The trigonometry of non-right triangles
So far, we've only dealt with right triangles, but trigonometry can be easily applied to non-right triangles because any non-right triangle can be divided by an altitude
Roll over or tap the triangle to see what that means →
Remember that an altitude is a line segment that has one endpoint at a vertex of a triangle intersects the opposite side (or an extension of it outside the triangle) at a right angle. See triangles.
Customary labeling of non-right triangles
This labeling scheme is commonly used for non-right triangles. Capital letters are angles and the corresponding lower-case letters go with the side opposite the angle: side a (with length of a units) is across from angle A (with a measure of A degrees or radians), and so on.
derivation
In mathematics, when we derive a formula, we basically invent it using one or more simpler, already-known principles. A derivation is essentially a proof that the formula we're creating works and can be relied upon.
Consider the triangle below. if we find the sines of angle $A$ and angle $C$ using their corresponding right triangles, we notice that they both contain the altitude, $x$.
The sine equations are
$$\text{sin}(A) = \frac{x}{c} \; \; \text{ and } \; \; \text{sin}(C) = \frac{x}{a}$$
We can rearrange those by solving each for $x$ (multiply by $c$ on both sides of the left equation, and by $a$ on both sides of the right):
$$x = c\cdot \text{sin}(A) \; \; \text{ and } \; \; x = a\cdot \text{sin}(C)$$
Now the transitive property says that if both $c\cdot \text{sin}(A)$ and $a\cdot \text{sin}(C)$ are equal to $x$, then they must be equal to each other:
$$c \, \text{sin}(A) = a \, \text{sin}(C)$$
We usually divide both sides by $ac$ to get the easy-to-remember expression of the law of sines:
$$\frac{\text{sin}(A)}{a} = \frac{\text{sin}(C)}{c}$$
We could do the same derivation with the other two altitudes, drawn from angles $A$ and $C$ to come up with similar relations for the other angle pairs. We call these together the law of sines. It's in the green box below.
The law of sines can be used to find the measure of an angle or a side of a non-right triangle if we know:
- two sides and an angle not between them or
- two angles and a side not between them.
Law of Sines
Although three fractions are related here with two equal signs, we only use them in pairs in practice.
Example 1
Find all of the missing measurements of this triangle:
$$180˚ - 76˚ - 34˚ = 70˚$$
Now set up one of the law of sines proportions and solve for the missing piece, in this case the length of the lower side:
$$ \begin{align} \frac{\text{sin}(76˚)}{x} &= \frac{\text{sin}(34˚)}{11} \\[5pt] x &= \frac{11 \cdot \text{sin}(76˚)}{\text{sin}(34˚)} \\[5pt] &= 19.1 \; \text{in.} \end{align}$$
Then do the same for the other missing side. It's best to use the original known angle and side so that round-off errors or mistakes don't add up.
$$ \begin{align} \frac{\text{sin}(70˚)}{y} &= \frac{\text{sin}(34˚)}{11} \\[5pt] y &= \frac{11 \cdot \text{sin}(70˚)}{\text{sin}(34˚)} \\[5pt] &= 18.5 \; \text{in.} \end{align}$$
Pro tip:
I've skipped a couple of steps in the algebra above. Remember that when the variable for which you're trying to solve is in a denominator, your first task is clear: Get it out of the denominator, usually by multiplying by that variable on both sides. Then the rest of the algebra to isolate that variable from everything else attached to it is fairly clear.
Example 2
Find all of the missing measurements of this triangle:
$$ \begin{align} \frac{\text{sin}(46˚)}{23} &= \frac{\text{sin}(Y˚)}{30} \\[5pt] \text{sin}(Y˚) &= \frac{30 \, \text{sin}(46˚)}{23} \\[5pt] Y &= \text{sin}^{-1}\left( \frac{30 \, \text{sin}(46˚)}{23} \right) \\[5pt] Y &= 69.8˚ \end{align}$$
Now it's easy to calculate the third angle:
$$180˚ - 46˚ - 69.8˚ = 64.2˚$$
Then apply the law of sines again for the missing side. We have two choices, we can solve
$$ \begin{align} \frac{\text{sin}(46˚)}{23} &= \frac{\text{sin}(64.2˚)}{x} \; \; \color{#E90F89}{\text{ or }}\\[5pt] \frac{\text{sin}(69.8˚)}{30} &= \frac{\text{sin}(64.2˚)}{x} \end{align}$$
Either gives the same answer,
$$x = 28.8 \; cm$$
Caution : Law of sines ambiguity
We have to be careful about the law of sines, because it can give ambiguous solutions when we use the inverse sine function $f(x) = \text{sin}^{-1}(x)$. Here's how it works. For a given triangle in which only one side is known, two possible triangles can be formed. Take a look:
The magenta sides mark out an obtuse triangle (small one on the left) and an acute triangle (outer triangle) using the same combination of two sides, where the third (bottom) side can be one of two lengths because it isn't initially known. The two possible angles are always related: their sum is 180˚.
