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.

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.

The trig functions all around the unit circle

In this picture, the angle in the first quadrant closes to the x-axis is labeled a°. It is the upper right hand point of the red rectangle.  (90 - a)° is also in the first quadrant and is the upper right hand corner of the blue rectangle. These two angles are complementary.  The other six angles are some "nice" angle away from a° or (90 - a)°, adding either 90°, 180° or 270°.  All the trig values of sine, cosine and tangent for the eight angles labeled here can be derived from the three values sina, cosa and tana.  We already seen the relationship between the trig values of complementary angles.

sin(90 - a)° = cosa
cos(90 - a)° = sina
tan(90 - a)° = 1/tana

When 90° is added, sine and cosine switch values and the cosine of the new angle is negated as follows.

sin(90 + a)° = cosa
cos(90 + a)° = -sina
tan(90 + a)° = -1/tana

90° more than the complementary angle is the same story with those values.


sin(180 - a)° = sina
cos(180 - a)° = -cosa
tan(180 - a)° = -tana

Adding 180° puts us at the antipode, the polar opposite of where we were.  Sine and cosine will just be negated, while tangent will remain exactly the same.


sin(180 + a)° = -sina
cos(180 + a)° = -cosa
tan(180 + a)° = tana

180° beyond the complement is a similar story.


sin(270 - a)° = -cosa
cos(270 - a)° = -sina
tan(270 - a)° = 1/tana

And then we have adding 270°. This is like adding 180° to the values we got when adding 90°.


sin(270 + a)° = -cosa
cos(270 + a)° = sina
tan(270 + a)° = -1/tana

270° more than the complementary angle is the same story with those values.


sin(360 - a)° = -sina
cos(360 - a)° = cosa
tan(360 - a)° = -tana

Example: 60° is one of our "famous" angles.

sin60° = sqrt(3)/2
cos60° = 1/2
tan60° = sqrt(3)

Find the following values using these three pieces of information.

a) sin150°
b) cos240°
c) tan330°
d) tan 120°
e) sin 210°
f) cos 300°

Answers in the comments.

The inverse trigonometric functions.

This week, we have been learning how to find all the trig functions of an angle if you only know sine, cosine or tangent.  Besides that, given any of these values we can find the angle itself by using the inverse trig function buttons.  On the TI-30X II s, the 2nd button is the blue button at the upper left hand corner.  We have already used it for square roots because the square root function is [2nd][x²].  The inverse trig functions of [sin], [cos], and [tan] are [2nd][sin], [2nd][cos] and [2nd][tan], respectively.

Almost none of the "nice" values of sine and cosine, the ones that are rational, give us "nice" angles.  The exception is ½. sin60° = cos30° = ½.

Here are some of the easy fractions for sine and the angles they represent, rounded to as close as the calculator will give them.

If sin alpha = 1/3, alpha  = 19.47122063°
If sin alpha = 2/3, alpha = 41.8103149°
If sin alpha =1/4, alpha = 14.47751219°
If sin alpha =3/4, alpha = 48.59037789°
If sin alpha =1/5, alpha = 11.53695903°
If sin alpha =2/5, alpha = 23.57817848°
If sin alpha =3/5, alpha = 36.86989765°

If sin alpha =4/5, alpha = 53.13010235°

Again, these are not exact.  The actual numbers are irrational.

One useful property of the inverse tangent law is that if we know tan alpha = x, then tan (90° - alpha) = 1/x.  This is to say if we have a line with slope 1/9 and we find the angle measure is beta, then the angle with slope 9 is (90° - beta).  In this particular case the approximate numbers are
tan 6.340191746° = 1/9 and
tan 83.659808254° = 9

Find the following angles for the given values of tangent, rounded to the nearest thousandth of a degree.  Answers in the comments.

a) If tan alpha = 2, alpha  =
b) If tan alpha = 3, alpha  =
c) If tan alpha = 4, alpha  =
d) If tan alpha = 5, alpha  =
e) If tan alpha = 6, alpha  =
f) If tan alpha = 7, alpha  =

Answers to lab 9/14

sin alpha = 1/10

(1/10)² + cos²alpha = 1
1/100 + cos²alpha = 1 
cos²alpha = 1 - 1/100 = (100-1)/100 = 99/100
sqrt(cos²alpha) = sqrt(99/100) = sqrt(99)/10 = 3sqrt(11)/10
cos alpha = 3sqrt(11)/10
tan alpha = sin/cos = 1/3sqrt(11) = sqrt(11)/33

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cos beta = 5/7


sin²beta + (5/7)² = 1

sin²beta + 25/49 = 1
sin²beta = 1 - 25/49 = (49-25)/49 = 24/49
sin beta = sqrt(24/49)

sin beta = sqrt(24)/7 = 2sqrt(6)/7
tan beta = sin/cos = 2sqrt(6)/5

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tan theta = 3
3²cos theta + cos² theta = 1
9cos² theta + cos² theta = 1
10cos² theta = 1

cos² theta = 1/10
cos theta = sqrt(1/10) = sqrt(10)/10
sin theta = tan theta cos theta = 3sqrt(10)/10


Don't worry if I didn't get around to correcting yours personally.  Everyone who was in class gets credit for the lab.




