For this week of Pre-Calculus 11, I learned about two important ideas in Trigonometry, which relate to obtuse triangles, known as the Sine and Cosine laws.
Usually we only use sine, cosine, and tangent with right triangles, often accompanied with the following formulae:
- sinθ = y/r or opposite/hypotenuse
- cosθ = x/r or adjacent/hypotenuse
- tanθ = y/x or opposite/adjacent
But this time, we can actually tweak the formulae and add in extra numbers to find the angles and sides of triangles which are not right triangles (no 90 degree angles in the triangle).
The Sine Law indicates that one side of a triangle (a) divided by the sine of angle opposite of the side (A) is equal to another side (b) divided by the sine of its opposite angle (B), and so forth with side c and angle C. This can be represented as:
- a/sinA = b/sinB = c/sinC
- This formula would be often used to find the side of a triangle
If you wanted to find the angle of a triangle, this would be the formula, albeit reversed:
- sinA/a = sinB/b = sinC/c
- Note: when solving for an angle of a triangle, make sure to check if the triangle is an ambiguous case (I will go further into this later)
I’d also like to mention that, in Sine Law, if you wanted to only solve for one variable (b or C for example), you only need to use two of three terms in the formula, and not all three. You should be only dealing with one variable at a time, or else things would get really complicated.
- An example could be that you know what a (11 cm), A (70 degrees), and C (50 degrees) are, and you are solving for c
- The formula would be rearranged to fit all of these values, which means excluding the one term that has b and B since they aren’t needed
- ex: 11/sin70 = c/sin50
- The formula would be rearranged to fit all of these values, which means excluding the one term that has b and B since they aren’t needed
You might be thinking: what’s the point of any other laws if the Sine Law can help us find both the angles and the sides? Well, there are two important things you have to keep in mind with Sine Law when solving a triangle, which will either make or break the formula. Firstly, the triangle should have a minimum of 3 given values: either 1 side and 2 angles, or 2 sides and 1 angle. Most questions would already have given you 3 exact values for any side or angle, but do keep this in mind in case you come across any trick questions, if they exist.
Secondly, you need a minimum of one term in the formula without any variables. Which means, either a and A have an exact value, or b and B, or c and C.
- You could have a equal 8 cm, A equal 40 degrees, and c equal 12 cm, and the triangle would be completely solvable; a and A are in the same term
- However, if you only know the exact values for a, B, and c, you would need to use another formula
And yes, this “other formula” is seen in the Cosine Law. Here is what you would use if you wanted to find the side of a triangle:
- a^2 = b^2 + c^2 – 2bc cosA
- One could say this formula almost looks like Pythagorean Theorem, where a^2 + b^2 = c^2. This and the basic Trig ratios create the foundation of this formula, which are explained in much more detail if you search up the history of Cosine Law!
- Note: this formula assumes you know the exact value of sides b, c, and angle A. If you wanted to find the value of b or c instead, swap the variables.
- b^2 = a^2 + c^2 – 2ac cosB
- c^2 = a^2 + b^2 – 2ab cosC
- The opposite angle of the side you want to find (b, for example) will always replace the other variable being multiplied by cos.
Like the Sine Law, there’s also a different formula if you were looking for the angle instead:
- cosA = (b^2 + c^2 – a^2)/(2bc)
- Note: like with solving for an angle with Sine Law, you will need to check if the triangle is an ambiguous case
- cosB = (a^2 + c^2 – b^2)/(2ac)
- cosC = (a^2 + b^2 – c^2)/(2ab)
Note that you could also use these formulae to find a side or angle of a right triangle. Of course, it would be more work to use the Sine or Cosine law all the time, so choose wisely! I’ll explain how you can choose what formula to use, given the triangle or a few hints.
When checking if a triangle could be solved with Sine Law, Cosine Law, or just basic Trig, there are a few things we need to ask:
- Are there three given values for the triangle? Preferably, if the values are for 2 sides and 1 angle, or 2 angles and 1 side. If so, Sine and Cosine Laws, and basic Trig could solve this. If not, well, this wouldn’t be solvable, or very difficult to solve.
- Is it a right triangle? If so, we can use basic Trig ratios to solve this question. If not, this should be either solvable with Sine or Cosine Law.
- Is there a value of both an angle and the side opposite to it? For example: a and A values are given. If so, we can use Sine Law to solve the triangle or variables. If not, we are forced to use Cosine Law.
For our current triangle, we do have 3 given values so this is solvable and it’s not impossible or incredibly difficult to do. But we do not have a 90 degree angle in the triangle, since we don’t have a little box nor a number saying ’90 degrees’ in any of the corners. So, we can easily cross off basic Trig on our list.
Then, we can see if we know the value of an angle and the side opposite of it.
Well, we know the values of a and A, being 15 units and 121 degrees, respectively. So, this should be solvable with Sine Law.