How do we handle this? Let's do an example.
First, we solve for angle $A$ using the LOS:
$$\frac{\text{sin}(30˚)}{7} = \frac{\text{sin}(A)}{10}$$
We can rearrange and use the inverse sine function to get the angle:
$$A = \text{sin}^{-1}\left( \frac{10 \, \text{sin}(30˚)}{7} \right) = 45.6˚$$
Now if there is an ambiguity, its measure will be 180˚ minus the angle we determined:
$$180˚ - 45.6˚ = 134.4˚$$
Now if that angle, added to the original angle (30˚) is less than 180˚, such a triangle can exist, and we have an ambiguous case. Here are the two possible triangles in this example:
In such a case, we have to know a little more about our triangle. Is it acute or obtuse? Is a given diagram drawn to scale (as these are)? If so, that might be enough to resolve the ambiguity.
In some cases, the quantity (180˚ minus our calculated angle) added back to the calculated angle will be greater than 180˚, and such a triangle cannot exist, so the solution is unique.
Make sure to be aware of this as you work problems. Check for ambiguities unless you're clear about the results from other clues in the problem.
The law of cosines
The law of cosines is used when we know:
- the lengths of two sides of a triangle and the measure of the angle between them, OR
- the lengths of all three sides of a triangle, but no angle measures.
The law of cosines is computationally a little more complicated to use than the law of sines, but fortunately, it only needs to be used once. After the law of cosines is applied to a triangle, the resulting information will always make it possible to use the law of sines to calculate further properties of the triangle.
Consider another non-right triangle, labeled as shown with side lengths $x$ and $y$. We can derive a useful law containing only the cosine function.
First use the Pythagorean theorem to derive two equations for each of the right triangles:
$$ \begin{align} c^2 &= y^2 + x^2 \; \; \text{ and} \\[5pt] a^2 &= (b - y)^2 + x^2 \end{align}$$
Notice that each contains and $x^2$, so we can eliminate $x^2$ between the two using the transitive property:
$$c^2 - y^2 = a^2 - (b - y)^2$$
Then expand the binomial $(b - y)^2$ to get the equation below, and note that the $y^2$ cancel:
$$c^2 - y^2 = a^2 - b^2 + 2by - y^2$$
Now we still have a y hanging around, but we can get rid of it using the cosine solution, notice that
$$\text{cos}(A) = \frac{y}{c}, \; \text{ so } \; y = c \, \text{cos}(A)$$
Substituting $c \cdot \text{cos}(A)$ for $y$, we get
$$a^2 = b^2 + c^2 - 2 bc \, \text{cos}(A)$$
which is the law of cosines
The law of cosines can be used to find the measure of an angle or a side of a non-right triangle if we know:
- two sides and the angle between them or
- three sides and no angles.
We could again do the same derivation using the other two altitudes of our triangle, to yield three versions of the law of cosines for any triangle. They are listed in the box below.
Law of Cosines
The Law of Cosines is just the Pythagorean relationship with a correction factor, e.g. $-2bc\cdot \text{cos}(A)$, to account for the fact that the triangle is not a right triangle. We can write three versions of the LOC, one for every angle/opposite side pair:
Pro tip
You'll only need to use the law of cosines once if you need it at all. After finding the missing angle or side with the LOC, the law of sines will work for the rest of the sides or angles of your triangle.
