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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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