asinus

 If BD = 3 and CD = 5, what is AC?

I

\(mx=\sqrt{8^2-(x-6)^2}\)

II

\(\frac{m}{2}\cdot x=\sqrt{3^2-(x-6)^2}\\ mx=2\cdot \sqrt{3^2-(x-6)^2}\)

III

\(\sqrt{8^2-(x-6)^2}=2\cdot \sqrt{3^2-(x-6)^2}\\ 64-x^2+12x-36=4\cdot(9-x^2+12x-36)\\ 64-x^2-12x-36=36-4x^2+48x-144\\ 3x^2-36x+136=0\\ x^2-12x+45.3333=0\\ x=6\pm \color{red}\sqrt {36-45.3333}\)

\(\color{blue}Under\ the\ given\ conditions\, no\ result\ is\ possible.\)

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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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