Example 3
Find all of the missing measurements of this triangle:
$$ \begin{align} a^2 &= 8^2 + 4^2 - 2(8)(4) \, \text{cos}(51˚) \\[5pt] a^2 &= 39.72 \; m \\[5pt] a &= 6.3 \; m \end{align}$$
Now using the new side, find one of the missing angles using the law of sines:
$$ \begin{align} \frac{\text{sin}(51˚)}{6.3} &= \frac{\text{sin}(B)}{8} \\[5pt] sin(B) &= \frac{8 \, sin(51˚)}{6.3} \\[5pt] B &= \text{sin}^{-1} \left( \frac{8 \, \text{sin}(51˚)}{6.3} \right) \\[5pt] &= 80.7˚ \end{align}$$
And then the third angle is
$$180˚ - 51˚ - 80.7˚ = 48.3˚$$
In general, try to use the law of sines first. It's easier and less prone to errors (but be careful of the ambiguity when solving for non-acute angles). But in this case, that wasn't possible, so the law of cosines was necessary.
Example 4
Find all of the missing measurements of this triangle:
$$3.5^2 = 8^2 + 8.5^2 - 2(8)(8.5) \, \text{cos}(A)$$
Rearrange to solve for A. You'll need an inverse cosine to get the angle.
$$ \begin{align} \text{cos}(A) &= \frac{3.5^2 - 8^2 - 8.8^2}{-2(8)(8.5)} \\[5pt] A &= \text{cos}^{-1} \left( \frac{3.5^2 - 8^2 - 8.8^2}{-2(8)(8.5)} \right) \\[5pt] &= 24.2˚ \end{align}$$
Use the law of sines to find a second angle
$$ \begin{align} \frac{\text{sin}(24.2˚)}{3.5} &= \frac{\text{sin}(B)}{8} \\[5pt] \text{sin}(B) &= \frac{8 \, \text{sin}(24.2˚)}{3.5} \\[5pt] B &= \text{sin}^{-1} \left( \frac{8 \, \text{sin}(24.2˚)}{3.5} \right) \\[5pt] &= 69.5˚ \end{align}$$
Finally, just calculate the third angle:
$$180˚ - 24.2˚ - 29.9˚ = 125.9˚$$
Practice problems
Find the measures of all missing sides and angles of these triangles:
(Hint: Always use the LOS first if you can. It's simpler. When that fails, use the LOC, but then you can usually use other means to fill in the rest of the measurements using the LOS or the Pythagorean theorem.)
-
Solution
We can easily calculate the missing angle:
$$180^{\circ} - 70^{\circ} = 80^{\circ} = 30^{\circ}$$
Now we have one angle-opposite side pair, so the LOS will be fine:
$$ \begin{align} \frac{\text{sin}(30˚)}{10} &= \frac{\text{sin}(70˚)}{y} \\[5pt] y &= \frac{10 \, \text{sin}(70˚)}{\text{sin}(30˚)} \\[5pt] y &= 18.8 \, cm \end{align}$$
Now again for side x:
$$ \begin{align} \frac{\text{sin}(30˚)}{10} &= \frac{\text{sin}(80˚)}{x} \\[5pt] x &= \frac{10 \, \text{sin}(80˚)}{\text{sin}(30˚)} \\[5pt] y &= 19.7 \, cm \end{align}$$
We're not calculating angles with $\text{sin}^{-1}(\theta)$, so we don't have to worry about ambiguity in the LOS. Here's the triangle:
-
Solution
Angle A is
$$180^{\circ} - 2(70^{\circ}) = 40^{\circ}$$
Notice that this is an isosceles triangle, so the length of side $x$ also has to be 18 cm. Use the LOS to find y:
$$ \begin{align} \frac{\text{sin}(70^{\circ})}{18} &= \frac{\text{sin}(40^{\circ})}{y} \\[5pt] y &= \frac{18 \cdot \text{sin}(40^{\circ})}{\text{sin}(70^{\circ})} \\[5pt] y &= 12.3 \, \text{cm} \end{align}$$
... and that's it. Here's the fully-labeled triangle:
-
Solution
We have an angle-opposite side pair, so we can use the LOS to find angle A (because we know the length of a side, $a = 5$ cm.
$$ \begin{align} \frac{\text{sin}(95^{\circ})}{100} &= \frac{\text{sin}(A)}{5} \\[5pt] \text{sin}(A) &= \frac{5 \cdot \text{sin}(95^{\circ})}{100} \\[5pt] \text{sin}(A) &= 0.04981 \\[5pt] A &= \text{sin}^{-1}(0.04981) \\[5pt] A &= 2.855^{\circ} \end{align}$$
We need to check for the ambiguous case: Angle A could be 180˚-2.855˚ = 177.14˚, but this angle couldn't form a triangle with our 95˚ angle, so we're OK.
