Solving Right Triangles with 3 Angles and Area - Warren

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A right triangle can be solved using three angles and the area by applying the Law of Sines to establish the relationship between the sides. The area can then be calculated using the formula A_triangle = 1/2 * a * b * sin(γ), where a and b are the sides and γ is the angle between them. This method allows for the determination of side lengths up to a constant factor. The discussion highlights the application of trigonometry in solving problems related to squares and rectangles. Ultimately, the approach provides a way to find the necessary dimensions for the triangle based on given angles and area.
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But I'm not getting it.

Is it possible to solve a right triangle give 3 angles and the area?

For instance 90-72-18, A=150.

What I'm trying to solve is a problem of squares to rectangles and vis versa using trigonometry. I'll use this to determine extension and contraction.

Thanks,

Warren
 
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Law of Sines will give you the relation between the sides, hence the sides up to a constant. Use the formula for area using two sides and their internal angle to determine this constant.
 
I was going in that direction but have only used this to solve SAS etc problems.

So far I have a/.31=b/.95=c/1


Could you elaborate please..

thanks,

Warren
 
Oh I got it!

Thanks!

Warren
 
Using A_{triangle} = \frac{1}{2}ab \sin(\gamma), where a and b are sides of the triangle, and gamma is the angle between them should give you enough information to solve the triangle.

Edit: too late...
 
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