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#### markosheehan

##### Member

- Jun 6, 2016

- 136

there is complicating way to do this however there is also a easy way and i am trying to find out the easy way to do it.

the answer is 120 degrees .

- Thread starter markosheehan
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- Thread starter
- #1

- Jun 6, 2016

- 136

there is complicating way to do this however there is also a easy way and i am trying to find out the easy way to do it.

the answer is 120 degrees .

- Admin
- #2

- Mar 5, 2012

- 9,491

Let's see... it usually helps to draw a picture...

there is complicating way to do this however there is also a easy way and i am trying to find out the easy way to do it.

the answer is 120 degrees .

\begin{tikzpicture}[scale=0.5,>=stealth]

\def\x{110};

\def\t{0.3};

\node (C) at (0,0) {};

\node (D) at (12,0) {};

\node (C1) at ({30*\t*cos(60)}, {30*\t*sin(60)}) {};

\node (D1) at ({12+15*\t*cos(\x)}, {15*\t*sin(\x)}) {};

\fill (C) circle (0.1) node[below] {C};

\fill (D) circle (0.1) node[below] {D};

\fill (C1) circle (0.1) node[above] {C'};

\fill (D1) circle (0.1) node[above right] {D'};

\draw (D) -- +(3,0);

\draw[->,thick] (C) +(1,0) node[above right] {$60^\circ$} arc (0:60:1);

\draw[->,thick] (D) +(1,0) node[above right] {$x$} arc (0:\x:1);

\draw (C) -- node[below] {12} (D);

\draw[->,ultra thick,blue] (C) -- node[above left] {30t} (C1);

\draw[->,ultra thick,blue] (D) -- node[above right] {15t} (D1);

\draw[->,ultra thick,red] (D1) -- node[above right] {d} (C1);

\end{tikzpicture}

The standard way to solve this, is to write $d^2$ as a function of both $x$ and $t$, and then find the critical point.

That is, set both partial derivatives to zero.

$$d^2=(30t\cos(60^\circ)-(12 + 15t\cos(x)))^2 + (30t\sin(60^\circ)-15t\sin(x))^2 \\ \pd{}t(d^2)=0 \\ \pd{}x(d^2)=0

$$

That's indeed a bit of work.

Wolfram says that the solution is $t=0.4$, $x=120^\circ$, and $d=6$.

\begin{tikzpicture}[scale=0.5,>=stealth]

\def\x{120};

\def\t{0.4};

\node (C) at (0,0) {};

\node (D) at (12,0) {};

\node (C1) at ({30*\t*cos(60)}, {30*\t*sin(60)}) {};

\node (D1) at ({12+15*\t*cos(\x)}, {15*\t*sin(\x)}) {};

\fill (C) circle (0.1) node[below] {C};

\fill (D) circle (0.1) node[below] {D};

\fill (C1) circle (0.1) node[above] {C'};

\fill (D1) circle (0.1) node[above right] {D'};

\draw (D) -- +(3,0);

\draw[->,thick] (C) +(1,0) node[above right] {$60^\circ$} arc (0:60:1);

\draw[->,thick] (D) +(1,0) node[above right] {$x$} arc (0:\x:1);

\draw (C) -- node[below] {12} (D);

\draw[->,ultra thick,blue] (C) -- node[above left] {30t} (C1);

\draw[->,ultra thick,blue] (D) -- node[above right] {15t} (D1);

\draw[->,ultra thick,red] (D1) -- node[above right] {d} (C1);

\end{tikzpicture}

Now here's an interesting observation.

At the minimum the vector $\vec {DD'}$ is parallel to the distance vector $\vec d$.

That makes sense, since as long as there is an angle between them, we can improve the distance by reducing that angle by changing $x$ appropriately.

It means we can eliminate either $x$ or $t$ from the expression for the distance, and then take its derivative and set it to zero.

Still some work though.