Trigonometry Angles & Degree Measure Trigonometric Ratios Right Triangle Problems Law of Sines Law...

Trigonometry

Angles & Degree MeasureTrigonometric RatiosRight Triangle ProblemsLaw of SinesLaw of CosinesProblem Solving

Angles and Degree Measure Angles are always measured from the right x-

axis to the terminal side of the angle.

Angles and Degree Measure Angles are often expressed in units of degrees,

but can be into smaller increments called minutes and seconds. These units are used to rename the decimal part of a degree measure.

1o = 60’ and 1’ = 60 “ 30.25o = 30o + (.25o)(60’/1o) = 30o 15’ 0” 30.29o = 30o + (.29o)(60’/1o) = 30o 17.4’ the

0.4’ is then multiplied again to get seconds; (.4’)(60”/1’) = 24”. This gives a final answer of

30o 17’ 24”

Angles and Degree Measure Solving for degrees from minutes and seconds is

also possible. Example: 30o 27’ 45” = the sum of each of the

values expressed in degree units. 30 + (27’)(1o /60’) + (45”)(1o /3600”) =30 + 0.45 + .0125 = 30.46250 = 30.46o

Angles and Degree Measure Coterminal angles are angles that have the same

starting and ending point but are expressed in different values based on the direction of the rotation or the number of rotations.

Going around CCW and returning to the terminal side results in (360 + ). The rotations can be any integer value. Even with increasing angle measure, each angle is still equal to the original angle.

Going around CW is (-) and results in (-360 + ).

Angles and Degree Measure Reference angles, denoted by , are always

measured from the x-axis of the quadrant that the terminal sides is located back to the terminal side. Reference angles are positive.

Trigonometric RatiosSine Function – sin Sine ratio is the value

of the length of the opposite side divided by the hypotenuse of a right triangle.

sin = a/c Values for sin

oscillate between 0 to 1 to 0 to –1 to 0

b

ac

Right Triangle a – opposite side b – adjacent side c – hypotenuse - theta (reference )

Trigonometric RatiosSine Function – sin Sine ratio works for

angle values between 0 and 360o and then repeats.

sin = a/c is also expressed as follows:

sin = opposite side hypotenuse

b

ac


Trigonometric RatiosCosine Function – cos Cosine ratio is the value of

the length of the adjacent side divided by the hypotenuse of a right triangle.

cos = b/c Values for cos oscillate

between 1 to 0 to -1 to 0 to 1

b

ac


Trigonometric RatiosCosine Function – cos Cosine ratio works for


cos = b/c is also expressed as follows:

cos = adjacent side hypotenuse

b

ac


Trigonometric RatiosTangent Function – tan Tangent ratio is the value

of the length of the opposite side divided by the adjacent side of a right triangle.

tan = a/b Values for tan oscillate

between and - within each 180o increment

b

ac


Trigonometric RatiosTangent Function – tan Tangent ratio works for


tan = a/b is also expressed as follows:

tan = opposite side adjacent side

b

ac


Trigonometric RatiosSine Function change in values

as changes: Sin 0 = 0.0000 Sin 30 = 0.5000 Sin 45 = 0.7071 Sin 60 = 0.8660 Sin 75 = 0.9659 Sin 90 = 1.0000 Sin 120 = 0.8660 Sin 135 = 0.7071 Sin 150 = 0.5000 Sin 180 = 0.0000 Afterwards, sin repeats with

negative values

b

ac


Trigonometric RatiosCosine Function change in

values as changes: Cos 0 = 1.0000 Cos 30 = 0.8660 Cos 45 = 0.7071 Cos 60 = 0.5000 Cos 75 = 0.2588 Cos 90 = 0.000 Cos 120 = -0.5000 Cos 135 = -0.7071 Cos 150 = -0.8660 Cos 180 = -1.0000 Afterwards, cos repeats

values

b

ac


Trigonometric RatiosTangent Function change in

values as changes: Tan 0 = 0.0000 Tan 30 = 0.5773 Tan 45 = 1.0000 Tan 60 = 1.7321 Tan 75 = 3.7321 Tan 90 = undefined Tan 120 = -1.7321 Tan 135 = -1.0000 Tan 150 = - 0.5773 Tan 180 = 0.0000 Afterwards, tan repeats

values

b

ac


Trigonometric RatiosNote that with each

function there is a relation between an angle measure and a ratio of sides of the right triangle.

The function of an angle is equal to the decimal value of the specific ratio of sides.

b

ac


Trigonometric RatiosLikewise, the specific ratio is

equal to the function of an angle.

Ex. Let sin = 6/10 = .6000Then = Sin –1 .6000 =

8

610


36.9o



Ex. Let sin = 1/2 = .5000Then = Sin –1 .5000 =

b = 1.732

a = 1C = 2


30.0o



Ex. Let cos = 8/10 = .8000Then = cos –1 .8000 =

8

610


36.9o



Ex. Let cos = 1.732/2 = .8660Then = cos –1 .8660 =

1.732

12


30.0o

Trigonometric RatiosWe can also solve for missing

values if we know two of the three variables in a ration.

Ex. sin 30o = 20/rThen r = 20/sin30o = 20/0.5 =

x

20r

30o

Tan 30o = 20/x

Then x = 20/tan30o =

is the complementary angle of and so = 90 – 30 or 60o

40

34.6

Right Triangle Problems

Any situation that includes a right triangle, becomes solvable with trigonometry.

Angle of Elevation

Angle of Depression


This diagram has an inscribed triangle whose hypotenuse is also the diameter of the circle.

A

B C Given: = 40o, AB = 6 cm

Determine the area of the shaded region.

