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Editorial Team
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Asked: 2026-05-15T21:58:55+00:00

I’m trying to calculate triangles base on the Area and the angles. If Angle-B

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I’m trying to calculate triangles base on the Area and the angles. If Angle-B is 90° then the formula works, but in my case, the angle can be from 0.1° to 179.8°. The formula assumes that the angle is 90, so I was thinking that there might be something that is hidden that could work for very angle. Here is the formula:

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The formula in code would be:

Height = sqrt((2 * Area) / (tan(Angle-A)));

I’m looking for the second half of the formula. Would the next part of the formula be something like this:

cos(sin(AngleB))
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1 Answer

  1. Editorial Team

    Okay, new try: If my calculations are correct, side B equals sqrt(2*area*sin(angle-B)/(sin(angle-A)*sin(angle-C))

    Since Area = 1/2 * A * B * sin(c) = 1/2 * C * B * sin(a) = 1/2 * A * C * sin(b) we get:

    A = 2 * area / (B * sin(c)) and using this we get:

    C = sin(c) * B / sin(b) and when we place that back into the equation of area, we get:

    B = sqrt(2*area*sin(angle-B)/(sin(angle-A)*sin(angle-C))

    When you know one side and all the angles, calculating the other sides should be easy using normal trigonometry.

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