What's up with SSA triangles? Why is there more than one possible solution?

Question

How do you know if there is one, two, or no possible solution for a side-side-angle triangle? Example: Angle A= 30, side a= 6, side b= 11.

Answer 1

It would be easier if I could draw a picture...

...okay, done. Check the link below.

Notice how the 6 side can swing two ways to make a 6-11-30º triangle. Because the angle is acute (30º), you are left with a situation where the side can go a couple different ways, thus the triangle is not fully specified with an S-S-A arrangement.

Now if the angle was 90º, then there would be 2 choices for creating the triangle, but both those would be the same. And if the angle was obtuse, then the side would only have one way to go. But with an acute angle, as shown in the picture, you have a couple arrangements that create a triangle.

It's best to just try drawing the triangle. Start with the angle. The baseline you can just represent as a dotted line going on for a while in both directions. Then extended out the side (e.g. 11 in this case). Now take your protractor and draw the last side. It will either intersect the baseline 0 times (side is too short), 1 time (side hits the baseline perfectly at a 90º angle), or 2 times (side is longer than that).

Answer 2

We always used the letters in the opposite order, which was how we remembered that they weren't conclusive, since we could be sure that that word wasn't in the textbook! Anyway, I'm assuming side a is opposite angle A and side b connects vertex A to side a, since that's the standard and you didn't say otherwise. Imagine that the side without a given length is horizontal, with angle A on the far left. The procedure is different for is A is acute or obtuse.

I first consider cases where A is acute. If side a is equal in length or longer than side b, there is only one solution, where side a extends further out to the right before hitting the horizontal. If side a went back towards the left, it would overshoot angle A and the given angle wouldn't even be inside the triangle. On the other hand, if side a is shorter than b*sin(A), it is shorter than the height of the triangle, and the given angle can't be formed; there is no solution. For a = b*sin(A), there is one solution, a right triangle; side a extends directly downwards. For values between b*sin(A) and b, there are two solutions: one where side a extends down and to the left, and one where it extends down and to the right. That includes your example, because it has a height of 11*sin(30) = 5.5, so the 6 length side can reach down on either side of the vertical.

If angle A is right or obtuse, it's all a little easier. There is no solution if side a is equal to or shorter than side b, and only one solution if a is longer than b. There can never be two solutions when the given angle is greater than or equal to 90 degrees.

Answer 3

Put side a and side b together to form an angle theta. Let angle A be formed by side a and side x, the third side of the triangle. Since theta can be an value without effecting the length of side a and side be and the measure of angle A, side x will also be any length depending on the measure of theta. Thus, there are infinitely many triangle that can be made with side-side-angle.

Answer 4

Thats the nature of trig values.

Graph a sine, cosine or tangent function; they all have the potential for positive and negative values - in SSA triangles it's possible that you are not actually calculating the inside angle, but the complementary angle.

Answer 5

experience approximately it in terms of geometry. if your stick is one meter long, and your cuts are the two 1cm from the ends, then the two short parts won't be long adequate to stretch the area to attach and close the triangle. The minimum length, with 2 acute angles drawing close the minimize of 0 levels and the obtuse perspective coming near a hundred and eighty levels (minimum field enclosed) may well be a decrease at 0.25 m and nil.25 m from the ends assuming a 1m finished stick length. So the two cuts would desire to equivalent a minimum of 0.5m whilst introduced jointly, although they are waiting to be longer. although as you're making them longer you will desire to save interior the previous suggested bounds. once you're incredibly arbitrary interior the region you decrease than your possibility of falling under this may well be a million/2 of. This assumes which you will desire to positioned each and every bit end to end, with no wasted stick at everywhere. in case you do not care approximately waste then the possibility authentic is a hundred%. Addition- this question grew to become into as quickly as actually bugging me as i grew to become into as quickly as attempting to sleep very final night, so I did some study this morning. I have been given here up with limitless web content that one and all exhibit the authentic answer to be 25%. So, in spite of what diverse SAT and GRE counsellors have informed me, my first instinct grew to become into as quickly as fallacious. it fairly is a repost of a surprising rationalization: to ascertain that the three parts to sort a triangle, all parts would desire to be shorter than a million/2 the scale of the unique stick. In a triangle, the sum of the lengths of any 2 sides is often longer than the scale of the relax component. If slightly finally ends up being longer than a million/2 the stick's length, the sum of the choice 2 parts may well be shorter and you are going to now not be effectual to sort a triangle. to dodge having slightly it incredibly is a minimum of a million/2 the scale of the stick, the two cuts would desire to be on opposite components of the middle component. the possibility of this happening is ½ (the two each and every cuts substitute into on the comparable a million/2 or not). Now anticipate that the often happening decrease is made on component p (the region p is a p.c. of the entire length of the stick). besides, the 2d minimize have have been given to fall under the p + ½ component.the possibility for that's of direction ½ . Now multiply the two possibilities and likewise you get the suitable respond: ½ &cases; ½ = ¼

Answer 6

Try using the law of sines.

6/sin(30) = 11/sin(B) , solve for B , B is 66.4 degrees

The remaining angle is 180 - 30 - 66.4 = 83.6 degrees

What is the length of the remaining side?

6/sin(30) = C/sin(83.6), solve for C, C = 11.9

Now use law of cosines to see if this agrees, there are 3 ways to use the law, so pick any one of them.

Maybe I rounded too much, as the length of C is I calculate using the law of cosines is 11.5 when it should be 11.9.