How To Find Sides Of Right Triangle With Known Angles And Area?

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SUMMARY

This discussion focuses on calculating the sides of a right triangle when the angles and area are known. The formulas provided are essential for determining the lengths of the sides based on the area (A) and angle (θ). The specific formulas include: \(c = \sqrt{\frac{2A}{\cos \theta \sin \theta}}\), \(a = \sqrt{2A \cot \theta}\), and \(b = \sqrt{2A \tan \theta}\). These equations enable users to find the dimensions of screens with different aspect ratios while maintaining the same area.

PREREQUISITES
  • Understanding of right triangle properties
  • Familiarity with trigonometric functions (sine, cosine, tangent)
  • Basic algebra for manipulating equations
  • Knowledge of area calculations for triangles
NEXT STEPS
  • Research trigonometric identities and their applications in geometry
  • Explore advanced triangle properties and the Law of Sines and Cosines
  • Learn about different aspect ratios and their implications for screen sizes
  • Investigate practical applications of triangle area calculations in design and engineering
USEFUL FOR

Mathematicians, engineers, graphic designers, and anyone involved in screen design or geometry-related calculations will benefit from this discussion.

DaleSwanson
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I'm comparing TV and monitor sizes, and I'm trying to figure out the formula that will let me find what size screen will have the same area as another screen (with different aspect ratios). This boils down to a right triangle with all the angles and the area known. Normally I'd find this on Google, but there are a large number of pages describing how to find the area with known sides or areas that are getting in the way.

So can someone post the formula for this?
 
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Let a be the adjacent, let b be the opposite, and let c be the hypotenuse and A the area. Then

<br /> \[<br /> c = \sqrt{\frac{2A}{\cos \theta \sin \theta}}<br /> \]<br /> \[<br /> a= \sqrt{2A \cot \theta}<br /> \]<br /> \[<br /> b = \sqrt{2A \tan \theta}<br /> \]<br />
 
Last edited:
Perfect, thanks.
 

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