 
 

Given one trig value for an angle, how to find the other two.

 

We have learned The Trigonometric Identity, 

sin² theta + cos² theta = 1. 

It's a simple application of the Pythagorean Theorem. a² + b² = c² where c = 1. This means if either the sine or cosine of an angle is known, we can find the other value easily enough. Simply square the value you know and subtract that value from 1, then take the square root of the value. This is where our practice with square roots and fractions will come in handy.

Example: sin theta = 1/5

(1/5)² + cos² theta = 1

cos² theta = 1 - 1/25 = 24/25

Taking square roots, we get

cos theta = sqrt(24/25) = sqrt(24)/5 = 2sqrt(6)/5

And, of course, if we know sine and cosine, tangent = sine/cosine.

sin theta = 1/5, cos theta = 2sqrt(6)/5, so tangent = 1/2sqrt(6) or sqrt(6)/12.

sub Example: cos theta = 1/5

It's the exact same work, just finding sine instead of cosine.

sin² theta + (1/5)²   = 1

sin² theta = 1 - 1/25 = 24/25

Taking square roots, we get

sin theta = sqrt(24/25) = sqrt(24)/5 = 2sqrt(6)/5

cos theta = 1/5, sin theta = 2sqrt(6)/5, so tangent = 2sqrt(6).

  

Finding the sine and cosine when we know tangent is different and a little trickier since sin theta/cos theta = tan theta, multiplying by cosine on both sides gives us

sin theta = tan thetacos theta.

We make a substitution into The Trigonometric Identity, do some algebraic manipulation and we can find cosine.  Once we have cosine and tangent, we multiply them together to get sine.

Example: tan theta = 2.

This says that 2²cos² theta + cos² theta = 1

5cos² theta = 1

cos² theta = 1/5

cos theta = sqrt(1/5) = sqrt(5)/5.

multiply cos and tan and we get sin theta = 2sqrt(5)/5.


Here are some practice problems.  Answers are in the comments.


a) sin theta = 1/3

b) cos theta = 1/3

c) tan theta = 1/3

The "famous" trig values between 0 and 90 degrees inclusive

0°
sin 0° = 0
cos 0° = 1
tan 0° = sin0°/cos0° = 0

30°
sin 30° = ½
cos 30° = sqrt(3)/2
tan 30° = sin 30°/cos 30° =sqrt(3)/3

45°
sin 45° =sqrt(2)/2
cos 45° =sqrt(2)/2
tan 45° = sin 45°/cos 45° =1

60°
sin 60° =sqrt(3)/2
cos 60° = ½
tan 60° = sin 60°/cos 60° =sqrt(3)

90°
sin 90° =1
cos 90° =0
tan 90° = sin 90°/cos 90° = UNDEFINED (can't divide by zero)

In the first quadrant, we have these three situations.

As the angle increases, sine increases, from a low of 0 to a high of 1.

As the angle increases, cosine decreases, from a high of 1 to a low of 0.

As the angle increases, tangent increases, from a low of 0 to a high of infinity.

A little preview of coming attractions.  We are not always going to think of sine and cosine as values on the Normal Standard Position Right Triangle, but instead as the x and y values on the unit circle, defined by the formula x² + y² = 1.

When the hypotenuse equals 1.

We know that the trigonometric functions are defined as ratios in a right triangle as follows:

sine (sin) = opposite/hypotenuse

cosine (cos) = adjacent/hypotenuse

tangent (tan) = opposite/adjacent

Consider all the similar right triangles that include the acute angle we call alpha in this picture.  If we think about the triangle where the hypotenuse = 1, our formulas get much simpler.

sin alpha = opposite

cos alpha = adjacent

tan alpha = opposite/adjacent = sin alpha/cos alpha

The tangent of alpha ALWAYS equals sin alpha/cos alpha, but to make sine and cosine equal to the lengths of sides of a triangle, we need the hypotenuse equal to 1.  Here are some other true statements given our picture.

