Hi, these are questions in grade 12 geometry course i'm taking. Please Help me out!

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- Jun 25th 2007, 10:57 AM #1

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- Jun 25th 2007, 11:19 AM #2

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## can u see it this time?

- Jun 25th 2007, 11:46 AM #3

- Jun 25th 2007, 11:48 AM #4
The length of the base of the triangle is $\displaystyle a$.

Now, using the trig ratio for tangent, we see that

$\displaystyle \tan B = \frac {|AC|}{a}$ ........$\displaystyle |AC|$ means the length of the line connecting A and C

$\displaystyle \Rightarrow \tan \left( \frac { \pi}{3} \right) = \frac {|AC|}{a}$

$\displaystyle \Rightarrow \sqrt {3} = \frac {|AC|}{a}$

$\displaystyle \Rightarrow a = \frac {|AC|}{ \sqrt {3}} = \frac { \sqrt {3} |AC|}{3}$

So we have $\displaystyle B \left( \frac { \sqrt {3} |AC|}{3}, 0 \right)$

question (b) says: BCi s a quarter circle with centre at (0,0). Line AD has equation y=x.

Also note that the angle the line y = x makes with the x-axis is $\displaystyle 45^{ \circ}$, or $\displaystyle \frac { \pi}{4}$

Using the trig ratio for sine we see that

$\displaystyle \sin \left( \frac { \pi}{4} \right) = \frac {|DE|}{|AD|} = \frac {|DE|}{r}$ .........E is the point where a vertical line from D cuts the x-axis, |DE| represents the length of that line

$\displaystyle \Rightarrow r = \frac {|DE|}{ \frac { \sqrt {2}}{2}} = \frac {2 |DE|}{ \sqrt {2}}$

So $\displaystyle B \left( 0, \frac {2 |DE|}{ \sqrt {2}} \right)$

These questions are a bit vague to me. I guess we could do the last one without introducing a new point, if we used the cosine rule or something like that

- Jun 25th 2007, 12:02 PM #5
It appears to me as if the question (a) is asking for the coordinates of C.

Because line*BC*has slope $\displaystyle \tan \left( {\frac{{2\pi }}{3}} \right) = - \sqrt 3 $ then say C(0,c) so $\displaystyle \frac{{c - 0}}{{0 - a}} = - \sqrt 3 \quad \Rightarrow \quad c = a\sqrt 3 \quad or\quad C:\left( {0,a\sqrt 3 } \right).$

Thus we have written the coordinates of C in terms of the given.

- Jun 25th 2007, 12:03 PM #6

- Jun 25th 2007, 12:20 PM #7

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- Jun 26th 2007, 08:40 AM #8

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- Jun 26th 2007, 09:14 AM #9
Coordinates prove:

We can assume that $\displaystyle A\in Oy,B,C\in Ox$ and $\displaystyle A(0,2a),B(2b,0),C(2c,0)$, for the simplicity of calculus.

Now $\displaystyle D(c,a),E(b,a)$. Then you can write the equations of $\displaystyle BD$ and $\displaystyle CE$, solve the system and you will find the coordinates of the point $\displaystyle F$. Then you can calculate the length of the segments $\displaystyle BF,FD$.

- Jun 26th 2007, 09:17 AM #10

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- Jun 26th 2007, 09:19 AM #11

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- Jun 26th 2007, 09:39 AM #12
Given a triangle $\displaystyle ABC$ in plane we can get the $\displaystyle Ox$ and $\displaystyle Oy$ axis anywhere in that plane. So we can get $\displaystyle Ox$ axis passing through $\displaystyle B$ and $\displaystyle C$ and $\displaystyle Oy$ axis passing through $\displaystyle A$.

Now, if there are two points $\displaystyle M_1(x_1,y_1),M_2(x_2,y_2)$, then the midpoint $\displaystyle M$ of the segment $\displaystyle M_1M_2$ has the coordinates $\displaystyle \displaystyle \frac{x_1+x_2}{2}$ and $\displaystyle \displaystyle \frac{y_1+y_2}{2}$. Then $\displaystyle D(c,a), \ E(b,a)$.

The equation of the line passing through $\displaystyle M_1(x_1,y_1),M_2(x_2,y_2)$ is $\displaystyle \displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$ or

$\displaystyle \left|\begin{array}{lll}

x & y & 1\\

x_1 & y_1 & 1\\

x_2 & y_2 & 1

\end{array}\right|=0$

Using one of these equations you can now write the equations of $\displaystyle BD$ and $\displaystyle CE$, then solve the system formed by these equations to find the coordinates of $\displaystyle F$.

Then you can use the distance formula between two points in plane:

$\displaystyle M_1M_2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

to calculate $\displaystyle BF$ and $\displaystyle FD$

It is more clear now?

- Jun 26th 2007, 10:38 AM #13
Vector prove:

Let $\displaystyle \displaystyle \frac{BF}{FD}=k$

Then $\displaystyle \displaystyle \overrightarrow{CF}=\frac{1}{1+k}\overrightarrow{C B}+\frac{k}{1+k}\overrightarrow{CD}$ and

$\displaystyle \displaystyle \overrightarrow{CE}=\frac{1}{2}(\overrightarrow{CB }+\overrightarrow{CA})=\displaystyle \frac{1}{2}\left(\overrightarrow{CB}+2\overrightar row{CD}\right)=\frac{1}{2}\overrightarrow{CB}+\ove rrightarrow{CD}$.

The vectors $\displaystyle \overrightarrow{CF}$ and $\displaystyle \overrightarrow{CE}$ are colinear, so

$\displaystyle \displaystyle \frac{\frac{1}{1+k}}{\frac{1}{2}}=\frac{\frac{k}{1 +k}}{1}\Rightarrow k=2$, so $\displaystyle \displaystyle \frac{BF}{FD}=2$

- Jun 27th 2007, 07:58 AM #14

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