Solution: Sin 40o = 6/AC; AC = 6/ sin40o = 9.33 cm

AC/2 = radius of the circle = 4.67 cm; BC = 6/tan 40o =7.15 cm

Ao = r2 = 68.4 cm2, A = ½bh = 21.5 cm2; A = 46.9 cm2


This diagram has a right triangle within a cone that can be used to solve for the surface area and volume of the cone.h

r

l

Given: = 55o, l = 6 cm

Determine the area and volume of the cone.

Solution: Sin 55o = h/l; h = 6 x sin55o = 4.91 cm

tan 55o = h/r; r = h/ tan 55o = 4.91/ tan 55o = 3.44 cm

Acone = r(r + l) = (3.44)(3.44 + 6)cm2; Acone = 102 cm2

Vcone =⅓ r2h = ⅓ (3.44)2(4.91)cm3; Vcone = 60.8 cm3


The angle of elevation of a ship at sea level to a neighboring lighthouse is 2o . The captain knows that the top of lighthouse is 165 ft above sea level. How far is the boat from the lighthouse?

h

x

Solution: tan 2o = h/x; x = h/tan 2o

x = 165/ tan 2o = 4725 ft; 1 mile = 5280 ft, so the ship is .89 mile away from the lighthouse.

Function Values – Unit CircleThe unit circle is a circle with a radius of 1 unit.

x

yr = 1 Sin = y; Cos = x; Tan = y/x

•The coordinate of P will lead to the value of x and y which in turn leads to the values for sine, cosine, and tangent.

•Use the reference angle in each quadrant and the coordinates to solve for the function value.

Function Values – Unit Circle

•Quadrant I – Angle =

•Quadrant II – Angle = (180 - )

•Quadrant III – Angle = (180 + )

•Quadrant IV – Angle = (360 - )

The reference angle is always measured in its quadrant from the x – axis.


•Quadrant I – P (x,y)

•Quadrant II – P (-x, y)

•Quadrant III – P (-x, -y)

•Quadrant IV – P (x, -y)

With the values of changing from (+) to (-) in each quadrant, and with the functions of sine, cosine, and tangent valued with ratios of x and y, the functions will also have the sign values of the variables in the quadrants.


DegreesSine

Cosine

Tangent

0 30 45 60 90 120 135 150

0 ½ 2/2 3/2 1 3/2 2/2 ½

180 210 225 240 270 300 315 330

0 -½ -2/2 -3/2 -1 -3/2 -2/2 -½

Degrees

Sine

Cosine

Tangent

1 3/2 2/2 ½ 0 -½ -2/2 -3/2

0 3/3 1 3 - -3 -1 -3/3

-1 -3/2 -2/2 -½ 0 ½ 2/2 3/2

0 3/3 1 3 - -3 -1 -3/3


Example: Cos 240 = _______

60

P (-½, -3/2)

- ½

Example: Sin 240 = _______

Example: Tan 240 = _______3

-3/2240

Law of SinesLaw of Sines – For a triangle with angles A, B, C and sides of lengths of a, b, c the ratio of the sine of each angle and its opposite side will be equal. Sin A = Sin B = Sin C

a b c

A

B

C

c

ba

hProof:Sin A = h/b; Sin B = h/a

h = b Sin A, h = a Sin B

b Sin A = a Sin B; Sin A = Sin Ba b

Law of Sines

Proof, continued:

A

B

C

c

ba

kk is the height of the same triangle from vertex A

Sin C = k/b and Sin B = k/c

k = b Sin C and k = c Sin B

b Sin C = c Sin B;

Sin B = Sin C b b c

Conclusion: Sin A = Sin B = Sin C a b c

Law of Sines

Let’s take a closer look:

Suppose A < 90o

A

c a

If a = c Sin A, one solution exists

a

Ac

c Sin A

If a < c Sin A, no solution exists

Law of Sines


Suppose A < 90o

A

ac

If a > c Sin A and a c, one solution exists

c Sin A

a

Ac

c Sin A

If c Sin A < a < c , two solution exists

a

Law of Sines


Suppose A 90o

A

a

c

If a < c, no solution exists

A

a

If a > c , one solution exists

c

Law of CosinesA

B CD

b

a

c

a -x x

h

b2 = h2 + x2; h2 = b2 - x2

Cos C = x/b ; x = b Cos C

c2 = h2 + (a-x)2; c2 = h2 + a2 –2ax + x2

Substitute and we get:

c2 = (b2 - x2)+ a2 –2a(bCos C) + x2

c2 = b2 + a2 – 2abCos C

For any triangle given two sides and an included angle,

a2 = b2 + c2 – 2bcCos A

b2 = a2 + c2 – 2acCos B

c2 = b2 + a2 – 2abCos C

PROOF:PROOF:

Law of Cosines - example

Given: A = 50o; AB = 8 cm, AC = 14 cm

x 2 = 8 2+ 14 2 – 2(8)(14)Cos50

Find x.

A

B

C

x

x 2 = 260 – 144; x 2 = 116

x = 10.8 cm

Law of Cosines – Hero’s Formula

For any triangle, the area of the triangle can be determined with Hero’s Formula.

s = ½(a + b + c) where s = semi-perimeter of the triangle.

A = s(s – a)(s – b)(s – c)

Law of Cosines – Hero’s Formula

Given: ABC; a = 7, b = 24, c = 25

A = 28(28 – 7)(28 – 24)(28 – 25) = 84 cm 2

7cm25 cm

24 cm

S = ½ (7 + 24 + 25) = 28

Check: A = ½(b)(h) = ½(7)(24) = 84 cm 2

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Trigonometry Angles & Degree Measure Trigonometric Ratios Right Triangle Problems Law of Sines Law...

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