1. sin alpha is the height of this triangle.
2. cos alpha is the base of the triangle, here labeled x.
3. If we have the triangle with the perpendicular lines at the horizontal and vertical and the angle alpha is on the left side, tan alpha is the slope of the hypotenuse.

And here is a statement that is always true for any angle alpha, a statement known as The Trigonometric Identity.

sin² alpha + cos² alpha = 1.

This is a very straightforward application of the Pythagorean Theorem, since all it is says is that

a²/c² + b²/c² = 1.

This is true because when we combine the two fractions, we get

(a² + b²)/c², which is the same as c²/c² = 1.

I say this statement is The Trigonometric Identity.  Other sources will say there are several trigonometric identities.  There are others, but we will see that all of them are just rearrangements of this one.  Make sure you understand this.  Quite often when you do not know the next step in a trig problem, The Trigonometric Identity is the correct path to take.

Practice problems for square roots in the denominator of fractions

Change these fractions with fractions in the denominators and write them in simplest form. Also write them rounded to the nearest thousandth.

a) 1/sqrt(5)

b) 10/sqrt(2)

c) 6/sqrt(12)

d) 5/sqrt(5)

Answers in the comments.

Practice problems for basic trig functions.

In this picture of a right triangle, let's call the hypotenuse c, the short leg opposite angle alpha a, and b will be the long leg opposite beta.  (Another way to say this is a is the height [or altitude, if you prefer], and b is the base.)  Here are some possible values for the lengths.  Given these lengths, find the requested trig functions for the given angles.  Write the values as fractions in lowest terms.

Let a = 8, b = 15 and c = 17.

a) sin alpha = __________

b) tan beta = ___________

c) cos beta = ___________

d) tan alpha = __________

Let a = 7, b = 24 and c = 25.

e) cos alpha = __________

f) tan beta = ___________

g) cos beta = ___________

h) tan alpha = __________

Answers in the comments.

Right triangles and trigonometric functions

Consider the 3-4-5 triangle.  Because 3² + 4² = 5², this is a right triangle. Any triangle similar to this, which means that all the side lengths are increased or decreased by the same scale, is also a right triangle.

This means a 3-4-5 triangle is similar to a 6-8-10 triangle or a 30-40-50 or a .3-.4-.5 triangle. It doesn't matter if we measure these distances in inches or feet or miles or centimeters or meters or kilometers, as long as we are consistent.  In all of these triangles, corresponding angles have NOT changed.  There is the 90° angle and two angles we will call alpha and beta.  We don't currently know the measure of these two angles, but because all three angles have to add up to 180°, alpha + beta = 90°.  Any two angles that add up to 90° are called complementary.


The picture above shows us two ways to look at the right triangle, one from the perspective of the alpha angle and the other from the perspective of beta.  The hypotenuse doesn't change, but the legs are given the labels Opposite and Adjacent.  If we think of the horizontal line segment as the base and the vertical as the height, the horizontal length is Adjacent to alpha and the vertical length is Opposite alpha.

From the point of view of beta, the Opposite and Adjacent are switched, but the Hypotenuse doesn't change.

The three major trigonometric functions, sine, cosine and tangent (sin, cos, tan) can be defined as ratios between the three sides.  Remember that if the angles don't change but we change the lengths, they change by some uniform multiple, so the ratios between the side lengths will remain the same.  Here are the ratios.

Sine = Opposite/Hypotenuse
Cosine = Adjacent/Hypotenuse
Tangent = Opposite/Adjacent

The mnemonic Soh-Cah-Toa is regularly used in classes now to remember the three fractions that define the three functions.  It kind of sounds like something from a Native American language, so the story is that Soh-Cah-Toa was an important chief of some tribe, but that is completely made up.

Another important thing to remember is that if two angles are complementary, that corresponds to the letters co. 

The sine of alpha is the cosine of beta.
The cosine of alpha is the sine of beta.  

 We will later learn about three more trig functions, cotangent, secant and cosecant.  The rule about adding the prefix "co" or getting rid of it will remain the same.


The tangent of alpha is the cotangent of beta.
The cotangent of alpha is the tangent of beta.  

The secant of alpha is the cosecant of beta.
The cosecant of alpha is the secant of beta.  

Heron's formula practice problems.

Here are all the possible triangles with integer side lengths and perimeter = 12, which means s = 12/2 = 6.  Find the areas using Heron's formula, simplifying the square roots.

a) 6, 5, 1

b) 6, 4, 2

c) 6, 3, 3

d) 5, 5, 2

e) 5, 4, 3

f) 4, 4, 4

Answers in the comments.