Practice with degrees and radians.

Here is a link to a blog post explaining degrees and radians with some practice problems.

Here are more practice problems for writing angles both in degrees and radians, where the degrees number is between 0° and 360° and the radians number is between 0 and 2pi.

This text editor can't write the superscript -1, so I'll use the "arc" prefix to discuss the inverse trig functions.

arccos(-.25): In degrees mode, the answer is 104.4775122...°, which I would round to 104.4775°. Changing to radians mode, the answer is 1.823476582..., which I would round to 1.8235 radians.
Dividing that answer by pi, I get .580430623..., which I would round to .5804pi.

arcsin(-.25): In degrees mode, the answer is -14.47751219...°. Adding 360° and rounding, the answer is 345.5225°.
Instead of changing to radians mode, I could just divide this by 180 to get 1.919569377... which is to say the reading rounds 1.9196pi radians.
Multiplying by pi, I get 6.030505052..., or 6.0305 radians.

Practice

a) arctan(-.25)

b) arccos(1/12)

c) arcsin(-1/12)

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.

Answers to homework 5

Reading left to right, top to bottom

cos120° = -1/2 ||| tan225° = 1 ||| sin330° = -1/2 ||| cot240° = sqrt(3)/3


csc150° = 2 ||| sec210° = -2sqrt(3)/3 ||| sin135° = sqrt(2)/2 ||| tan300° = -sqrt(3)


cot315° = -1 ||| sin225° = -sqrt(2)/2 ||| sec270° = not defined ||| csc240° = -2sqrt(3)/3


a = 2, b = 3, gamma = 30°
Answer: c = sqrt(13 - 6sqrt(3)) ~= 1.6148
 a = 2, b = 3, gamma = 45°
Answer: c = sqrt(13 - 6sqrt(2)) ~= 2.1248

a = 2, b = 3, gamma = 60°
Answer: c = sqrt(7) ~= 2.6458

a = 2, b = 3, gamma = 90°
Answer: c = sqrt(13) ~= 3.6056

a = 2, b = 3, gamma = 120°
Answer: c = sqrt(19) ~= 4.3589

a = 2, b = 3, gamma = 135°
Answer: c = sqrt(13 + 6sqrt(2)) ~= 4.6352

a = 2, b = 3, gamma = 150°
Answer: c = sqrt(13 + 6sqrt(3)) ~= 4.8366

 

Three sides of a triangle:Area from Heron's formula, angles from the Law of Cosines

The Law of Cosine has three different statements, given side lengths a, b and c and angles opposite alpha, beta and gamma, respectively.

a² = b² + c² - 2bccosalpha
b² = a² + c² - 2accosbeta
c² = a² + b² - 2abcosgamma

On earlier homework, we figured out the third side from two lengths and the measure of the angle between them.  With a little algebraic manipulation, we can figure out the cosines of the three angles is we are given the three side lengths, and using the cosine inverse function, written here as arccos(number), we can find the three angles, or at least approximations, since the angles will often be given as irrational numbers.

cosalpha = [b² + c² - a²]/[2bc]
cosbeta = [a² + c² - b²]/[2ac]
cosgamma = [a² + b² - c²]/[2ab]

Examples

If we know the perimeter of a triangle is 7 and the sides are all whole numbers, there are only two possible answers, <3, 3, 1> and <3, 2, 2>. Let's answer all the questions from the quiz base on these numbers

c = 3, b = 3, a = 1
==============
perimeter = 7, so semi-perimeter = 7/2

The triangle is isosceles and acute.

Area = sqrt(7/2[7/2 - 3][7/2 - 3][7/2 - 1]) = sqrt(7/2 × 1/2 × 1/2 × 5/2) = sqrt(35/16) = sqrt(35)/4
approximation: 1.4790

Because c = b, gamma = beta, so we only need to do one calculation to find both.
cosgamma = [1 + 9 - 9]/[2 × 1 × 3] = 1/6
arccos(1/6) ~= 80.4059°
So gamma = beta ~= 80.4059°

If we calculate alpha, we get cosalpha = [9 + 9 - 1]/[2 × 3 × 3] = 17/18
arccos[17/18] = 19.1881°
So alpha ~= 19.1881°

You get the same answer if you subtract 2gamma from 180°.

--
c = 3, b = 2, a =2
==============
perimeter = 7, so semi-perimeter = 7/2

The triangle is isosceles and obtuse.

