**4.8 Applications and Models**

Solving a Right Triangle:

Recall:

Formulas: given triangle ABC with angle C = 90º

Recall: Sin A = (opposite/hypotenuse), Cos A = (adjacent/hypotenuse),

Tan A = (opposite/ adjacent) or Tan A = (Sin A)/ (Cos A)

Sin A = (CB)/ (AB)

Cos A =(AC)/(AB)

Tan A = (CB)/(AC)

**Angle of Elevation and Angle of Depression**:

http://www.mathsteacher.com.au/year10/ch15_trigonometry/12_elevation_depression/23elevdep.htm

The

**angle of elevation**of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (the line of sight).

If the object is below the level of the observer, then the angle between the horizontal and the observer's line of sight is called the

**angle of depression**.

**Example 1**: A shadow of length L is created by an 850 foot building when the sun is θº above the horizon. Write L as a function of θ.

Drawing your triangle, you use the opposite and the adjacent sides so you would use Tangent.

Tan θ = 850 / L so writing this to have L as a function of θ, means to solve for L.

L = 850 / tan θ

What are the values of L when θ = {10º, 20º, 30º, 40º, 50º, 60º}?

Answer: {4820.6, 2335.4, 1472.2, 1013.0, 713.23, 490.75} respectively.

In

**surveying and navigation**, directions are generally given in the terms of BEARINGS. A bearing measures the acute angle of a path or line of sight makes with a fixed north-south line.

**Check out this web-site:**

http://www.mathsteacher.com.au/year7/ch08_angles/07_bear/bearing.htm

**Example 1:**An airplane flying at 550 mph has a bearing of N 58º E. After flying 1.5 hours, how far north and how far east has the plane traveled from its point of departure?

1.5 hours times 550 mph = 825 mi.

Sin 58º = x/ 825

x = 699.6396793 miles ≈ 700 miles East

Cos 58º = y/ 825

x = 437.183393 miles ≈ 437 miles North

**Harmonic Motion**– check out this web site:

http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html

**Amplitude**– the maximum distance the ball moves vertically upward or downward from its equilibrium (or resting position) – displacement.

**Example 1:**lets say it took 4 seconds for the ball to move from resting position to resting position. This means the period is 4. The maximum displacement is 10 cm.

Using the formula: d = a sin ωt

The absolute value of a = 10

Period = (2π/ω) = 4

4ω = 2 π

ω = π / 2

Therefore: d = 10 sin (πt)/2

**Definition**: A point that moves on a coordinate line is said to be in Simple Harmonic Motion if its distance “d” from the origin at time “t” is given by either

d = a sin ωt or d = a cos ωt where “a” and “ω” are real numbers such that ω is greater than 0. The motion has the amplitude equal to the absolute value of a, period is equal to (2π)/ω and the frequency is equal to ω/ (2π)

**Example 1:**d = .5 cos 20πt

a. find the displacement: a = .5

b. find the frequency: ω/ (2π) = 20π/ (2π) = 10

c. least positive value of t when d = 0

0 = .5 cos 20πt

0 = cos 20πt now a thought: when does 0 = cos x x = π/2 so

20π t = π/2

t = 1/40

**Example 2**: d = (1/64) sin 792πt

a. find the displacement: a = 1/64

b. find the frequency: ω/ (2π) = 792π/ (2π) = 396

c. least positive value of t when d = 0

0 = (1/64) sin 792πt

0 = sin 792πt now a thought: when does 0 = cos x x = π/2 so

792πt = π/2

t = 1/792

**Example 3:**find a model for simple harmonic motion given:

Displacement (t = 0) of 2 feet

Amplitude of 2 feet

Period of 10 seconds

d = a cos ωt

a = 2

period = (2π/ω) = 10

(2π/10) = ω

π/5 = ω

Therefore: d = 2 cos πt/5

**Section 4.8 Homework #41**; pg. 361; #1 – 21 odd, 25, 29, 33, 37, 45, 49 – 59 odd, 65

pg. 351; #8, 18, 34, 59, 91, 93

Review Homework due the day after 4.8, Homework #42; pg. 370; # 111 – 123 odd, 127, 135, 143, 155, 159 – 171 odd.