdot cos\ \ P=64+100-25\\ cos\ P=\frac{64+100-25}{160}\\ P=29.686° \)

\(f_{PX}(x)=tan\ (\frac{29.686}{2})\cdot x\\ f_{PX}(x)=0.265x\\ sin\ R:sin\ P=8:5\\ sin\ \ R= \dfrac{8\cdot sin\ P }{5}=\dfrac{8\cdot sin\ 29.686° }{5}\\ R=52.4105°\\ f_{RQ}(x)=tan\ (-R)\cdot (x-10)\\ f_{RQ}(x)=tan\ (-52,4105°)\cdot (x-10)\\ f_{RQ}(x)=-1,3\cdot (x-10)\\\)

\(f_{PX}(x)=f_{PQ}(x)\\ 0.265x=-1.3(x-10)\\ 0.265x=-1.3x+13\\ x=\dfrac{13}{1.565}\\ \color{blue}x_X=8.3067\\ \color{blue}y_X=2.2013\)

\(f_{XY}(x)=m(x-x_X)+y_X\\ m=tan\ (90°-R)=tan\ (90°-52.4105°)\\ m=0.7698\\ f_{XY}(x)=0.7698(x-8.3067)+2.2013\\ f_{XY}(x)=0.7698x-6.3946+2.2013=0\\ \color{blue}x_Y=5.4472\\ \color{blue}y_Y=0\)

\(L_{XY}=\sqrt{y_X^2+(x_X-x_Y)^2}=\sqrt{2.2013^2+(8.3067-5.4472)^2}\\ \color{blue}L_{XY}=3.6087\)

\(\color{blue}The\ length\ of\ \overline{XY}\ is\ 3.6807.\)

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The question can be solved by specifying PR=10 instead of PR=1.

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