A = 180 – 75.5 – 75.5
A = 29
O =
O=
The distance from O to P’ is 0.501, (as shown on the graph to the left).
This triangle has been constructed by the dots (midpoints) of the circles in the diagram above. Side’s A to P and O to P are both the same, they are 3. I know this because points A and O are both the radiuses of P making the lengths of those lines equal. Side’s A to O and A to P’ are both 1, because they are the radiuses of point A (midpoint of triangle C3), I am told that r=1(r being the radius).
In the second part of this problem I have to find the distance from O to P’, considering that O to P = 3.
A = 180 – 80.4 – 80.4
A = 19.2
O = 80.4
O to P’ = 0.334
This triangle has been constructed by the dots (midpoints) of the circles in the diagram above. Side’s A to P and O to P are both the same, they are 4. I know this because points A and O are both the radiuses of P making the lengths of those lines equal. Side’s A to O and A to P’ are both 1, because they are the radiuses of point A (midpoint of triangle C3), I am told that r=1(r being the radius).
In the third part of this problem I have to find the distance from O to P’, considering that O to P = 4.
A = 180 – 82.8 – 82.
A = 14.4
O = 82.8
O to P’ = 0 .25
General statement
Using the three different values for O to P (OP=2, OP=3, and OP=4) I was able to come up with a general statement as I saw a pattern developing in my results for the distance from O to P’. Each time that the length of O to P increased the fraction decreased:
When, O to P = 2
cosO =
When, O to P = 3
cosO =
Each time that the value of O to P is increased the fraction decreases because the denominator increases each time while the numerator stays the same.
The lower the fraction that I
, the larger the angle becomes:
O =
O = 75.5
O = 80.4
And so on…
This pattern continues as I keep increasing the distance from O to P.
As the side length O to P increases the side length of O to P’ decreases:
OP = 2 is 0.501 for OP’
OP = 3 is 0.334 for OP’
OP = 4 is 0.25 for OP’
From finding out more and more results I started to see a pattern, the numerator of the fraction stays the same. The numerator of the fractions above is 1, the radius is also 1.
,
,
,
,
…
Then I looked at the values of O to P that were increasing.
I took the radius and divided it by the length of OP:
Now to test this on a real example I will use OP = 2, OP = 3, and OP = 4.
=0.5
=0.333
=0.25
This method gives an accurate enough measurement of the length of the side O to P’.
Now I am given a set distance for O to P which is 2. I have to find the length of OP’ when the radius is 2, 3 and 4. From what I notice I have to write a general statement.
When constructing a diagram for r = 2 when OP = 2, I ran into a problem. This is because I am no longer able to have two triangles. The triangle with points AOP’ has now become equilateral meaning all three sides and all three angles are equal. If O to P is 2 this means that A to P is also 2.
What I found out is that because all of the side lengths are 2 the two triangles have almost overlapped each other forming one single triangle. This is because when the radius is increased from r=1 to r=2, while the distance from O to P is also 2, both points P and P’ line up right next to each other. Therefore making the length from O to P’ 2 the same as the length as O to P as they are on the same co-ordinate. Moving the point P’ closer to O along the boundaries of the triangle will decrease the side length of A to P’ and I cant do that because I am told that the radius (or A to P’) has to be 2.
When OP=2 considering that r=2 the length of O to P’ is 2.
In the next triangle I am asked to investigate r=3 while OP stays the same (2).
Because the radius is changed to 3, point P’ can no longer be between O and P as the distance from O to P is only 2. This causes point P’ to cross over point P now being further from point O than P.
0 = 41.41
To find angle A of triangle APP’ I added sides P and P’ and took away the sum of those two from 180 (which is the sum of all the angles of a triangle). This gave me angle A which is 55.8
.
Now that I possess all three angles of the triangle I am focusing on I can figure out the length of P to P’ using the sine law.
P to P’ = 2.501
In the third part of this problem I am told that the radius is 4, while OP = 2. I tried constructing a triangle given those figures but I wasn’t able to do so. This is because the side A to P is too short to reach the base of the triangle.
I know that OP=2 therefore AP= must also be 2 (as they are the radiuses of circle C2). I know that A to O is 4 so is A to P’, (as they are the radiuses of circle C3). Like seen in the previous problem where r=3, point P’ overlaps point P as line A to P’ is made longer than A to P. This leaves the side length AP (which is 2) hanging unable to join to the base of the triangle (because it is too short).
General Statement
Using the same procedure as I used last time I was able to come up with the results. Only this time, because the radius changed and not the length of OP, things were complicated as side lengths became either too short or too long to be able to maintain more or less the triangle that I initially started with. The shape of the triangle was now changed. This meant that some sides may not necessarily reach the base of the triangle, (like seen in the diagram above).
I had to modify my general statement formula in order for it to be able to adapt to the new type of problem, where the radius changes.
The length of O to P’ can be determined by
.
For example (using r=2 and r=3):
(when r=2)
=2
OP’ =
(when r=3)
=4.5
Using my previous procedure of sine and cosine I will now demonstrate their use through other values of r and OP.
When r=1 and OP=5
A = 180-84.26-84.26
A = 11.48
O = 84.26
When r=5 and OP=2
Because the radius is too big for the side length of OP to reach the base, it means that the triangle will be invalid (the side OP will not reach the base of the triangle).
General Statement
Through solving the two problems presented I was able to modify my initial formula from the first general statement in in order for it to work for the second problem in the second general statement.
OP’=
- This formula for the general statement works for solving both part one and two of this problem. To solve the first problem we can substitute the numbers into
as
will still be 1.
Testing the validity of the general statement
Find O to P’ when r=1, OP=6 and OP=7 Find O to P’ when OP=2 and r=6
OP’ =
= 0.167 therefore O to P’ is 0.167. OP’ =
= 18 therefore O to P’ is 18.
OP’ =
= 0.143 therefore O to P’ is 0.143.
The limitations of the general statement
When using cosine and sine to figure out the angles and side lengths the calculator gives of more numbers. I put those numbers to three significant figures cutting of the rest. Yet when I use the general statement formula the numbers I am given are more basic which could result in a small, (not very significant) calculation error. Apart from that I found my general statement to work well in solving these problems.
Explaining how I arrived at the general statement
From looking at my results for the first problem I created a general statement formula. I noticed that the radius stayed the same so I placed it on top of my equation and divided it by the distance from O to P giving my length of O to P’
Although when I reached the second problem the general statement did not work therefore it meant that I would have to modify it somehow to make it work. I squared the radius and divided it by the length O to P.
When I looked back on the first problem the modified general statement formula worked, as squaring 1 (
) will still give me 1 therefore not changing the values I received through the use of technology through sine and cosine.
by
kangypoo202 (student)
