Math: Triangles, angles, lengths.

Loren Michael

I have an obtuse triangle, but I don't know the perpendicular height or the angles, only the length of the three sides. I want to find the perpendicular height and the angles given only knowledge of the side lengths.

How do I do this?

Savant

Law of cosines can find you the angles: c^2 = a^2 + b^2 - 2ab cos (C), where C is the angle opposite the side c, so just solve for it for each of the angles.

The perpendicular height depends on which side you are going perpendicular from. But once you find the angles, you are essentially forming some more right triangles and you'll know one of the angles in one of the new triangles you are drawing (it'll be A, B, or C from the original triangle). You could then use the law of sines on the new triangle, since you'll know the length of the side opposite the right angle and the angle opposite the perpendicular height. If you draw it out you'll see this easier.

The law of sines is a / sin (A) = b / sin (B) = c / sin (C) for all sides and angles of a triangle.

Edit:

Wikipedia actually has a nice picture of what I'm taking about:

Smasher

Given the lengths of the three sides you can use the law of cosines to get the three angles.

To get the perpendicular height, using

as a reference, you'll already have b and alpha, from which you can calculate the bottom side of the triangle to the right with b*cos(alpha) as shown. From there you can use the law of cosines again to get the third side, which is the perpendicular height of the main triangle.

Loren Michael

Savant wrote: »

Law of cosines can find you the angles: c^2 = a^2 + b^2 - 2ab cos (C), where C is the angle opposite the side c, so just solve for it for each of the angles.

The perpendicular height depends on which side you are going perpendicular from. But once you find the angles, you are essentially forming some more right triangles and you'll know one of the angles in one of the new triangles you are drawing (it'll be A, B, or C from the original triangle). You could then use the law of sines on the new triangle, since you'll know the length of the side opposite the right angle and the angle opposite the perpendicular height. If you draw it out you'll see this easier.

The law of sines is a / sin (A) = b / sin (B) = c / sin (C) for all sides and angles of a triangle.

Am I reading you wrong? That first one implies that I already know angle C, doesn't it?

With respect to that wiki link:



I think I get that, but what's arccos?

Trig has been forever for me, sorry and thanks.

Smasher

With respect to that wiki link:



I think I get that, but what's arccos?

Trig has been forever for me, sorry and thanks.

Arccos is basically the inverse function of cosine. You'll need either a calculator or lookup table to evaluate it, but that's pretty much true of most regular trig functions anyway. It's also printed as cos^-1.

Loren Michael wrote: »

Savant wrote: »

Law of cosines can find you the angles: c^2 = a^2 + b^2 - 2ab cos (C), where C is the angle opposite the side c, so just solve for it for each of the angles.

The perpendicular height depends on which side you are going perpendicular from. But once you find the angles, you are essentially forming some more right triangles and you'll know one of the angles in one of the new triangles you are drawing (it'll be A, B, or C from the original triangle). You could then use the law of sines on the new triangle, since you'll know the length of the side opposite the right angle and the angle opposite the perpendicular height. If you draw it out you'll see this easier.

The law of sines is a / sin (A) = b / sin (B) = c / sin (C) for all sides and angles of a triangle.

Am I reading you wrong? That first one implies that I already know angle C, doesn't it?

You'll have to rearrange the equation to isolate and solve for C. So that would be C = arccos((c^2 - b^2 - a^2)/(-2ab)) assuming I didn't mess up.

Loren Michael

Oh. Duh.

Thanks, I should have known that, A bunch of shit just came back to me. Thanks, you guys triggered an important epiphany!