Inverse cosine with varriables

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    Cosine Inverse
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SUMMARY

The discussion revolves around finding the area of a sector in a triangle defined by variables rather than fixed values. Specifically, the triangle ABC has sides AC and BC as radii equal to x, and side AB defined as x - 2y. While traditional methods involve trigonometric functions, the conversation suggests using the Pythagorean theorem to derive the height of the triangle, allowing for the area to be expressed in terms of x and y without relying on trigonometric functions. The area of the triangle can be calculated using the formula (base × height) / 2, which can be adapted to find the sector's area.

PREREQUISITES
  • Understanding of basic trigonometric functions and their applications
  • Familiarity with the Pythagorean theorem
  • Knowledge of geometric principles related to circles and sectors
  • Ability to manipulate algebraic expressions involving variables
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  • Explore methods for calculating areas of sectors without trigonometric functions
  • Study the application of the Pythagorean theorem in variable-based geometry
  • Research advanced geometric formulas for sectors and triangles
  • Learn about the implications of variable definitions in geometric calculations
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Mathematicians, geometry enthusiasts, and students studying trigonometry and algebra who are interested in alternative methods for calculating areas in variable-defined geometric figures.

kootromics
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Is there any equation or method that can be used in place of a trigonomic function when side values of a triangle are known only as varriables? For example: triangle abc where c is the center of a circle and AC and BC are radiui who's value = x, and AB = x-2y. So far as I know, the area of this sector (ACB) can be found in terms of x and y but must contain a trigonomic function of x and y as well. ( such as (cos-1( x..y/x..y))(pie.r^2) . Is there any way to know the area of this sector simply in values of x and y without a trigonomic function?
 
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Think of AB as the base the triangle. Use Pythagoras to determine the height of the triangle (draw a perpendicular from the center of the circle to the base - which bisects it!). Then you can write the area of the triangle (base X height / 2) in terms of x and y without the need for trig functions.
 
I did something like that as well with A = ½ab.sinC and the cosine rule, but it's the area of the sector, rather than of the triangle he wants.
 

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