MATH 1113 PreCalculus

Day #32, Arbitrary Triangles

Someplace back in a geometry class, you may have learned the different ways of proving that two arbitrary triangles are congruent (have same angle measures and same side lengths -- meaning that you could pick one of them up, cut it out, flip it, rotate it, whatever, and set it exactly on top of the other one.)

From our point of view, we want to consider when we have enough information about a triangle to deduce all of the side lengths and all of the angle measures.

With right triangles, we only needed two sides, or one side and one of the acute angles.  With arbitrary triangles, we need at least one more piece of information.

We could know:

S-S-S (know all 3 sides)
S-A-S (two sides and the angle between them)
A-S-A (two angles, and the side between them)
A-A-S (two angles and the side not between them)

and any of these gives enough information to uniquely determine the triangle (or determine that no triangle is possible.)

There's also a special case, called the "ambiguous case:"

S-S-A (two sides and the angle that's NOT between them)

where it turns out that there may be one unique triangle, or no triangles, or two possible triangles.

If all you know are the angles, then there are an infinite number of possible triangles (all similar to each other.)

The keys for solving the triangles are the two new laws:

The Law of Sines and

The Law of Cosines

Next time we'll look at the law of cosines, which I always remember as "c² = a² + b² -- almost."  But this time, we'll start with the Law of Sines.

For any triangle, the ratio of the length of any side to the sine of the angle opposite remains invariant for the triangle.

Here's a demonstration:

Geogebra Applet:

So, the Law of Sines will let you solve a triangle, any time you know an angle along with the opposite side, as well as one other side or angle.

Example:  Solve the triangle (A-A-S):

Angle A = 40^o, Angle C = 70^o, side c = 15 cm

Here's a Geogebra applet that looks at the A-A-S situation.  See if you can find an approximate answer first by using the sliders and getting a triangle like you want.

Geogebra Applet (First adjust the sliders and the side length, then slide point "C" until a triangle is formed, then read the values):

How do you solve this, using the Law of Sines?

First, we can find the other angle:  B = 180 - (40+70) = 70^o.  (Note that, in this special case, we automatically know that Side b is also 15, since it's an isosceles triangle, but usually that won't work.)

Now find the invariant for the triangle:  c/sin(C) = 15.9627.

Then, a/sin(40) = 15.9627, so a = sin(40)*15.9627 = 10.26

and b/sin(70) = 15.9627, so b = sin(70)*15.9627 = 15 (which we already knew, this time.)

Example: Solve the triangle (A-S-A)

Suppose angle A = 27^o, side c = 22.3 and angle B = 61^o.  Find the rest of the values.

Here's the A-S-A Geogebra applet:

Applet

Notice that this is about as close as we can get to 22.3 on the applet.

To solve it using trigonometry, first note that Angle C = 180 - (27+61) = 92^o.

Then form 22.3/sin(92) = 22.3136

side a = sin(27^o)*22.3136 = 10.13

side b = sin(61^o)*22.3136 = 19.52

The only way (in the above two cases) where you may fail to have a triangle is when the two given angles add up to more than 180^o.

Page Last Modified: 25 March, 2007

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