Area = sqrt(7/2[7/2 - 3][7/2 - 2][7/2 - 2]) = sqrt(7/2 × 1/2 × 3/2 × 3/2) = sqrt(63/16) = 3sqrt(7)/4
approximation: 1.9843

Because a = b, alpha = beta, so we only need to do one calculation to find both.
cosalpha = [9 + 4 - 4]/[2 × 3 × 2] = 9/12 = 3/4
arccos(1/6) ~= 80.4059°
So alpha = beta ~= 41.4096°

If we calculate gamma, we get cosgamma = [4 + 4 - 9]/[2 × 2 × 2] = -1/8
arccos[-1/8] = 19.1881°
So gamma ~= 97.1808°

You get the same answer if you subtract 2alpha from 180°.

Practice problems posted on Wednesday.

Find the classifications, exact and approximate area and approximate all three angles of the following triangles with perimeter = 11.

a) side lengths c = 5, b = 5, a = 1
b) side lengths c = 5, b = 4, a = 2
c) side lengths c = 5, b = 3, a = 3

Answers in the comments.


 

Link to a post about the Law of Cosines and some practice problems

Here is a post about Law of Cosines from 2011.

Here are examples with two side lengths a and b and the measure of between them, gamma, in degrees.  Use these to find the length of c exactly and rounded to four places after the decimal.

a = 3, b = 4, gamma = 45°

a = 3, b = 4, gamma = 90°

a = 3, b = 4, gamma = 135°

Answers in the comments.

The trig functions at perpendicular angles and at the opposite side of the circle (the antipode).

If you know the values of sine and cosine at some angle, call it alpha, you also know all the other trig values at that angle

tangent (a.k.a tan) = sine/cosine
cotangent (a.k.a cot) = cosine/sine
secant (a.k.a sec) = 1/cosine
cosecant (a.k.a csc) = 1/sine

More than that, we know the values of sine and cosine at alpha+90°, alpha+180° and alpha+270°. Here are the rules for sine and cosine, and from those all the rest fall into place.

perpendicular clockwise
sin(alpha+90°) = cos(alpha)
cos(alpha+90°) = -sin(alpha)

antipodal (going the opposite direction on the same line)
sin(alpha+180°) = -sin(alpha)
cos(alpha+180°) = -cos(alpha)

perpendicular counter-clockwise
sin(alpha+270°) = -cos(alpha)
cos(alpha+270°) = sin(alpha)

These rules will answer all the questions about the trig values at the angles listed on this homework. Make sure to give the answers in exact values with no square roots in the denominators.

Practice for finding the other trig functions and the angle from the value of sine, cosine or tangent, assuming all are positive

Example #1: cosalpha = 2/5

(2/5)² + sin²alpha = 1
4/25 + sin²alpha = 1
sin²alpha = 21/25
sinalpha = sqrt(21)/5

since we have sine and cosine, tan = sin/cos.

tanalpha = [sqrt(21)/5]/[2/5] = sqrt(21)/2

With this editor, I can't write -1 as a superscript, so instead I will use the words arcsine, arccosine and arc tantangent. On your calculator, arccos(2/5) = 66.42182152...°, which we round to 66.4218°.

If you take the unrounded value and change it to degrees-minutes-seconds, rounding the seconds to the nearest whole number, we get 66° 25' 19".

Example #1.1: If we had sinalpha = 2/5, the work would look nearly identical, except cosalpha would equal sqrt(21)/5.  Tangent of this angle is the reciprocal of sqrt(21)/2, which is 2sqrt(21)/21 when written in rational denominator form.  The angle is the complement of our original angle, which means they add up to 90°. arcsin(2/5) = 23.57817848...°, which rounds to 23.5782°.

The DMS version of the unrounded value, rounded to the nearest whole second is 23° 34' 41".

Example #2: tanbeta = 2/5

Since tan = sin/cos,
tan × cos = sin

Using this, cos²beta + (2/5)²cos²beta = 1

cos²beta + 4/25cos²beta = 1
29/25cos²beta = 1
cos²beta =25/29
cosbeta = sqrt(25/29) = 5/sqrt(29) = 5sqrt(29)/29

arctan(2/5) = 21.80140949...°, which rounds to 21.8014°, and in DMS is 21° 48' 5".

Here are practice problems. Get the other two trig values, the angle rounded to the nearest ten thousandth of a degree and rounded to the nearest second in DMS mode.

1) cosalpha = 1/10

2) tanbeta = 1/10

3) singamma = 1/9

4) tandelta = 1/9

Answers in the comments.