# Find the Lengths of the Sides of a Triangle

Suppose you have a right triangle ABC, with angle C being the right angle. Thismeans the base of the triangle is a, the height is b and the hypotenuse is c.  You can find the length of a side of the right triangle if the lengths of the other 2 sides are given by using the Pythagorean Theorem, which states that a^2 + b^2 = c^2 (a^2 means a raised to the second power, or a squared).

For example, suppose a = 3 and b = 5. To find c, we take 3^2 + 5^2 = c^2.  9 + 25 = c^2, 34 = c^2.  By taking the square root of 34, we find that c is approximately 5.83.

If only 1 side of the right triangle is given but we know the measure of angle A or B, we can find the lengths of the other sides using trigonometry. The trigonometric function sine = opposite side divided by hypotenuse. Cosine = adjacent side divided by hypotenuse and tangent = opposite side divided by adjacent side.

For example, suppose in the same right triangle above, we know the measure of angle B = 35 and we know side a = 4.  We can find the other sides using the trigonometric functions above.  The side opposite angle B is b and the side adjacent to angle B is a with c being the hypotenuse.

Therefore we can use tangent B = opposite/adjacent = b/a.

Tangent 35 = b/4

4(tangent 35) = b.

Using a scientific calculator to solve for tangent 35, we get b = 2.8.  We can get c by using cosine b. Recall that cosine = adjacent/hypotenuse.

Therefore cosine B = a/c

Cosine 35 = 4/c

c(Cosine 35) = 4

c = 4/cosine 35.

Using a scientific calculator to solve for cosine 35, we get c = 4.88.

Suppose we have the same triangle ABC, but this time it is not a right triangle. We can still solve for the lengths of the sides of the triangle by using either the Law of Sines or the Law of Cosines, depending on the situation.

If you are given the length of side and two angles (SAA) or two angles and the side in between them (ASA), you can use the Law of Sines, which states:

If A, B and C are the measures of the angles of a triangle and a, b and c are the lengths of the sides, then SinA/a = SinB/b = SinC/c.

Suppose we know A = 46 degree, B = 53 degrees and c = 14 inches. We can find a and b as follows:

SinA/a = SinC/c

We know C = 180 – (46 + 53) since the sum of the angles of a triangle equals 180. Therefore, C = 81.

Sin46/a = Sin81/14

14Sin46 = a(Sin81)

a = 14Sin46/Sin81, which is approximately 10.2 inches.

To get b, we use

SinB/b = SinC/c

Sin53/b = Sin81/14

14Sin53 = b(Sin81)

b = 14Sin53/Sin81, which is approximately 11.3 inches.

If you are given two sides and the angle opposite one of the sides (SSA), then it’s tricky to determine the length of the other side because there might be two triangles, one triangle or no triangle. This is known as the “ambiguous case”.  The Law of Sines will determine the number of triangles and gives the solution for each.

For example, suppose A = 43, a = 90 and b = 55. We can find angle B by using

SinA/a = SinB/b

Sin43/90 = SinB/55

55Sin43 = 90SinB

SinB = 55Sin43/90

SinB is approximately 0.4167, which gives angles of 25 degrees and 155 degrees.  If B is 155 degrees, then we already have more than 180 degrees when adding this to angle A (155 + 43 = 198).  The only possibility then is for B to be 25 degrees. Therefore, angle C is 180 – (25 + 43) = 112. Now we can get c by using

SinA/a = SinC/c

Sin43/90 = Sin112/c

c(Sin43) = 90Sin112

c = 90Sin112/Sin43, which is approximately 122.

Here’s an example where the Law of Sines will show there is no solution.

Suppose A = 70 degrees, a = 55 and b = 74.

SinA/a = SinB/b

Sin70/55 = SinB/74

74Sin70 = 55SinB

SinB = 74Sin70/55

SinB = 1.26

The sine of an angle can never exceed 1, therefore there is no angle B for which SinB = 1.26.  The triangle with the given measurements in the problem does not exist.

A third case is there there are 2 possible triangles.  Suppose in triangle ABC, A = 38 degrees, a = 57 and b = 60.  Find angle B using SinA/a = SinB/b

Sin38/57 = SinB/60

60Sin38 = 57(SinB)

SinB = 60Sin38/57

SinB = 0.648

There are two possible angles for B for which SinB = 0.648, 40 degrees and 140 degrees. If you add either angle to angle A, you will not exceed 180 degrees. Therefore there are two triangles possible. In the first triangle, angle C = 180 – (40 + 37) = 103 and in the second triangle angle C = 180 – (140 + 37) = 3.

In the first possible triangle, we can solve for c using SinA/a = SinC/c

Sin38/57 = Sin103/c

cSin38 = 57Sin103

c = 57Sin103/Sin38, which is approximately 90.2.

In the second possible triangle,

Sin38/57 = Sin3/c

cSin38 = 57Sin3

c = 57Sin3/Sin38, which is approximately 4.8.

The Law of Cosines is used when given two sides and the included angle (SAS) or given the lengths of all 3 sides.  Since this article is about finding the lengths of the sides, we can exclude SSS and focus on SAS.

The Law of Cosines states that if A,B and C are the measures of angles of a right triangle and a, b and c are the sides opposite those angles, then a^2 = b^2 + c^2 – 2bcCosA,  b^2 =a^2 + c^2 – 2acCosb, c^2 = a^2 + b^2 – 2abCosc.

Suppose we have triangle ABC and we are given A = 65 degrees, b = 18 and c = 27.  We can find “a” by using the Law of Cosines.

a^2 = b^2 + c^2 – 2bcCosA

a^2 = 18^2 + 27^2 – 2(18)(27)Cos65

a^2 = 324 + 729 – 972Cos65

a^2 = 1053 – 411

a^2 = 642

a is approximately 25.3.

It’s easy to determine which of the 3 versions of the Law of Cosines to use. If you are solving for side a, use a^2 = b^2 + c^2 – 2bcCosA. If you are solving for side b, use b^2 = a^2 + c^2 – 2acCosB and if solving for side c, use c^2 = a^2 + b^2 – 2abCosC.

I hope this article showing how to find the sides of a triangle using the Pythagorean Theorem, Law of Sines and Law of Cosines is helpful.