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I currently have a lot of things going on for college, so i cant get my brain settled into figuring this out. (I stay up until 4am almost every morning studying). My question is interesting but not so hard to achieve.

Im attaching the diagram here:


Task 1:
A farmer has a field as shown in the diagram, where AB,BC,CD and AD are all boundary fences. There are two straight line pathways in the field which run from A to C and B to D. There is also a quad bike track which is in the shape of an arc, also from B to D.

Your task is to identify all the missing key features (and their dimensions) of the field.

Task 2:
The farmer intends to build a new fence line across the field (labelled ABCD), to subdivide it to better manage the livestock and pasture ,which will start at point B and connect with the existing fence line AD. This new fence will then divide the field into a triangular section, towards the North, and a quadrilateral section, towards the South, of the original field.

The farmer would like the area of the triangular section to be between a half and a third of the area of the whole field.

Your task is to recommend to the farmer where the new fence should be built. Identify how far along the boundary fence AD that the new fence should be connected. The farmer would like to give you the reasons for your recommendation(s)

$\frac{a}{\sin a} = \frac{b}{\sin b} = \frac{c}{\sin c}$
$a^2 = b^2 + c^2 - 2bc \cos a$
$\cos A = \frac{b^2+c^2-a^2}{2bc}$
$\text{Area of triangle} = \frac{1}{2} bc \sin A$
$\text{Arc Length} = (\theta/360) \times 2\pi r$
$\text{Area of length} = (\theta /360) \times 2\pi r^2$

Thanks guys!

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@André Nicolas ; Im sorry BCD is 100º I added it to the diagram, once again sorry about that! – Ivan Sep 23 '12 at 7:58
A non-mathematical aside: If you're consistently up until 4 AM studying (and need be awake for classes significantly before noon), then you're getting too little sleep to learn anything. That's far beyond the point where moving time from "sleep" to "study" will have a negative effect on how much you learn. If you make a strict habit of being in bed by midnight, you will be so much more efficient during the day that it will easily make up for spending 4 hours less on studying. – Henning Makholm Sep 23 '12 at 12:25

Strange problem! We can find the length of $AC$, since by the Cosine Law (second in your formula list), we have $$(AC)^2=120^2+60^2-2(120)(60)\cos(110^{\circ}).$$ Now that we have $AC$, we can find the remaining angles of $\triangle ADC$, using the Sine Law (first in your formula list). In fact, we only need one of these angles, since the other can be obtained from the fact that the angles of a triangle add up to $180^\circ$. So if $\theta=\angle DAC$, then $$\frac{\sin\theta}{60}=\frac{\sin(110^\circ)}{AC},$$ and since we know $AC$, we know $\sin\theta$, so we know $\theta$.

Unfortunately, progress ends here, unless, for example, we know $\angle DCB$, whose size is not indicated in the diagram. For note that if that angle is unspecified, we can rotate the side $CB$ a little bit around $C$, changing thereby the lengths of $AB$, $DB$, and several angles. Thus we cannot determine all the missing dimensions and angles.

Added: It turned out that there was inadvertently missing information, and that it was the measure of $\angle DCB$, which turns out to be $100^\circ$. Because $\triangle CDB$ is isosceles, the angles $CDB$ and $CBD$ are then each $40^\circ$. Also, we can now find $DB$ by the Cosine Law, since $(DB)^2=60^2+60^2-2(60)(60)\cos(100^\circ)$.

Since we know $\angle DCA$ from an earlier calculation, by subtracting from the known $100^\circ$ of $\angle DCB$, we find $\angle ACB$. And now since we know $\angle ACB$ and the sides $CA$ and $CB$, we find $AB$ by the Cosine Law. And now we can find $\angle BAC$ using the Sine Law. Any missing angles can be easily filled in.

We presumably also want the sides of the four "little" triangles that the diagonals divide the field into. Since we know all the angles and one side, this canbe done by the Sine Law.

If we care to, we can find the areas of various triangles, by using the supplied formula. The length of the track is also easy, for it is the fraction $\frac{100}{360}$ of the full circumference ($120\pi$) of a circle of radius $60$. Actually, there is some ambiguity here, because nothing really says that the radius is $60$. So, properly speaking, the length of the track is not determined.

**Note: Could you please simplify this?

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Yes, the diagram as labeled is clearly unrigid, since as you say, the length $\overline{AC}$ is known, but the placement of the point $B$ can be anywhere that’s the right distance from $C$. – Lubin Sep 23 '12 at 4:36
@André Nicolas ; Im sorry BCD is 100º I added it to the diagram, once again sorry about that! – Ivan Sep 23 '12 at 7:57

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