I was trying to make some triangles when $2$ sides and the non-included angle was given. When the angle given was below $90$ degrees, there were two possible triangles which could be made. But, when it went above $90$ degrees, only one possibility came. I was using a compass to determine the third side's possibilities. I checked the internet if there were questions like this, but there were none. Also, no site said specifically that it works for $90+$ angles.
Does this mean that the SSA congruence can work for angles greater than $90$ degrees? If so, I was taught very wrong in school.
Does SSA congruence criterion work if the non-included angle is obtuse
geometrytriangles
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I think there are a few points to be made here. First of all, let me clarify that you are correct, but perhaps not directly.
You are given a triangle with two base angles identical. That means the triangle is isosceles. You are also given a median of the triangle, which means that the median partitions the isosceles triangle into two right triangles. The subtlety here is that the above facts are not freely given. They require proof. If you have not shown that your median intersects your base at a right angle, then you cannot just assume the triangles are congruent.
Here is a picture with two triangles, $\triangle ABD$ and $\triangle ABE$. They share two sides of the same length, namely $\overline{AB} = \overline{AB}$ and $\overline{EB} = \overline{DB}$. They also share angle $\angle DAB = \angle EAB$. The triangles are not congruent.
There is sometimes a danger in geometry where too much is assumed from a given picture. Certainly pictures are indispensable in geometry, but always keep in mind that they are merely guides. This becomes especially true in higher geometry, where it's no longer possible (or feasible) to accurately illustrate all aspects of the problem (for example, inverse geometry or hyperbolic geometry becomes a little more difficult to picture).
When you solved your problem, you implicitly made use of side-angle-side congruence (note the angle has to be in between the two sides you use) using the base, the right angle and the shared median. That is perhaps the more natural congruence to use, but if you have not shown that the two triangles are indeed both right triangles, then you have skipped steps.
Use the Cosine Law.
Let $\triangle ABC$ have sides $a$, $b$, and $c$. We are using the usual convention that the length of the side opposite vertex $A$ is called $a$, and so on.
Let $\theta=\angle C$. Then the Cosine Law says that $$c^2=a^2+b^2-2ab\cos \theta.$$ Since we know $a$, $b$, and $c$, we can use the above formula to calculate $\cos\theta$. Then we can use the $\cos^{-1}$ button on the calculator to find $\theta$ to excellent accuracy.
We can use the Cosine Law three times to get the three angles. But we only need to do the calculation for two of the angles: If we have them, the third can be easily found.
Best Answer
Here is a trigonometric proof. Say $\triangle ABC$ has an angle at $A$ with a measure of $x$, and we know side $AC$ has length $b$ and side $BC$ has length $a$. We now want to find the side length of $AB$, or $c$. Using the law of cosines, $$b^2 + c^2 -2bc(\cos(x)) = a^2,$$ and rearranging gives us a quadratic in $c$: $$c^2 - 2b(\cos(x))(c) + (b^2- a^2) = 0.$$ Using the quadratic formula, we get c = 2acos(x) +/- sqrt(4a^2cos^2(x)-4(b^2-a^2))/2. Taking the 4s out of the square root and dividing, we get c = acos(x) +/- sqrt(a^2cos^2(x) - (b^2-a^2)) We know that if x > 90, or triangle ABC is obtuse, then cos (x) must be negative. Therefore, when triangle ABC is obtuse, we have that c is a negative number plus or minus a positive number. (Square roots are always positive. ) This means that only the positive sign works, since the negative sign means a negative minus a positive, or a negative, and side lengths can't be negative, so there's only one solution. You're right! It does work. Following this logic, we know that there is only one solution when sqrt(a^2cos^2(x) - (b^2-a^2)) > acos(x), since then the negative sign is invalid. Doing a bunch of algebra yields b < a, which is the general formula for when SSA works.