Math Portfolio Ib Circles

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  • Topic: Triangle, Law of cosines, Trigonometry
  • Pages : 6 (1194 words )
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  • Published : October 22, 2012
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MATHS PORTFORLIO SL TYPE I

CIRCLES


In this portfolio I am investigating the positions of points in intersecting circles. (These are shown on the following page. The following diagram shows a circle C1 with centre O and radius r, and any point P.

The circle C2 has centre P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has the centre A, and radius r. The point Pʹ is the intersection of C3 with (OP). This is shown in the diagram below.

As shown on the assignment sheet, r=OA. We therefore need to find the values of OPʹ when r=OA=1 for the following of the values of OP: OP=2, OP=3 and OP=4. We first of all extract the triangle OPA from the above diagram and since we have the values for all the three sides we can finds the angle AOP which will later on help to get the value of length OPʹ. The circle C2 and triangle OPA are shown below with all side of OPA indicated. OP=AP since they are the radii of the same circle, C2. Having all the three sides, we can now calubulate the angle AOP using the cosine rule. Angle AOP is calculated below: Cos AOP=(2^(2 _ ) 2^(2 _ ) 1^2)/(-2×2×1)

Cos AOP=0.25
∴ AOP =COS-10.25
=75.52248781
≈75.5˚
Since we now having the triangle AOP, we can extract the triangle AOPʹ from the diagram shown on the previous page which in return will help us to find OPʹ using the sine rule. The triangle AOPʹ is shown below:

O P

For accuracy the value of angle AOP will be used as cos-10.25 instead of 75.5˚. Since triangle AOPʹ is an isosceles triangle, AOP=APʹO=cos-10.25. Therefore OAPʹ= (180-(2×cos-10.25)). The calculation of the value of OPʹ is shown below: (OPʹ)/(sinOAPʹ)=(APʹ)/(sinAOPʹ)

(OPʹ)/sin⁡(180-(2×〖cos〗^(-1) 0.25) ) =1/sin⁡(〖cos〗^(-1) 0.25) opʹ= 1/sin⁡(0.25) ×(180-(2×〖cos〗^(-1) 0.25) )
opʹ=1/2
When OP=3, the triangle OPA and the calculation of OPʹ are as follows:

Cos AOP = (3^2-3^2-1^2)/(-2×3×1)
Cos AOP = 1/6
∴ AOP = cos-1 1/6
=84.4˚
From the triangle AOPʹ we can now calculate the length of OPʹ using the sine rule as before. The triangle AOPʹ and the calculator of OPʹ is shown below:
A

16

OP'

APʹO=AOPʹ=cos-11/6
OAPʹ= (180-(2×〖cos〗^(-1) 1/6)

(OPʹ)/(sinOAP ʹ)= (APʹ)/(sinAOPʹ) (OPʹ)/sin⁡(180-(2×〖cos〗^(-1) 1/6) ) =1/sin⁡(〖cos〗^(-1) 1/6)

opʹ= 1/sin⁡(1/6) ×(180-(2×〖cos〗^(-1) 1/6) ) = 1/3
When OP=4;
Cos AOP =(4^2-4^2-1^2)/(-2×4×1)
Cos AOP = 1/8
∴ AOP =cos-1 1/8
=82.4˚
Using the sine rule;
(OPʹ)/(sinOAP ʹ)= (APʹ)/(sinAOPʹ) (OPʹ)/sin⁡(180-(2×〖cos〗^(-1) 1/8) ) =1/sin⁡(〖cos〗^(-1) 1/8)

opʹ= 1/(sin⁡(〖cos〗^(-1 ) 1/8))×(180-(2×〖cos〗^(-1) 1/8) ) = 1/4

When OP=2, OPʹ= 1/2;when OP=3 , OPʹ= 1/3 and when OP=4, OPʹ= 1/4 . This indicates that the value of OPʹ is dependent on the value of OP. In fact it is inversely proportional to the value of OP. To arrive at the value of OPʹ, 1 is divided by the value of OP. Therefore generally, the value of OP' can be written as:

OP=r/OP
Moreover, from the values of OPʹ calculated above, it is observed that the value of OPʹ is twice Cos AOP. The general statement therefore can be written as: OPʹ= 2 Cosθ
Let OP=2. Find OPʹ when r=2, r=3 and r=4. Describe what you notice and write a general statement to represent this. Comment on whether or not this statement is consistent with your earlier statement. First of all we need to calculate the value...
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