Then B = 180˚ - 2.855˚ - 95˚ = 82.14˚. Finally, we use the LOS to find $x$
$$ \begin{align} \frac{\text{sin}(95^{\circ})}{100} &= \frac{\text{sin}(82.14^{\circ})}{x} \\[5pt] x &= \frac{100 \cdot \text{sin}(82.14^{\circ})}{\text{sin}(95^{\circ})} \\[5pt] x &= 99.44 \end{align}$$
Here's everything we know about this triangle now:
-
Solution
In this case we have no angle-opposite side pairs, so we'll need to use the law of cosines (LOC). Let's call the side opposite the angle 152˚ side $c$.
$$ \begin{align} c^2 &= 10^2+12^2-2(10)(12)\text{cos}(152^{\circ}) \\[5pt] c^2 &= 455.91 \; \rightarrow \; c = 21.4 \; \text{cm} \end{align}$$
Notice that this length is greater than 10+12, so by the triangle inequality, this is a valid triangle.
Now we can use the LOS for a second angle (you could pick either):
$$ \begin{align} \frac{\text{sin}(152^{\circ})}{21.4} &= \frac{\text{sin}(B)}{10} \\[5pt] \text{sin}(B) &= \frac{10\cdot \text{sin}(152^{\circ})}{21.4} \\[5pt] \text{sin}(B) &= 0.21938 \; \rightarrow \; B = 12.7^{\circ} \end{align}$$
So angle A = 180˚ - 152˚ - 12.7˚ = 15.3˚. Here's the completed triangle:
-
Solution
The LOS will work fine here:
$$ \begin{align} \frac{\text{sin}(30^{\circ})}{6.1} &= \frac{\text{sin}(50^{\circ})}{x} \\[5pt] x &= \frac{6.1 \cdot \text{sin}(50^{\circ})}{\text{sin}(30^{\circ})} = 9.34 \; \text{cm} \end{align}$$
Then the other side is:
$$ \begin{align} \frac{\text{sin}(30^{\circ})}{6.1} &= \frac{\text{sin}(100^{\circ})}{y} \\[5pt] y &= \frac{6.1 \cdot \text{sin}(100^{\circ})}{\text{sin}(30^{\circ})} = 12.01 \; \text{cm} \end{align}$$
Here is the complete triangle:
-
Solution
Three sides and no angles means we need the LOC first. Let's solve for angle B:
$$ \begin{align} b^2 &= a^2 + c^2 - 2ac \, \text{cos}(B) \\[5pt] \text{cos}(B) &= \frac{b^2-a^2-c^2}{-2ac} \\[5pt] \text{cos}(B) &= \frac{6^2-12^2-11.75^2}{-2(12)(11.75)} = 0.87256 \\[5pt] B &= \text{cos}^{-1}(0.87256) = 29.24^{\circ} \end{align}$$
Now we can use the LOS to find A (or C), then the other angle by subtraction:
$$ \begin{align} \frac{\text{sin}(29.24^{\circ})}{6} &= \frac{\text{sin}(A)}{12} \\[5pt] \text{sin}(A) &= \frac{12 \cdot \text{sin}(29.24^{\circ})}{6} \\[5pt] A &= \text{sin}^{-1}(0.9977) \\[5pt] A &= 77.68^{\circ} \end{align}$$
Then C = 180˚ - 77.68˚ - 29.24˚ = 73.08˚. Here's the full triangle:
Problem 2 solution
Problem 3 solution
Problem 4 solution
Problem 5 solution
Problem 6 solution
Video examples
1. Derivation of the law of sines
The law of sines is an important relationship between the angle measures and side lengths of non-right triangles. In this video you'll see where it comes from.
Minutes of your life: 2:04
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