asinus

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

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:

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

To be continued

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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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What is the probability that BD < 4?

\(\color{blue}P=\dfrac{A_{DBC}}{A_{ABD}}\)

\(f_{AC}(x)=\sqrt{5^2-x^2}\\ f_{BC}(x)=\sqrt{3^2-(x-7)^2}\\ 25-x^2=9-x^2+14x-49\\ 14x=65\\ x_C=4.6429 \\ y_C=1.8558\\ C(4.6429,1.8558)\\ A(0,0)\\ B(7.0)\)

\(f_{AD}=\frac{1.8558}{4.6429}x\\ f_{BD}=\sqrt{4^2-(x-7)^2}\\ \frac{1.8558}{4.6429}x=4^2-(x-7)^2\\ \frac{1.8558}{4.6429}x=16-x^2+14x-49 \)

\(x^2-13.6003x+33=0\\ x=6.8001\pm \sqrt{46.2420-33}\\ x_D=3.1661\\ y_D=1.2635\\ D(3.1661,1.2635)\)

\(A_{ABC}=\frac{7}{2}\cdot 1.8558\\ A_{ABC}=6.4953\\ A_{ABD}=\frac{7}{2}\cdot 1.2635\\ \color{blue}A_{ABD}=4.4224\\ A_{DBC}=A_{ABC}-A_{ABD}=6.4953-4.4224\\ \color{blue}A_{DBC}=2,0729\)

\(P=\dfrac{A_{DBC}}{A_{ABD}}=\frac{2.0729}{4.4224}:\frac{2.0729}{2.0729}=1:2.1334\)

The probability that BD < 4 is  \(\color{blue }P=1:2.1334\)

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What is distance from A to B along the (minor) arc of a great circle?

\(s=2\\ r=2\\ b=2\cdot \frac{\alpha}{2}\cdot \frac{2\pi r}{360^{\circ}} \\sin\frac{\alpha}{2}=\frac{1}{2}\\ b=2\cdot arcsin \frac{1}{2}\cdot\frac{2\pi \cdot 2}{360^{\circ}}\\ \color{blue}b=2.0944\)

\(1.11111 + 0.11111 + 0.01111 + 0.00111 +0.00011 + 0.00001\color{blue} = 1.23456\)

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What is b?

\(b=\sqrt{(12-7)^2+(5+2)^2}\cdot cos(30^{\circ }+arctan\frac{5}{5+2})+5\\ \color{blue}b=8.562\)

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Where is the picture?

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The equation of the reflection of the line y = 2x + 1 across the line y = 5x - 3.

Intersection:

\(f_1(x)=f_2(x)\\ 2x+1=5x-3\\ 3x=4\\ x_{1,2}=\frac{4}{3}\\ y_{1,2}=\frac{11}{3}=3\frac{2}{3}\\ P_{1,2}(1\frac{1}{3},3\frac{2}{3})\)

gradient

\(m_{1,2}=tan\ [\frac{1}{2}(\arctan 2+\arctan 5)]\\ m_{1,2}=2.9145\)

Point direction equation

\(y=m(x-x_P)+y_P \\ y=2.914536(x-1.\overline {33})+3.\overline{66}\\ y=2.914536x-3.886053+3.666667 \)

The equation of the reflection is

\(\color{blue}f_{1,2}(x)= y=2.914536x-0.219381\)

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If M is the midpoint of BC, then find AM^2.

\(sin\ ( 22.2^{\circ})=\frac{\overline {BM}}{4}\\ \overline {BM}=4\ sin\ ( 22.5^{\circ})\\ \overline {AM}^2 =4^2-\overline {BM}^2=16- 16\ sin^2\ (22.5^{\circ })=16\ (1-sin^2\ (22.5^{\circ }))\\ \color{blue}\overline {AM}^2=13.657\)

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What does that mean?

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 If CE = 3 and DF = 3, then what is BD?

Im Dreieck ABC liegen die Punkte D und F auf {AB} und E auf {AC}, sodass {DE}\parallel {BC} und {EF} parallel {CD} sind. Wenn CE = 3 und DF = 3, was ist dann BD?

Let AF = AE = c and BD = x. 

Then, according to the first ray theorem:

\(\dfrac{s+3}{s}=\dfrac{s+3+x}{s+3}\\ s^2+6x+9=s^2+3s+sx\\ sx=3s+9\\ x=3+\dfrac{9}{s}\)

To solve the question, it is necessary to specify an additional value, for example AE.

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