Let’s put down all our values first, before condensing the formula.
We can see that the variable we are looking for, b, is in the same fraction as one of the angles we know, being B. The first fraction is also completely filled (no variables), so we can directly begin solving for b by just using these two terms. We can neglect the 3rd term with Cs, since they aren’t necessary.
The side length b, is equal to 8.7 units.
For this triangle, it seems like we need to find the value of c. And again, we need to decide which formula to use to get to our answer.
- Are there 3 given values? Yep.
- Does the triangle have any 90 degree angles? Certainly not – basic Trig shouldn’t be able to solve this.
- But do we know the value of any angle with a side length opposite of it? Yes, we do!
I can see that I know the exact values of a and A, being 7 and 44 degrees, respectively. This means that I can solve for c using the Sine Law.
Let’s put all our values into the original 3-term formula, to double-check.
But look at this! What makes this interesting, is that we cannot shorten this formula down without having more than 1 variable. In some cases with Sine Law, we will encounter trying to solve for a variable, but will force us to solve other variables in order to find the actual value of the variable the question is asking for.
Therefor, we need to find all the angles of the triangle, before finding the value for c. This will get rid of the variable C in the 3rd term, replacing it with a number.
If you’re solving for an angle, double-check if the angle could also be a Quadrant II angle, making the triangle an ambiguous case.
This just means whether the angle in question could be greater than 90 degrees. This often happens when we do not know what the triangle originally looked like (the question didn’t draw the triangle for you), and so there could be 2 possible answers in 2 different triangles. In this case, you’ll see that the triangle does not include another angle, because the triangle was already drawn for us from the start. We only needed to solve for c, and even then, all the angles don’t seem to be greater than 90 degrees.
If the question told me to draw the triangle myself, then there would be the possibility of having 2 answers. But for now, we can neglect the extra work.
- The question would often be worded like this: In △ABC, ∠A = 44 degrees, a = 7, and b = 9. Solve for c.
- This would allow us to draw our triangle in two different ways: either with ∠B being 63 or 117 degrees
Now, let’s find the angle for C. This would be easy, since we just have to subtract angle A and B from 180 degrees. All angles in a triangle add up to 180 degrees, so it would be common sense that angle C is the difference of the total and the combined amount of A and B.
Finally, we can solve for c, since the term with c and C only have one variable without a value.
c equals 9.6 units.
Let’s take a look at one final example.
In △ABC, ∠B = 94 degrees, a = 10, and c = 13. Solve for △ABC.
Firstly, we have to know what this question is asking for.
- It seems like the angle, B, is 94 degrees
- The side lengths a and c are 10 and 13 units, respectively
- We are solving for all unknown angles and sides of the triangle (solve △ABC is the indication)
Since this example doesn’t have any triangle drawn for us, we will need to draw it ourselves. A big hint for drawing triangles, is that you don’t really need to pay attention to what the triangle is supposed to look like. By just having a visual of the problem, it helps a ton. So, until you have all the information you need to correctly draw the triangle, just get something on paper for reference.
Here’s my triangle; I like starting these word problems out with a non-right angle triangle. Now I can just add the rest of the details we know.
…and here’s the complete reference triangle. Let’s begin by asking ourselves a few questions to figure out if we need to use basic Trig, Sine Law, or Cosine Law to find what we need. I like to start with any missing side lengths, first.
So,
- Are there 3 given values? Yes.
- Does the triangle have any 90 degree angles? Nope.
- But do we know the value of any angle with a side length opposite of it? This time around, no.
With these questions answered, we know that we can’t use basic Trig nor Sine Law to figure out what b is. This leaves us with using Cosine Law.
Let’s write down all the values we know in the formula for side lengths for Cosine Law.
Well, it seems like we can find b with this. All what’s left is the usual algebra.
b equals 17 units. But that is not all, since we need to find the other unknown values, namely the other two angles, to complete the question.
Let’s start with solving for angle A. Make sure to use the other formula used for solving angles, and check for ambiguous cases!
A equals 36 degrees.
Note that this triangle, regardless of how I would’ve drawn it, is not an ambiguous case. If I use the Quadrant II angle instead of the original angle I found (36 degrees), the QII angle added to angle B will be way higher than 180. This will not form a triangle at all, so I rejected the possibility of a 2nd answer.
All I have to find now, is angle C. Which is simple, like before when we had 2 angles. Just subtract angle A and B from 180 to get angle C!
C equals 50 degrees.
Let’s add our new values into the triangle I drew earlier.
Sine and Cosine Law definitely bring something to the table: a way to solve non-right angle triangles! While it does take much more work compared to using basic Trig ratios, there’s actually not much new in these laws. We are just extending our knowledge in basic Trigonometry and Pythagorean Theorem, changing it up a bit so it relates to these triangles.