Hello,

Suppose there's a circle with radius r and centre C. Let there be a point outside the circle P, such that the length PC is known. Now, lets take a point X on the circumference of the circle, such that the inclination of CX with respect to PC is a. Is it possible to find PX? If so, how??

By inclination do you mean the tangent of the angle XCP? If so you have side-angle-side and you can determine the third side PX. https://www.mathsisfun.com/algebra/trig-solving-sas-triangles.html

 

In other words we know PC, and we know that CX = r, and we know the angle XCP as the arctangent of the inclination.

 

Am I understanding inclination correctly? As the slope of the line with respect to CP, in other words imagining CP as the positive x-axis and then the inclination is the slope of CX?

Edited September 28, 20169 yr by wtf

There are four possibilities conforming to your description.

 

For the first 3 wtf has told you how to find PX by solving the triangle PCX.

Note that PX does not need to be a tangent it may cut the circle.

 

In the fourth possibility PCX are colinear so PX = PC + r and a = 180^o

 

Note that for

 

case 1 the triangle is a right angled triangle so a simple formula is used

 

case 2 and case 3 uses the cosine rule.

Note that for acute angles ( a<90) cos(a) is edit (oops beware the double negative) negative positive so the last term is negative, but for obtuse angles (a>90) cos(a) is negative so the last term is positive.

Note also that since we are using the cosine rule there are no positions of P that cannot be solved, unlike for a sine rule problem.

 

case 4 is again a simpler equation.

 

Edited September 28, 20169 yr by studiot

OK

Thanks.

Note that for acute angles ...( a<90)

A small quibble. OP said that a is the inclination, not the angle. The inclination is the slope of CX with respect to PC.

 

In other words the angle is the arctan of a. OP did not provide a clarification so I assume this is what is meant.

 

(In your pictures this would be more clear if you drew CX as having a positive angle with PC. In your pictures, the inclination is negative.)

Edited September 28, 20169 yr by wtf

A small quibble. OP said that a is the inclination, not the angle. The inclination is the slope of CX with respect to PC.

 

In other words the angle is the arctan of a. OP did not provide a clarification so I assume this is what is meant.

 

(In your pictures this would be more clear if you drew CX as having a positive angle with PC. In your pictures, the inclination is negative.)

 

I don't suppose your visits overlapped.

 

I think formal study is difficult in India and the OP is trying his best with the English language, so I took it to mean to angle

 

I'm glad it was a small quibble since inclinometers (I have several) measure angle.

 

https://en.wikipedia.org/wiki/Inclinometer

 

They were much used in the survey of India.

I don't suppose your visits overlapped.

 

I think formal study is difficult in India and the OP is trying his best with the English language, so I took it to mean to angle

 

I'm glad it was a small quibble since inclinometers (I have several) measure angle.

 

https://en.wikipedia.org/wiki/Inclinometer

 

They were much used in the survey of India.

You are right. I looked up inclination last night and for some reason I thought it said the slope, but actually it said the angle. My mistake. Inclination is the angle.

Are there not two colinear solutions?

 

[math] a=180^o [/math]

 

and

 

[math] a=0^o [/math]

Edited December 1, 20169 yr by AbstractDreamer

Are there not two colinear solutions?

 

[math] a=180^o [/math]

 

and

 

[math] a=0^o [/math]

 

Yes but a = 0 is the limiting situation for my case 1, with a=0, PX = PCsin(a) = PCsin(0) = 0.

 

Does this calculation work for a = 180? (case 4)

 

when a=0

PX=PC-r

 

Can you get this solution from case 1?

Edited December 2, 20169 yr by AbstractDreamer

when a=0

PX=PC-r

 

Can you get this solution from case 1?

 

You are absolutely right, I was too quick there.

 

Well done for spotting that. +1