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.

The exact trig values "famous" angles from 0° to 90°

Here is a handy reference for the major trigonometric function values for the "famous" angles in the first quadrant and at 0° and 90°, which technically are on the x-axis and y-axis respectively, not in the first quadrant but instead on the borders.

Here are a few well known properties of these functions over this range.

1. Sine and tangent are increasing.  While sine increases, it is bounded by 1 as the highest possible value. Tangent is unbounded and can get as large as any positive number you can name.  As I said in class, tangent is the same as the slope of the line that connects the origin (0, 0) to the point (cos alpha, sin alpha). (The editor in Blogger is bound by the symbols available in HTML, so I will have to spell out Greek letters when used as variables.)

2. Cosine is decreasing over the range.

3. Cosine is short for "complementary sine", which means cos(90°-alpha)= sin alpha, when alpha is an angle measured in degrees. In the other direction, sin(90°-alpha)= cos alpha.

Link to an explanation of The Trigonometric Identity

There are many trigonometric identities, but except for some definitions of terms, nearly all of them are re-arrangements of what I call The Trigonometric Identity

sin²alpha + cos²alpha = 1

Here is a link to the post from last year.

Euclid's proof and similar triangles

Using the handout from today, we know Euclid's proof is based on similar triangles.  With two similar triangles, the ratio between corresponding sides remains constant. If we list the sides of each triangle in the order short-leg-long leg-hypotenuse, here are the three triangles.

The big triangle: a-b-c
The triangle with b as hypotenuse: h-y-b
The triangle with a as hypotenuse: x-h-a

Using this information, given the lengths of one triangle we can find all the others.

Example: a-b-c are 3-4-5

The triangle with b as hypotenuse: h-y-b
h/3 = y/4 = 4/5

h = 12/5, y = 16/5


The triangle with a as hypotenuse: x-h-a
x/3 = h/4 = 3/5

x = 9/5, h = 12/5

Not surprisingly, we get the same value for h in both similar triangles it belongs to.

practice problems:

a) h-y-b are 3-4-5

b) x-h-a are 3-4-5

c) a-b-c are 8-15-17

Answers in the comments.

Slope, distance and the area of a triangle defined by three points

Here is a link to practice problems for slope, distance and the area of a triangle defined by three points.

Here are another three points and the origin: (1, 7) (4, 6) (3,-3) and (0,0)

Slope problems.
Find the slope of the line connecting the points. Write them as fractions in lowest terms. (Note: if the answer is a whole number, leave it as a whole number. 3 is simpler than 3/1.)

a) (1,7) and (4,6)
b) (1,7) and (3,-3)
c) (1,7) and (0,0)
d) (4,6) and (3,-3)
e) (4,6) and (0,0)
f) (3,-3) and (0,0)

Distance problems
Find the distance between the pairs of points and write the answer as a simplified square root.
a) (1,7) and (4,6)
b) (1,7) and (3,-3)
c) (1,7) and (0,0)
d) (4,6) and (3,-3)
e) (4,6) and (0,0)
f) (3,-3) and (0,0)

Areas of triangles
Find the areas of the triangles defined by the following three points.
a) (1,7), (4,6) and (0,0)
b) (1,7), (3,-3) and (0,0)
c) (4,6), (3,-3) and (0,0)
d) (1,7), (4,6) and(3,-3)

Answers in the comments.