To this point, we have only seen how the trigonometric functions relate to the geometry of circles. As the name suggests, however, there are important connections between trigonometric functions and measurements in triangles. In this section, we will develop five different formulas that relate different length and angle measurements in triangles. The first three formulas only apply to triangles in which one of the angles is a right angle (i.e., radian measure p/2), while the other two formulas can be applied to any triangle.
From High School Geometry, you may remember that the shape of a triangle is generally determined by three measurements. This is often expressed formally in terms of the standard Congruence Theorems: typically referred to as the Angle-Side-Angle Theorem, the Side-Angle-Side Theorem, and the Side-Side-Side Theorem. That is, once we know the measurements of two angles and a side, two sides and an angle, or three sides, the measurements of the remaining angles and sides in the triangle are determined. Using the formulas in this section, we will see how we can solve for those remaining measurements.
Say we have a right triangle with sides a, b, and c, with opposite angles A, B, and C, as in the following picture:
that is, C = p/2. Since the sum of all three angles in any triangle is always p, we know that p = A + B + C = A + B + p/2. In particular:
B = p - p/2 - A = p/2 - A.
That is, once we know one of the other angles in a right triangle, we actually know all three angles.
If we let t = A in the definition of sin and cos, we have the following picture:
where this triangle has all the same angles as our original triangle. This means that this triangle is similar to our original triangle, so that the ratios of corresponding sides are equal. This means, for example, that sin(A) = sin(t) = sin(t)/1 = a/c. Likewise, cos(A) = cos(t)/1 = b/c, and tan(A) = sin(t)/cos(t) = a/b.
This is our first group of formulas, which apply to right triangles:
If a triangle has the indicated length and angle
measurements:
where C = p/2, then the trigonometric
functions give the ratios of pairs of sides:
sin(A) = a/c, cos(A) = b/c, and
tan(A) = a/b.
Notice how all three formulas relate the angle A to the other three sides a, b, and c.
The side opposite angle C has been known as the hypotenuse of the right triangle, at least since the time of Pythagoras (around 540 BC). The side opposite angle A is naturally referred to as the "opposite" side, while the remaining side is commonly called the "adjacent" side (relative to angle A). With this terminology, we can remember these formulas by the mnemonics:
sin = opposite/hypotenuse,
cos = adjacent/hypotenuse,
tan = opposite/adjacent,
using the acronym, SohCahToa. Note: It is a common misconception that these formulas define the trigonometric functions, but these formulas only hold when 0 < A < p/2, while the trigonometric functions are defined for all values of A.
Since we already know one of the angles (i.e., C) in a right triangle, the triangle will be determined once we know either two other sides or one other side and another angle (cf. the discussion in the Introduction). We can use all the formulas:
A + B = p/2,
a2 + b2 = c2,
sin(A) = a/c,
cos(A) = b/c, and
tan(A) = a/b,
to solve for the remaining measurements in the triangle. For example, if we know that a = 5 in. and B = 1.2 radians, then we can conclude that:
A = p/2 - B = p/2
- 1.2 » 0.37 radians,
c = a/sin(A) » 5/sin(0.37)
» 13.83 in., and
b = a/tan(A) » 5/tan(0.37)
» 12.89 in.
You should notice that we could "flip" the triangle around so that the roles of angles A and B, as well sides a and b, in the formulas are reversed. This would allow us to us the exact value of the angle B, as opposed to the approximate value for A:
c = a/cos(B) = 5/cos(1.2) » 13.80
in., and
b = a tan(B) = 5tan(1.2) » 12.86
in.
This approach minimizes the accumulation of round-off error, and so gives slightly more accurate results.
Practice working with these formulas to determine the measurements in a right
triangle, by
completing the following Exercises.
While the formulas from the previous section are helpful in certain circumstances, most triangles do not contain a right angle, so we need some more powerful tools. There are two such formulas, known as the Law of Cosines and the Law of Sines. These will allow us to infer from (most) any choice of three measurements the remaining three measurements in any triangle.
The Law of Cosines can be solved using basic geometry in the complex plane and the definition of cos. However, we will need one basic fact relating the geometry of a complex number, z = x + y i with the trigonometric functions. We know that the distance from z to the origin is given by its absolute value |z|. We also know that multiplication by a real number, like 1/|z|, corresponds to a stretch/shrink around the origin. In particular, z and z/|z| will be on the same line through the origin, as in the following picture:
Since the point z/|z| is |z/|z|| = |z|/|z| = 1 unit away from the origin, it lies on the unit circle and is given by z/|z| = C(t) = et i = cos(t) + sin(t) i, where t is the angle of rotation (in radians) from the positive, horizontal axis to z:
Multiplying this equation through by |z|, show that:
We can express any complex number, z, in terms of its distance from the origin, |z|, and its angle of rotation, t, by z = |z|(cos(t) + sin(t) i).
Note: This is known as the polar form of the complex number, z.
Given any triangle ABC:
we can now obtain a formula for c in terms of a, b, and C, known as the Law of Cosines, using a little bit of complex algebra and geometry. First, we place vertex C at the origin in the complex plane, vertex A will land at the point, b, on the real axis, and vertex B will land at some point, z, as in the following picture:
However, we can see that z is a units from the origin, making an angle of C radians with the positive real axis; that is, in polar form, z = a(cos(C) + sin(C) i).
