# Using the Law of Sines to find all triangles with given values of two sides and an angle

Our teacher skimmed over this and we have homework over it. Textbook is mostly unhelpful. I'm confused on how ambiguous case works, and everything I see online just confuses me more. I'm not quite sure WHEN and WHY you know there are 2 triangles, and then I'm not sure WHICH angle you use to figure that part out. (I vaguely understand that you subtract one of the angles you have from 180 to find the complement, but that's about it.)

Current question is: Use the Law of Sines to solve for all possible triangles that satisfy the given conditions.

$$a = 73, \quad b = 100, \quad \angle A = 26^\circ$$

I'm not necessarily looking for a spoonfed answer, I'm more curious as to where I go from here and why I do that. And, like I said, how to know when there are 2 triangles and WHICH of the angles you subtract from 180 to find the complementary angle if you know there are 2 triangles.

## 4 Answers

Hint: Sides $c$ and $a$ are given, and angle $A$. Notice that there are two possible triangles with side length $a$, and angle $C'$ is the supplement of $C$. Hopes this helps.

• I believe you mean the angles $\ C \$ and $\ C' \$ are supplementary. That's why the Law of Sines can give two results... – colormegone Jan 28 '14 at 2:13

If you have some paper, I would DEFINITELY draw a generic case triangle for this.

The question is asking you to prove that the side-angle representation of the triangle (like SAS or SSA (don't reverse the order of letters, please)) case for known values of angles and sides on a triangle is valid to prove one triangle, except as you said sometimes there are two triangles and sometimes there is no way to prove which triangle you have because there is not enough information (ie infinite triangles). This would not be one of those cases, as it would appear from what the question is asking.

To put this mumbo jumbo in parameters of the problem: This is the SSA (side,side,angle) case for a known triangle. Side one a=73, side two b=100, and angle A=26 degrees. The SSA case is solvable and is one triangle or two triangles. I believe this is a two case, and I think as there will always be two as long as a=b is false. In other words, only one triangle when it is isosceles, but two scalene triangles otherwise. Check geometrically by fixing side a horizantally and swinging side b. Recall that angle A is fixed, so you will see that side b touches in two places when you make a mental arc with it. The two places are points on side c, which is of unknown length, and we can swing side b because Angle C is unknown.

The SSA case uses the law of sines as you mentioned. You have to find all the angles first, the first with law of sines and the second with the triangle sum theorem (i think I just made up that name). That is, angle C=180-A-B. You don't want a spoonfed answer so I'll stop here. Make sure you start with law of sines!

EDIT: I haven't solved a problem like this in a while, but basically just prove it in the case for both triangles. I always think of this kind of problem like some kind of real-world linkage, where you can rotate the sides that don't hae fixed angles at either end, ie side c.

EDIT: If you're really stumped, try http://www.mathsisfun.com/algebra/trig-solving-ssa-triangles.html It's not the same problem but it is the same solving process.

EDIT: If you are given $a,b,$ and $B$ in $\triangle ABC$ (using the standard trig naming conventions) you can use the Sine Law to obtain at least one solution $A_1$ for $A$ where $A_1=\sin^{-1}(\frac{a \cdot \sin B}{b})$.

If $A_1\ne 90^\circ$ and $b\lt a$, then there exists another solution $A_2=180^\circ - A_1$.

Referring to qwr's figure, the SSA triangle ("ambiguous case") can be resolved using an "altitude test". The green segment is the altitude of the triangle(s) $\ ABC \$ because it is perpendicular to the unknown base $\ b \$ . The trig we've had for right triangles tells us that $\ h \ = \ c \ \sin A \ ,$ for which we have been given values.

We now compare our "loose side" $\ a \ ,$ which can be thought of as if it were on a "hinge" at point $\ B \ ,$ to this altitude:

If $\ a \ = \ h \ ,$ then that side is the altitude of the triangle, making $\ ABC \$ a right triangle; we can then use the trigonometry of right triangles to finish solving for the remaining unknown angle $\ B \$ and the base $\ b \ ;$

if $\ a \ < \ h \ ,$ then it is simply too short to reach the "base" of this figure, and thus cannot form a closed triangle: in this case, there is no triangle to be solved;

if $\ h \ < \ a \ < \ c \ ,$ then the side $\ a \$ is more than long enough to reach the base and form a closed triangle; however, there are two possible places where it could contact the base (at $\ C \$ and $\ C' \$ ) in qwr's diagram; we know $\ a \$ and $\ \sin A \ ,$ so we can use the Law of Sines to work out angle $\ C \$ , since we know the length of $\ c \$ ; the trouble is that there are two possible angles $\ C \$ that both have the same value of $\ \sin C \ ,$ one in the first quadrant, the other in the second; in this case, both solutions are valid and we have two possible triangles, one with side $\ a \$ extended away from side $\ c \$ (making $\ C \$ an acute angle), the other where it is "tucked under" side $\ c \$ (for which $\ C \$ is an obtuse angle) ;

finally, if $\ c \ < \ a \ ,$ then side $\ a \$ is again quite long enough to reach the base, but too long to "tuck under" side $\ c \$ ; in this case, we can only have the solution for angle $\ C \$ where it is acute (side $\ a \$ being "extended" away from side $\ c \$ ) .

Believe me, this is the hardest portion of learning "solutions of triangles" (having taught trig a number of times); by comparison, all the other triangle types are fairly simple to work with.

For the problem you show, $\ h \ = \ 100 \ \sin 26º \ \approx \ 43.8 \ .$ Side $\ a \$ is longer than that, but shorter than side $\ b \ ,$ so this is a "two-triangle" case, with two values for angle $\ B \$ . I take it you can proceed from there...