Relation between ##d## and ##\theta##

  • Context: High School 
  • Thread starter Thread starter Adel Makram
  • Start date Start date
  • Tags Tags
    Relation
Click For Summary
SUMMARY

The discussion focuses on deriving a closed-form relationship between the variables ##\theta## and ##d## as functions of the radius ##r## and the distance ##s## from point ##p## to the center of a circle. The key equation presented is ##s^2 = r^2 + d^2 + 2dr \cos(a)##, with the relationship between angles given by ##a + \theta = \pi/2##. By substituting ##a## with ##\pi/2 - \theta##, a new equation is formed that establishes the connection between ##d## and ##\theta##.

PREREQUISITES
  • Understanding of trigonometric identities and relationships
  • Familiarity with the Law of Cosines
  • Basic knowledge of geometry involving circles
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the Law of Cosines in depth to understand its applications
  • Explore trigonometric identities, particularly those involving angle sums
  • Learn about geometric properties of circles and their relevance in problem-solving
  • Practice algebraic manipulation techniques for rearranging equations
USEFUL FOR

Mathematicians, physics students, and anyone interested in geometric relationships and trigonometric applications in problem-solving.

Adel Makram
Messages
632
Reaction score
15
I wish to find a closed form relation between ##\theta## and ##d## in the attached figure as a function of ##r##, the radius of the circle and ##s## the distance from the point ##p## to the center.
Thank you.
circle.png
 
Mathematics news on Phys.org
Can you please show your work so far? What part are you having trouble with?

Is this question for schoolwork?
 
  • Like
Likes   Reactions: Adel Makram
berkeman said:
Can you please show your work so far? What part are you having trouble with?

Is this question for schoolwork?
Just right now, I had insight for how to solve it :)
##s^2=r^2+d^2+2 d r cos (a)##
But ##a+\theta=\pi/2##.
##s^2=r^2+d^2+2 d r cos (\pi/2 -\theta)##
rearranding, gives a relation between ##d## and ##\theta##.
circle.png
 
  • Like
Likes   Reactions: mfb

Similar threads

  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 59 ·
2
Replies
59
Views
230K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 70 ·
3
Replies
70
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K