Next, we add this complex number with -b. Algebraically, this gives z + -b = z - b = a(cos(C) + sin(C) i) - b = (acos(C) - b) + asin(C) i. Geometrically, the parallelogram rule for complex addition, shows that the distance from this point to the origin is equal to c, as shown in the following picture:
In other words, . Using the formula for the absolute value gives:
Squaring both sides and simplifying gives the desired formula:
c2 = (acos(C) - b)2
+ (asin(C))2 = (acos(C))2
- 2(acos(C))b
+ b2
+ (asin(C))2
= a2cos2(C) -
2abcos(C)
+ b2 + a2sin2(C)
= a2(cos2(C) +
sin2(C)) + b2 -
2abcos(C)
= a2 + b2
- 2abcos(C)
where we use the Pythagorean identity, cos2(C) + sin2(C) = 1, at the last step.
This formula is called the Law of Cosines:
If a triangle has the indicated length and angle
measurements:
then c2 = a2 + b2 - 2abcos(C).
In words, this relates the length one side to its opposite angle, in terms of the lengths of the other two sides. By taking different choices of side and opposite angle, we also have a2 = b2 + c2 - 2bccos(A) and b2 = a2 + c2 - 2accos(B). You should observe that this looks like the Pythagorean Theorem (i.e., c2 = a2 + b2), but with a "correction" term (that is, -2abcos(C)) to handle the case when C is not a right angle. Note: When C is a right angle, cos(C) = cos(p/2) = 0, so that this includes the Pythagorean Theorem as a special case.
The second general formula, known the Law of Sines, follows from standard Euclidean geometry. The key idea is that any triangle may be inscribed in a circle. The center of the circle is the common point of intersection of the perpendicular bisections of the three sides of the triangle, as shown in the following picture:
We can express the radius of the circle, R, in terms of length of a side, a, and its opposite angle, A. To do this, we simply need to remember from High School geometry that:
the angle formed by any three points on a circle is the half of the measure of the arc subtended; in particular, it does not change when we move the vertex of the angle.
This means that, we have the following picture:
so that the new triangle drawn, with side equal to a diameter (i.e., with length 2R), has the same angle, A, opposite a common side of length a. But this new triangle contains a right angle; the angle opposite the diameter has measure p/2, since it subtends an arc which is half the circle (i.e., with measure p). Using the formulas for right triangles, we can say that sin(A) = a/2R, or 2R = a/sin(A).
You should stop and be amazed by this formula, since it holds for every pair of side and opposite angle! This means that:
2R
= a/sin(A),
and
2R = b/sin(B),
and
2R = c/sin(C).
This is usually stated as the Law of Sines:
If a triangle has the indicated length and angle
measurements:
then 2R = a/sin(A) = b/sin(B) = c/sin(C).
In words, the ratio of the length of a side and the sine of its opposite angle is a constant (i.e., the same for all three sides). Notice how this relates the length one side to its opposite angle, in terms of another length and opposite angle. From another perspective, the ratio of any two sides, is equal to the ratios of the opposite angles; for example, a/b = sin(A)/sin(B).
Taken together, the angle sum formula, the Law of Cosines, and the Law of Sines:
A + B
+ C = p,
c2 = a2 + b2 - 2abcos(C)
(and a2 = b2 + c2 - 2bccos(A)
and b2 = a2 + c2 - 2accos(B)),
and
a/b = sin(A)/sin(B) (and a/c = sin(A)/sin(C)
and b/c = sin(B)/sin(C)),
allow us to solve for the remaining measurements in the triangle, once we know either two sides and one angle, or two angles and one side, or three sides (cf. the discussion in the Introduction).
It is easy to determine when to apply each formula, by looking at what you know about each pair of side/opposite angle. For example, say we know that a = 5 in., b = 3 in. and B = 1.2 radians, as in the following picture:
Since we only know one angle, we cannot use the angle sum formula. We can hope to use one of the other two formulas to determine an equation to solve for A from a = 5. Since we do not know both of the other two sides, we cannot use the Law of Cosines. However, since we know both values, b = 3 in. and B = 0.6 radians, we can use the Law of Sines to deduce that:
5/3 = sin(A)/sin(0.6), so that A = sin-1(5/ 3 sin(0.6)) » 1.226.
Now we repeat the process, looking through each of the three formulas to see which can be applied. Since we now know two angles, we can use the angle sum formula to solve for the remaining angle:
C » p - 1.226 - 0.6 » 1.316 radians.
The only other unknown is c, with opposite angle, C » 1.316 radians. Since we know another side/opposite angle pair, the Law of Sines (say, c/ 3 » sin(1.316)/sin(0.6), so c » 3sin(1.316)/sin(0.6) » 5.14) would be the easier formula to use. However, we also know the two other sides, so we can also use the Law of Cosines:
c2 » 52 + 32 - 2·5·3cos(1.316) » 26.44,
Taking square roots again gives c » 5.14.
If you know two angles and need to find the third angle, use the Angle Sum formula.
If you know a side but not its opposite angle, or vice versa:
If you know the two other sides, you can use the Law of Cosines to solve for the unknown.
If you know both another side and its opposite angle, you can use the Law of Sines to solve for the unknown.
Practice working with these three formulas by completing
the following Exercises.
Go to Algebraic Functions
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