Trigonometry: Solving for Cosine Using Segment Angle and Triangle Properties

In summary, the conversation is about finding cos(BETA) based on given quantities in a segment of a circle. There is a discussion about using the Pythagorean Theorem to find the third side, but it is noted that there is too much information and cos(BETA) can be expressed in multiple ways. The conversation also touches on naming the hypotenuse as R and using the Pythagorean Theorem to find its value. Ultimately, the conversation ends with the individual figuring out how to find cos(BETA) using the given information.
  • #1
S_BX
4
0
Hi,
I have a segment of a circle, and know the following quantities.
I need to find cos(BETA), and I know that it's supposed to be = (L^2 - 4h^2)/(L^2 + 4h^2), and I've been trying this for hours.. I suppose it has something to do with the triangle - as you can find the 3rd side (=radius-h), but I just can't get it.. Can somebody please help me out?

http://www.stevenbetts.btinternet.co.uk/bangle.gif
 
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  • #2
You could use the Pythagorean Theorem to find the 3rd side. But, there is too much information, so [tex]\cos\beta[/tex] can be expressed many ways.
 
  • #3
lotrgreengrapes7926 said:
You could use the Pythagorean Theorem to find the 3rd side. But, there is too much information, so [tex]\cos\beta[/tex] can be expressed many ways.
I just took the 3rd side to be R-h.
 
  • #4
Could you show me how you came up with

[tex] \frac {L^2 + 4h^2} {8h}[/tex] for the hypotenuse. Why not just call it what it is R?
 
  • #5
Integral said:
Could you show me how you came up with

[tex] \frac {L^2 + 4h^2} {8h}[/tex] for the hypotenuse. Why not just call it what it is R?

Because the 1st part of the question asked me to show that R = that. I got it by Pythagoras theorem...

R^2 = (L/2)^2 + (R-h)^2 (pythag)
R^2 = L^2/4 + R^2 - 2hR + h^2 (expand square terms)
2hR = L^2/4 + h^2 (gather like terms)
8hR = L^2 + 4h^2 (multiply through by 4)
R = (L^2 + 4h^2)/8h (rearrange for R)


The second part of the question is to find cos(BETA) where beta is the half angle, using the above result, and I'm totally stuck on that part.
 
  • #6
Solved

Never mind guys - I got it eventually! :) Thanks for trying to help though! :)
 

Related to Trigonometry: Solving for Cosine Using Segment Angle and Triangle Properties

1. What is a segment angle in trigonometry?

A segment angle in trigonometry refers to the angle formed by two intersecting lines or segments. It is measured in degrees or radians and is an important concept in understanding geometric shapes and solving trigonometric equations.

2. How do I find the segment angle in a right triangle?

In a right triangle, the segment angle can be found using the inverse trigonometric functions (sine, cosine, and tangent). For example, if you know the lengths of the sides of the triangle, you can use the inverse sine function to find the angle opposite to the side.

3. Can the segment angle be negative?

Yes, the segment angle can be negative. In trigonometry, angles are measured in a counterclockwise direction from the positive x-axis. Angles in the clockwise direction are considered negative.

4. How is the segment angle related to the unit circle?

The unit circle is a circle with a radius of 1, centered at the origin on a coordinate plane. The segment angle can be expressed in terms of the coordinates on the unit circle. For example, the x-coordinate of a point on the unit circle is equal to the cosine of the segment angle, and the y-coordinate is equal to the sine of the segment angle.

5. Can the segment angle be larger than 360 degrees?

Yes, the segment angle can be larger than 360 degrees. In trigonometry, angles can be measured beyond a full rotation, and they are referred to as coterminal angles. For example, an angle of 450 degrees is equivalent to an angle of 90 degrees.

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