Deciding between The Law of Sines and The Law of Cosines.

To completely describe a triangle we should give all three side lengths and all three angles.  If we get the right three pieces of information, we can use either The Law of Sines or The Law of Cosines to find the other three.

Angle-Side-Angle or Side-Angle-Angle (ASA OR SAA):  If we have two angle measurements, the third angle is easily found by subtracting the sum of the two known angles from 180°.  We also have one side length and we know what angle is opposite that length so The Law of Sines is the easiest choice.

Example #1: beta = 30°, gamma = 70°, b = 10"

By subtraction, alpha = (180-30-70)° = 80°, so we can find a and c as follows.

10/sin 30° = a/sin 80°, which becomes sin 80°(10)/sin 30° = 19.69615506...  Let's round this to the nearest thousandth and say a ~= 19.696".


10/sin 30° = c/sin 70°, which becomes sin 70°(10)/sin 30° = 18.79385242...  Let's round this to the nearest thousandth and say c ~= 18.794".

Notice that alpha is the largest angle and a is the longest side.  This is always true that the order of the size of the angles will be the same as the order of the size of the opposite sides.  This helps to check your answers roughly to catch big mistakes like dividing when you should have multiplied or vice versa.
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Example #2: beta = 30°, gamma = 70°, a = 10 m


As before, alpha = 80°, so we can find b and c as follows.


b/sin 30° = 10/sin 80°, which becomes sin 30°(10)/sin 80° =  5.077133059... Let's round this to the nearest thousandth and say b ~= 5.077 m.

10/sin 80° = c/sin 70°, which becomes sin 70°(10)/sin 80° = 9.541888941...  Round this to the nearest thousandth and c ~= 9.542 m.
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Side-Angle-Side (SAS): Find the third side with The Law of Cosines, and then go to the SSS instructions.


Example #3: a = 5', c = 6', beta = 120°

b² =5² + 6² - 2(5)(6)cos120° = 25 + 36 - -30 = 91, so b = sqrt(91) ~= 9.539'
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Side-Side-Side (SSS): rearranging The Law of Cosines, we can get any of the cosines of the angles as combination of a, b and c.

Example #4: a = 5', b = sqrt(91)', c = 6'

We already know that beta is 120° from Example #3, but let's check to see how exact we get it using the Law of Cosines and the inverse cosine function.



cosbeta = (a² + c² - b²)/2ac = (25 + 36 - 91)/2(5)(6) = -30/60 = -1/2

The inverse cosine of -1/2 is exactly 120°.

Now let's find alpha

cosalpha = (b² + c² - a²)/2bc = (91 + 36 - 25)/2(6)sqrt(91) = 102/12sqrt(91) ~= .891042111...

On the calculator, we can [2nd][cos][2nd][(-)] and get 26.9955084.... which is to say alpha ~= 26.996°.

Instead of using The Law of Cosines to find gamma, let's just subtract our two angles from 180° and get gamma = (180-120-26.996)° = 33.004°.

If we use Law of Cosines, the answer is the same to the nearest thousandth of a degree.
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Example #5: a = 5', b = 9.539', c = 6'

We already know that beta is 120° from Example #3, but we are using the approximation of b instead of the exact number.  Let's see if it changes our answer.


cosbeta = (a² + c² - b²)/2ac = (25 + 36 - 9.539²)/2(5)(6) ~= -0.49987535...

The inverse cosine of the answer is 119.9917536...°.  Rounding this to the nearest thousandth of a degree gives us 119.992°, so we are off by a little bit.


Now let's find alpha

cosalpha = (b² + c² - a²)/2bc = (9.539² + 36 - 25)/2(6)(9.539) ~= .891013392...


On the calculator, we can [2nd][cos][2nd][(-)] and get 26.99913319.... which is to say alpha ~= 26.999°.

Instead of using The Law of Cosines to find gamma, subtract the two angles from 180° and get gamma = (180-119.992-26.999)° = 33.009°.

If we use Law of Cosines, the answer is the same to the nearest thousandth of a degree.
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Practice problems. Round inexact answers to a thousandth of a unit.

1) alpha = 75°, beta = 45°, c = 12"

Find a, b and gamma.

2) a = 7 m, b = 8 m, c = 9 m


Find alpha, beta and gamma.


3) alpha = 120°, b = 6', c = 4'

Find a, beta and gamma.

Answers in the comments.

Law of Cosines for third side length of a triangle, Law of Sines for area of the triangle

If we know the length of two sides of a triangle, call them a and b, and we have some way to find the angle between them, call it gamma, here is how we can use the Law of Cosines and the Law of Sines best. (Here we assume gamma is an angle between 0° and 180°.)

The third side length can be found using

c² = a² + b² - 2abcosgamma

and the area is given by

Area = ½absingamma

We need a and b as  numbers, but we can know gamma a few different ways.


1) gamma is given. Then we go to the calculator and find cosgamma and singamma.

2) cosgamma is given. We can find sine using The Trigonometric Identity. If we are asked for the measure of gamma, use the inverse cosine function.

3) singamma is given and the quadrant for gamma is given or we are told gamma is acute or obtuse.
We can find cosine using The Trigonometric Identity, where cosine will be positive if we are in the first quadrant (gamma is acute) or cosine will be negative if in the second quadrant (gamma is obtuse). If we are asked for the measure of gamma, use the inverse cosine function.


Example 1.

gamma = 135°, a = 6, b = 2.
sin135° = sqrt(2)/2, cos135° = -sqrt(2)/2

c² = 6² + 2² - 2(6)(2)(-sqrt(2)/2) = 40 + 12sqrt(2), so c = sqrt(40 + 12sqrt(2)) or approximately 7.5479 units.

