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# IB Math SL Portfolio The aim of this task is to investigate positions of points in intersecting circles.

Extracts from this document...

Introduction

Aim: The aim of this task is to investigate positions of points in intersecting circles.

Let r = 1. Use an analytic approach to find OP’ when

Radius equals 1; OP is equal to 2, so the other side length must be equal to 2 because this is an isosceles triangle.

Use Law of Cosines To Find Angle:

Cos O =        ------  Where A= 2, O = 2, P = 1.    Where A=A, B=O, C=P

Cosine O=          O=

O = 75. 522 Degrees

Use Law of Sines to Find Other Angles:

SO

Sin A= .9682

A=                             A= 75.522 degrees

Sum of Triangle = 180 degrees                     Where x is the angle measurement of P

75.522 + x degrees = 180

151.044 + x degrees = 180

151.044-151.044 + x degrees = 180 – 151.044

x degrees = 28. 956

P = 28.956

The angle and side length of A and O are the same. This makes perfect sense because the triangle is an isosceles and since A and O have the same side length, it makes perfect sense that they have the same angle measurement. Hence the sum of the angles of A and O has to be greater than P.

Angle Measurement:

A= 75.522 degrees

O= 75.522 degrees

P= 28.956 degrees

Radius equals 1, OP is equal to 3, so the other side length must be equal to 3 because this is an isosceles triangle.

Use Law of Cosines To Find Angle:

Cos O =        ------  Where A= 3, O = 3, P = 1.    Where A=A, B=O, C=P

Cosine O=          O=

O = 83.621 Degrees

Use Law of Sines to Find Other Angles:

SO

Sin A= .9994

A=                             A= 83.621 degrees

Middle

and O have the same side length, it makes perfect sense that they have the same angle measurement. Hence the sum of the angles of A and O has to be greater than P.

Angle Measurement:

A= 86.417 degrees

O= 86.417 degrees

P= 7.166 degrees

After strenuous thinking, I was able to arrive at the formula for OP Prime:

OP Prime=

As you increase the side lengths and keep the radius the same you limitate how big the angles can be. As you increase the side lengths and keep the raidus the same, the angles of O AND P get larger and angle P get smaller.

Let us analyze:

Radius of 1 and OP equal to 2:       OP Prime =         OP Prime =

Radius of 1 and OP equal to 3:       OP Prime =         OP Prime =

Radius of 1 and OP equal to 4:       OP Prime =          OP Prime =

As you can see: as you increase the side lengths and keep the radius the same OP Prime gets smaller.

Radius equals 2, OP is equal to 2, so the other side length must be equal to 2 because this is an equalateral triangle.

Use Law of Cosines To Find Angle:

Cos O =        ------  Where A= 2, O = 2, P = 2.    Where A=A, B=O , C=P

Cosine O=          O=

O = 60 Degrees

Use Law of Sines to Find Other Angles:

SO

Sin A= .866

A=                             A= 60 degrees

Sum of Triangle = 180 degrees                     Where x is the angle measurement of P

60+ x degrees = 180

120 + x degrees = 180

120-120 + x degrees = 180 – 120

x degrees = 60

P = 60

Conclusion

Scopes and Limitations:

The different scopes and limitations is that the radius can never be bigger than the product of the other side lengths because you would have to end taking the cosine inverse of a number over one which never works and results in no triangle. The second thing is this formula only applies to when you are able to form a triangle so this formula does not hold true every time. So before you do anything you have to find your angles measurements of the triangle and make sure everything is in order to proceed form there. I arrived at my general statement by deriving my angles and comparing them. Also, I was able to work with my fellow classmates so I was bale to produce a general statement with everyone’s brains combined. as you increase the side lengths and keep the radius the same OP Prime gets smaller and my second was as you increase the radius and keep the side lengths the same OP Prime gets larger, so changing the radius or side lengths affect what OP Prime is going to be. It would be either smaller or bigger than side P.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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