asinus

Find AB.

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\(AB=\sqrt{409+(\frac{15}{sin(120°-\arctan{\frac{20}{3}})})^2-2*\sqrt{409}*\frac{15}{sin(120^°-\arctan{\frac{20}{3}})}*\cos{60°}}=22.4\)

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The path to the farthest corner is \sqrt (2^2+1^2)= \sqrt 5.

Franklin reaches 100% of the cube surface.

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If K is the area of ​​the triangle, what should be determined if K is the area of ​​the triangle?

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Find the area of the shaded region.

Where can I find the shaded region?

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Let cos A = x

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\(\sin 4A = \sin A + \sin 2A\)     Find acute angle A.

\( \sin 4A=8\sin A\cos^3 A-4\sin A\cos A\\ \cos^3 A=\\ \sin 2A=2\sin A\cos A\\ 8\sin A\cos^3 A-4\sin A\cos A=\sin A+2\sin A\cos A\)

will be continued

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Find length PS.

PS is half the base of an equilateral triangle OSA with side length 5.

\(\color{blue}\overline{PS}=2.5\)

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\(\color{blac}Find\ \angle CAD\ in\ degrees.\\\)

Thanks webbinsight for the useful suggestions to solve this tricky task!

\(\angle ADB =\angle ADE=\alpha\\ \angle ACB=\angle ACE=\beta\)

The angles intersection AD/EC are twice:

\((120°-\alpha )\ and\ (60°+\alpha )\)

The angles intersection AC/BD are twice:

\((30°+\beta)\ and\ (150°-\beta)\\ \angle CAD=60°+\alpha -\beta\)

In the triangle (intersection AC/BD) - A-D the following applies:

\(\alpha =180°-(150°-\beta )-(60°+\alpha -\beta)\\ \color{blue}\beta = 15°\)

The angles intersection BD/CE are twice:

\((180°-30°-2\beta=120°)\ and\ (60°)\)

In the triangle (intersection AD/CE) -  (intersection CE/BD) - D the following applies:

\((120°)+(60°+\alpha )+(\alpha )=180°\\ \color{blue} \alpha =0\ ?\ ?\ ?\)

I ask everyone for help.

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Find WX

\(U(-3,0), V(3,0), Y(-2,2), Z(2,2)\\ UY(-2.5,1)\\ m=\frac{y_Y-y_U}{x_Y-x_U}=\frac{2-0}{(-2)-(-3)}\\ m=2\)

\(f(x)=m(x-x_U)+y_U=2(x-(-3))+0\\ f(x)=2x+6\\ f(x_M)=-\frac{1}{m}(x-x_{UY})+y_{UY}=-\frac{1}{2}(x-(-2.5))+1\\ f(x_M)=-0.5x-0.25\\ x=0\\ \color{blue}M(0,-0.25)\ ,\ Center\ of\ the\ enclosing\ circle\)

\(r=\sqrt{x_Z\ ^2+(y_Z-y_M)^2}=\sqrt{2^2+(2-(-0.25))^2}\\ \color{blue}r=3.0104\ ,\ Radius\ of\ the\ enclosing\ circle\)

\(f_{circ}(x)=y_M\pm \sqrt{r^2-x^2}\\ 1=-0.25+\sqrt{3.0104^2-x^2}\\ (1+0.25)^2=3.0104^2-x^2\\ x_{WX}=\pm \sqrt{3.0104^2-1.25^2}\\ x_{WX}=\pm \sqrt{7.5}=\pm 2.7386 \\ \color{blue}\overline{WX}=5.4772 \)

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\(-\sqrt{3}sec(\theta)=2\\ -\sqrt{3}\cdot \dfrac{1}{cos(\theta)}=2\\ cos(\theta)=\dfrac{-\sqrt{3}}{2}\\ \theta =\dfrac{\pi}{180^{\circ}}\cdot (\arccos\dfrac{-\sqrt{3}}{2})+2\pi n\\ \theta =\dfrac{\pi}{180^{\circ}}\cdot (150^{\circ})+2\pi n\\ \color{blue}\theta=\dfrac{5\pi}{6}+2\pi n\)

Multiply the DEG-result of  \(\arccos \dfrac{-\sqrt{3}}{2} \ by\ \dfrac{\pi}{180^{\circ}}(=1)\) to express it as a fraction. To express the multiple result of \(\arccos \dfrac{-\sqrt{3}}{2}\)  add \(2\pi n.\)

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What is sin C in an acute triangle?

\(A=\arcsin \dfrac{2}{5}\\ A=23.5782^{\circ}\\ B=\arcsin \dfrac{2}{\sqrt{7}}\\ B=49.1066^\circ \\ C=180^{\circ }-23.5782^{\circ}-49.1066^\circ\\ \color{red }C=107.3152^{\circ}\\ The\ triangle\ is\ obtuse-angled!\\ \color{blue}\sin C=0.9547\)

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 What is the area of the triangle with side lengths tan A, tan B, and tan C?

\(f_{AC}(x)=\sqrt{9^2-x^2}\\ f_{BC}(x)=\sqrt{6^2-x^2+10x-5^2}\\ 81-x^2=36-x^2+10x-25\\ 10x=81-36+25\\ x_C=\dfrac{81-36+25}{10}\\ x_C=7\\ y_C=\sqrt{9^2-7^2}\\ C\ (7,5.6569)\)

\(\angle B=180^\circ -atan\ \frac{y_c}{x_C-x_B}=180^\circ-atan\ \frac{5.6569}{7-5}\\ \angle B=109.4712^\circ\\ \color{red}tan\ \angle B=-2.8284\\ The\ sides\ of\ a\ real\ triangle\ are\ greater\ than\ zero. \)

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My question to CPhill:

How can one formulate the equation AB / BD = AC / DC without knowing the angles of triangle ABC?

Please describe how I can arrive at this final solution.

Thanks in advance for your answer!

I will not be available until April 27, 2025.

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