Oblique triangle: a triangle without a right angle. It can be made up of three acute angles or one obtuse angle and two acute angles.

You have three angles (A,B,C) and opposite of each angle there is a side length (a,b,c). In some problems you are only given a couple side lengths and an angle (SSA), two angles and a side (AAS), or two angles with a side length in between (ASA). If this is the case than you can use the Law of Sines. The Law of Sines sets up a ratio between the three angles and their opposite side lengths.

Law of Sines:

$SinA/a=SinB/b=SinC/c$

SSA: the Ambiguous Case
This case involves two triangles that aren’t congruent. There is a possibility of being no triangles, one triangle, or two triangles when you are given only two sides and an angle.
You usually get zero triangles when sin-1(x>1) this means that the number in the parenthesis is greater than one and the opposite leg is larger than the hypotenuse. This can’t be true since the hypotenuse is always the longest leg.
For two triangles, you have to test the triangle in all the quadrants that apply. If the triangle is 180° when in all quadrants than that means there are two triangles. So first, use the reference angle to find a second solution in the second quadrant. Then subtract both the given angle and the angle you found from 180° in order to test see if it is correct. Do this for both angles you found. If there are leftovers after solving with both angles than you have two triangles.
For one triangle, follow the same steps as you would in order to find two triangles. The only difference is that only one of the angles calculated will work.

Example 1: 2 Triangles
Angle A= 40
Side a= 5
Side b= 7

First find what the measure of angle B is by setting up the sine ratio.
$sin40/5 = sinB/7$

Isolate SinB so you can find the measure of angle B.
$7sin40/5 = sinB$

$sinB = .8999$

$sin-1(.8999) = B$

In order to find “B” we need to take the reciprocal of sine.
$B’ = 64.145°$

Plug the expression above into your calculator in order to find B’, make sure the mode is degrees and not radians.

$Ⅰ = 64.145° Ⅱ = 115.855°$ To find quadrant Ⅱ you need to subtract 64.145° from 180°.

In the end, you should have two triangles. One that includes 64.145° and another that includes 115.855°. The triangles will look different since 115.855° is obtuse and 64.145° is acute.

(1)
\begin{equation} (64.145°) (115.855°) \end{equation}

Example 2: 1 Triangle
Angle A= 40
Side a= 3
Side b= 2

Again we are finding the measure of angle B by setting up the sine ratio.
$sin40/3 = sinB/2$

Isolate SinB so you can find the measure of angle B.
$2sin40/3 = sinB$

$sinB = .4285$

Take the reciprocal of sine in order to find the measure of angle B.
$sin-1(.4285) = B$

Plug expression above into your calculator in order to find the measure of B’.
$B = 25.4°$

$Ⅰ = 25.4° Ⅱ = 154.6°$

You test each angle measure to see which one works.
$180°-40°-25.4° = 114.6°$

Only the angle from quadrant Ⅰ generates a triangle.
$180°-40°-154.6° = -14.6°$

Example 3: 0 Triangles
Angle C= 50
Side a= 2
Side c= 1

Set up the sine ratio to find the measure of angle A.
$sin50°/1=sinA/2$

Plug the left side into the calculator to find sinA.
$2sin50°/1=sinA$

$1.5321=sinA$

Find the inverse of sin in order to find A.
$sin^^-1^^(1.5321)=A$

A should equal error because sin can't be greater than one. The hypotenuse has to be the longest side of the triangle and if sin is greater than one, that means that the hypotenuse is shorter than the opposite leg.
$A=error$

Law of Cosines:

Side lengths-
$a^^2^^=b^^2^^+c^^2^^-2bc*cosA$
$b^^2^^=a^^2^^+c^^2^^-2ac*cosB$
$c^^2^^=a^^2^^+b^^2^^-2ab*cosC$

Angle Measures-
$A=cos^^-1^^((a^^2^^-b^^2^^-c^^2^^)/-2bc)$
$B=cos^^-1^^((b^^2^^-a^^2^^-c^^2^^)/-2ac)$
$C=cos^^-1^^((c^^2^^-a^^2^^-b^^2^^)/-2ab)$

You can use these equations to solve all parts of different triangles or the area of a triangle. It works best when you are given all three sides of the triangles or two sides with an angle measure in between. You can use these given measurements to find the missing values and solve the triangle.

Area of Triangles:

$Area=(1/2)bh$

SAS (Side-Angle-Side)-
$Area=(1/2)bc*sinA$
$Area=(1/2)ac*sinB$
$Area=(1/2)ab*sinC$

Use this when you need to find the area and are given two side lengths with an angle measure between them. The angle measure has to be between the side lengths in order for this to work.

SSS (Side-Side-Side)-
Heron's Formula
$Area^^2^^=s(s-a)(s-b)(s-c)$
$S=(1/2)(a+b+c)$

Use this when you have all three side lengths and need to find the area.