Solving Trig Problems Precal – Sections 5.1-5.5. Reference Angles p. 280 Associated with every...

Solving Trig Problems

Precal – Sections 5.1-5.5

Reference Angles p. 280

Associated with every angle drawn in standard position there is another angle called the reference angle.

The reference angle is the acute angle formed by the terminal side of the given angle and the x-axis. Reference angles may appear in all four quadrants. Angles in quadrant I are their own reference angles.

Calculating Reference Angles

The values of trigonometric functions of angles greater than90° (or less than 0°) can be found using corresponding acuteangles called reference angles.

Let be an angle in standard position. Its reference angle is the acute angle (read theta prime) formed by the terminal side of and the x-axis.

00'

0

Reference Angles

90° < < 180°;0

ππ

2< <0

0

Radians: = π 0–0'

0'

= 180°Degrees: 0' – 0

0

0'

– 180°0=Degrees: 0'

Radians: = 0 – π0'

180° < < 270°;

π3π

2< <

0

0

0

0'

– 0360°=Degrees: 0'

Radians: = 0–2π0'

270° < < 360°;

2π3π

2< <

0

0

Using reference angles

Use reference angles and the unit circle to find the six trig functions for a 135 angle:

0 = 320°

Because 270° < < 360°, the reference angle is = 360° – 320° = 40°.

00'

0 = – 5π6

SOLUTION

Find the reference angle for each angle .00'

Because is coterminal with and π < < , the

reference angle is = – π = .

07π6

7π6

π6

7π6

3π2

0'

0

Let (3, – 4) be a point on the terminal side of an angle instandard position. Evaluate the six trigonometric functions of .

0

0

SOLUTION

Use the Pythagorean theorem to find the value of r.

r = x 2 + y

2

= 3 2 + (– 4 )

2

= 25

= 5

Evaluating Trigonometric Functions Given a Point

r

(3, – 4)

Using x = 3, y = – 4, and r = 5,you can write the following:

45

sin = = – yr0

cos = =xr

350

tan = = – 0yx

43

csc = = – ry

54

0

sec = =rx

53

0

cot = = – 0xy

34

Evaluating Trigonometric Functions Given a Point

0r

(3, – 4)

EVALUATING TRIGONOMETRIC FUNCTIONSCONCEPT

SUMMARY

Use these steps to evaluate a trigonometric function ofany angle .0

2

Use the quadrant in which lies to determine thesign of the trigonometric function value of .

00

3

1 Find the reference angle .0'

Evaluate the trigonometric function for angle .0'

EVALUATING TRIGONOMETRIC FUNCTIONSCONCEPT

SUMMARY

Signs of Function Values

Quadrant IQuadrant II

Quadrant III Quadrant IV

sin , csc : +0 0

cos , sec : –0 0

tan , cot : –0 0

sin , csc : +0 0

cos , sec : +0 0

tan , cot : +0 0

sin , csc : –0 0

cos , sec : –0 0

tan , cot : +0 0

sin , csc : –0 0

cos , sec : +0 0

tan , cot : –0 0

Evaluate tan (– 210°).

SOLUTION

The angle – 210° is coterminal with 150°.

The tangent function is negative in Quadrant II,so you can write:

tan (– 210°) = – tan 30° = – 33

0' = 30°

0 = – 210°

The reference angle is = 180° – 150° = 30° .0'

HW:

Page 296 (1-3, 5-12, 14-21)

Angle of Elevation and Depression

The angle of elevation is measured from the horizontal up to the object.

Imagine you are standing here.

Angle of Elevation and DepressionThe angle of depression is measured from the horizontal down to the object.

Constructing a right triangle, we are able to use trig to solve the triangle.

A second similar triangle may also be formed.


Example #1

Angle of Elevation and DepressionSuppose the angle of depression from a lighthouse to a sailboat is 5.7o. If the lighthouse is 150 ft tall, how far away is the sailboat?

Construct a triangle and label the known parts. Use a variable for the unknown value.

5.7o

5.7o

150 ft.

x

Angle of Elevation and DepressionSuppose the angle of depression from a lighthouse to a sailboat is 5.7o. If the lighthouse is 150 ft tall, how far away is the sailboat?

5.7o

5.7o

150 ft.

x

Set up an equation and solve.


5.7o

150 ft.

x

150tan(5.7 )o

x

tan(5.7 ) 150ox

150

tan(5.7 )ox

Remember to use degree mode!

x is approximately 1,503 ft.


Example #2

Angle of Elevation and DepressionA spire sits on top of the top floor of a building. From a point 500 ft. from the base of a building, the angle of elevation to the top floor of the building is 35o. The angle of elevation to the top of the spire is 38o. How tall is the spire?

Construct the required triangles and label.

500 ft.

38o 35o

Angle of Elevation and DepressionWrite an equation and solve.

Total height (t) = building height (b) + spire height (s)

500 ft.

38o 35o

Solve for the spire height.

t

b

s

Total Height

tan(38 )500

o t

500 tan(38 )o t

Angle of Elevation and DepressionWrite an equation and solve.

500 ft.

38o 35o

Building Height

tan(35 )500

o b

500 tan(35 )o b t

b

s


5050 0 t0 tan an(3(38 ) 5 )o o s

Write an equation and solve.

500 ft.

38o 35o

500 tan(38 )o t 500 tan(35 )o b

5050 0 t0 tan an(3(38 ) 5 )o o s

The height of the spire is approximately 41 feet.

t

b

s

Total height (t) = building height (b) + spire height (s)


Example #3


A hiker measures the angle of elevation to a mountain peak in the distance at 28o. Moving 1,500 ft closer on a level surface, the angle of elevation is measured to be 29o. How much higher is the mountain peak than the hiker?

Construct a diagram and label.

1st measurement 28o.

2nd measurement 1,500 ft closer is 29o.


Adding labels to the diagram, we need to find h.

28o 29o

1500 ft x ft

h ft

Write an equation for each triangle. Remember, we can only solve right triangles. The base of the triangle with an angle of 28o is 1500 + x.

tan 281500

o h

x

tan 29o h

x


tan 29o h

x

tan(29 )ox h

Now we have two equations with two variables.Solve by substitution.

tan 281500

o h

x

(1500 ) tan(28 )ox h

(1500 ) tan(28 ) tan(29 )o ox x

Solve each equation for h.

Substitute.


1500 tan(28 ) tan(28 ) tan(29 )o o ox x

(1500 ) tan(28 ) tan(29 )o ox x Solve for x. Distribute.

Get the x’s on one side and factor out the x.

Divide.

1500 tan(28 ) tan(29 ) tan(28 )o o ox x

1500 tan(28 ) tan(29 ) tan(28 )o o ox

1500 tan(28 )

tan(29 ) tan(28 )

o

o ox

x = 35,291 ft.


tan(29 )ox h

However, we were to find the height of the mountain. Use one of the equations solved for “h” to solve for the height.

x = 35,291 ft.

(35,291) tan(29 ) 19,562o

The height of the mountain above the hiker is 19,562 ft.


Start homework on a new page.Assignment 4.8 / 1, 3, 8, 11, 14-16, 19, 26

Quiz 4.1-4.5 FridayHomework up to today is due Friday.

Remember to change your calculator between radians and degrees when required. All graphing of trig functions is done in radians.

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