Find all values.

Mar 2019
181
1
TTF area
So the question basically just asks to find all the dimensions etc....

The picture is attached.

I firstly halved it into the two big triangles with the line AC, I then found the length of the line AC=151.41, rounding = 151.4m, angle A in ACD was equal to 22.862 degrees so about 22degrees. Moving on to the other triangle ACB, I got length AB=172.279, 172m.

My questions if the above is also correct is how do I find the two smaller triangles' dimensions and the arc's dimensions.

Thanks.
 

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Mar 2015
182
68
Universe 2.71828i3.14159
If 2 segments and the angle between them are given then the third segment is $c^2=a^2+b^2-2ab \cos \gamma$.

Length of an arc is $C' = r \cdot \gamma$.

You also have to adjust results for the curvature of the Earth.
 
Mar 2019
181
1
TTF area
Wait so if I call the place where all the lines seem to intersect X than line dx is just the height from the base of triangle ACD and respectively for ABC so I found that with ACD height base being 56.4 I got line AX= 145. Using angle C in triangle ABC which is 20degrees I get line XB=61m How do I get the radius?
 

skeeter

Math Team
Jul 2011
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the arc radius is 60m according to the diagram

arc length, $s = r \cdot \theta$ where $\theta$ is measured in radians
 
Mar 2019
181
1
TTF area
Well I suppose I use my angle of 100degrees so the central angle is 100degrees, the radius as you stated is 60m (how did you find this?), the diameter is 120m and arc length is 104.7198m. Note if it matters according to all my calculation the area is 3141.59m^2
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
Well I suppose I use my angle of 100degrees so the central angle is 100degrees, the radius as you stated is 60m (how did you find this?), the diameter is 120m and arc length is 104.7198m. Note if it matters according to all my calculation the area is 3141.59m^2
see attached ...
 

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Mar 2019
181
1
TTF area
The second part is..... they want to build a new fence across line ABCD. The fence will start at point B and connect with the existing line AD. The new fence will then divide the field into a triangular section, towards the north, and a quadrilateral section, towards the south. They want the area of the triangular section to be between a third and half of the area of the whole field .

How do I go about this?
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
The second part is..... they want to build a new fence across line ABCD. The fence will start at point B and connect with the existing line AD. The new fence will then divide the field into a triangular section, towards the north, and a quadrilateral section, towards the south. They want the area of the triangular section to be between a third and half of the area of the whole field .

How do I go about this?
I would calculate the perpendicular distance, $h$, between point B and line segment AD. (see attached)

If you set point D as the origin, point B has coordinates $x = |BD|\cos(40)$ and $y=|BD|\sin(40)$. Line DA has equation $y = mx$ where $m = \tan(110)$

In general, the distance from a point $(x_1,y_1)$ to a line $ax+by+c=0$ is $h = \dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$

With the calculated height, $h$, you can set up the following inequality ...

$\dfrac{A}{3} < \dfrac{b \cdot h}{2} < \dfrac{A}{2}$

... then determine the interval of values for the triangle's base, $b$, which will be the distances from point A$
 

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Mar 2019
181
1
TTF area
Not gonna lie I don't fully understand what you're saying here