Area =  ½(6)(2)sqrt(2)/2 = 3sqrt(2) or approximately 4.246 square units.


Example 2.

gamma = 45°, a = 6, b = 2.
sin45° = sqrt(2)/2, cos45° = sqrt(2)/2

c² = 6² + 2² - 2(6)(2)(sqrt(2)/2) = 40 - 12sqrt(2), so c = sqrt(40 - 12sqrt(2)) or approximately 4.7989 units.

Area =  ½(6)(2)sqrt(2)/2 = 3sqrt(2) or approximately 4.246 square units.

(Notice the area for the first two examples is the same.)


Example 3.

singamma = .2, gamma in first quadrant, a = 6, b = 2.

We can also write singamma as 1/5, so 1/5² + cos²gamma = 1
1/25 + cos²gamma = 1
cos²gamma = 24/25
cosgamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(2sqrt(6)/5) = 40 - 48sqrt(6)/5, so c = sqrt(40 - 48sqrt(6)/5) or approximately 4.0602 units.

Area =  ½(6)(2).2 = 1.2 square units exactly.





Example 4.

singamma = .2, gamma in second quadrant, a = 6, b = 2.


We can also write singamma as 1/5, so 1/5² + cos²gamma = 1
1/25 + cos²gamma = 1
cos²gamma = 24/25
cosgamma = -sqrt(24/25) = -2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(-2sqrt(6)/5) = 40 + 48sqrt(6)/5, so c = sqrt(40 + 48sqrt(6)/5) or approximately 7.9696 units.

Area =  ½(6)(2).2 = 1.2 square units exactly.
(Notice the area for examples 3 and 4 is the same.)

Example 5.

cosgamma = .2, a = 6, b = 2.

We can also write cosgamma as 1/5, so 1/5² + sin²gamma = 1
1/25 + sin²gamma = 1
sin²gamma = 24/25
singamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(.2) = 40 - 4.8, so c = sqrt(35.2) or approximately 5.9330 units.

Area =  ½(6)(2)2sqrt(6)/5 or approximately 5.8788 square units.


Example 6.

cosgamma = -.2, a = 6, b = 2.


We can also write cosgamma as -1/5, so (-1/5)² + sin²gamma = 1
1/25 + sin²gamma = 1
sin²gamma = 24/25
singamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(-.2) = 40 + 4.8, so c = sqrt(44.8) or approximately 6.6933 units.

Area =  ½(6)(2)2sqrt(6)/5 or approximately 5.8788 square units.

Yet again, the areas of examples 5 and 6 are the same.


Note: this was updated on Oct. 11 to correct errors in the lengths of the third side. Thanks to my student Fred Johnson for pointing out the errors.

The Law of Cosines

The Law of Sines is a relationship between side lengths of a triangle and the sine of the opposite angle. It is derived from information we get trying to find the height of a triangle relative to one of the sides being called the base.

The Law of Cosines is used to find a third side length when we have two side lengths and the measure of the angle that lies between them, referred to in geometry as SAS or side-angle-side.  At 90°, the cosine is zero and the extra term - 2abcosC becomes 0 and we get the Pythagorean Theorem.

Besides SAS, the Law of Cosines is useful if we know the all three sides (SSS) and want to find the measure of the angles.  By rearranging the variable with algebraic manipulation we get

cosA = (b² + c² - a²)/2bc
cosB = (a² + c² - b²)/2ac
cosC = (a² + b² - c²)/2ab  

The numerator might be 0 or less, but that is not a problem since cosine can take any value from -1 to 1, including 0 when the angle is 90°.  Since no side of a legitimate can be of length 0, we do not have to worry about the denominator being 0, so all the  possible examples we have of triangles will give use three cosine values, and we can use the inverse cosine function to find the three angles.

The Law of Sines and finding the area of a triangle given two side lengths and the angle between them.

 We have several different ways to get the area of a triangle.  The simplest is Area = ½bh,which assumes we know the length of one of the sides and the length of the perpendicular line segment that connects the opposite vertex to the known side length.  If we know all three side lengths, we have Heron's Formula, which we have discussed in previous blog posts and in class.  Let's assume instead we know two side lengths AND the measure of the angle that lies between them.  In the picture above, it is assumed we know the length of side b, and either we know angle A and side c, or we know angle C and side a.  In either case, the height is opposite the known angle in the right triangle formed by the dotted line.  Since sin = opp/hyp we get

sinA = height/c
or
sinC = height/a

This means we have two ways to represent this height,

height = csinA
or
height = asinC

What this gives us is two ways to represent the area that assumes b is the base

Area = ½basinC
or

Area = ½bcsinA

By rearranging the letters we have a similar formula with sin B


Area = ½acsinB

From here, the algebraic manipulation goes as follows.
 

In the pictures I found online the angles are labeled A, B and C.  In class, I usually use alpha, beta and gamma.  The math hasn't changed, just the labels.

An angle in a triangle has to be between 0° and 180°, which means in Quadrant I or Quadrant II, or possibly on the x or y axis.  The value of sine is greater than or equal to zero in that range of angles, so using this method to find the area will always give us a non-negative number.

Examples:  Let's say a = 3 and b = 4.  If we have the measure of angle C, we can use our formula to find the area.  Round the answer to the nearest thousandth.

a) Angle C = 0°
b) Angle C = 10°
c) Angle C = 20°
d) Angle C = 30°
e) Angle C = 40°
f) Angle C = 50°
g) Angle C = 60°
h) Angle C = 70°
i) Angle C = 80°
j) Angle C = 90°

Answers in the comments.

Notes for Homework #8, due Tues., March 19

Notes on the law of sines and how it differs from the law of cosines.

Notes for Homework 8, due March 20

Notes on Law of Sines and Law of Cosines