Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-1 Solving Right Triangles 2.4...

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-1

Solving Right Triangles2.4Significant Digits ▪ Solving Triangles ▪ Angles of Elevation or Depression


Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm.

2.4 Example 1 Solving a Right Triangle Given an Angle and a Side (page 69)


2.4 Example 1 Solving a Right Triangle Given an Angle and a Side (cont.)

Three significant digits


Solve right triangle ABC, if a = 44.25 cm and b = 55.87 cm.

2.4 Example 2 Solving a Right Triangle Given Two Sides (page 70)

Use the Pythagorean theorem to find c:

Four significant digits


2.4 Example 2 Solving a Right Triangle Given Two Sides (cont.)


The angle of depression from the top of a tree to a point on the ground 15.5 m from the base of the tree is 60.4°. Find the height of the tree.

2.4 Extra Example Finding a Length When the Angle of Elevation is Known

The measure of equals the measure of the angle of depression because the two angles are alternate interior angles, so


2.4 Extra Example Finding a Length When the Angle of Elevation is Known (cont.)

The tree is about 27.3 m tall.



The length of a shadow of a flagpole 55.20 ft tall is 27.65 ft. Find the angle of elevation of the sun.

2.4 Example 3 Finding a Length When the Angle of Elevation is Known (page 71)

The angle of elevation of the sun is 63.39°.

Four significant digits


Further Applications of Right Triangles2.5Bearing ▪ Further Applications


Marla needs to find the height of a building. From a given point on the ground, she finds that the angle of elevation to the top of the building is 74.2°. She then walks back 35 feet. From the second point, the angle of elevation to the top of the building is 51.8°. Find the height of the building.

2.5 Example 4 Solving a Problem Involving Angles of Elevation (page 85) – similar to Homework #28


There are two unknowns, the distance from the base of the building, x, and the height of the building, h.

2.5 Example 4 Solving a Problem Involving Angles of Elevation (cont.)

In triangle ABC

In triangle BCD


Set the two expressions for h equal and solve for x.

2.5 Example 4 Solving a Problem Involving Angles of Elevation (cont.)

Since h = x tan 74.2°, substitute the expression for x to find h.

The building is about 69 feet tall.


• See section 2.5 #15 and #25• Help with sketch on #26

2.5 Examples for homework


7.1 Oblique Triangles and the Law of Sines (Unit 2)

7.2 The Ambiguous Case of the Law of Sines (Unit 2)

7.3 The Law of Cosines (Unit 2)

7.4 Vectors, Operations, and the Dot Product (Unit 4)

7.5Applications of Vectors (Unit 4)

Applications of Trigonometry and Vectors7


Oblique Triangles and the Law of Sines

7.1

Congruency and Oblique Triangles ▪ Derivation of the Law of Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a Triangle


Solve triangle ABC if A = 28.8°, C = 102.6°, and c = 25.3 in.

7.1 Example 1 Using the Law of Sines to Solve a Triangle (AAS)


Use the Law of Sines to find the lengths of the missing sides.

7.1 Example 1 Using the Law of Sines to Solve a Triangle (AAS) (cont.)


Jerry wishes to measure the distance across the Big Muddy River. He determines that C = 117.2°, A = 28.8°, and b = 75.6 ft. Find the distance a across the river.

7.1 Example 2 Using the Law of Sines in an Application (ASA)


7.1 Example 2 Using the Law of Sines in an Application (ASA) (cont.)

Use the Law of Sines to find the length of side a.

The distance across the river is about 65.1 ft.


The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W. Find the distance between the ship and the lighthouse at each location.

7.1 Example 3 Using the Law of Sines in an Application (ASA)

Let x = the distance to the lighthouse at bearing N 52° W andy = the distance to the lighthouse at bearing N 23° W.



The lighthouse is located at Z, and the ship is first located at Y and then at X.



The distance between the ship and the second location is about 9.4 km.

The distance between the ship and the first location is about 4.7 km.


7.1 Example 4 Finding the Area of a Triangle (SAS)

Find the area of triangle DEF in the figure.


7.1 Example 5 Finding the Area of a Triangle (ASA)

Find the area of triangle ABC if B = 58°10′, a = 32.5 cm, and C = 73°30′.

We must find AC (side b) or AB (side c) in order to find the area of the triangle.


7.1 Example 5 Finding the Area of a Triangle (ASA) (cont.)


7.1 Example 5 Finding the Area of a Triangle (ASA) (cont.)

37.0 cm

48°20′


The Ambiguous Case of the Law of Sines7.2Description of the Ambiguous Case ▪ Solving SSA Triangles (Case 2) ▪ Analyzing Data for Possible Number of Triangles


Solve triangle ABC if a = 17.9 cm, c = 13.2 cm, and C = 75°30′.

7.2 Example 1 Solving the Ambiguous Case (No Such Triangle)

Use the Law of Sines to find A.

Since sin A > 1 is impossible, no such triangle exists.


An attempt to sketch the triangle leads to this figure.

7.2 Example 1 Solving the Ambiguous Case (No Such Triangle) (cont.)


Solve triangle ABC if A = 61.4°, a = 35.5 cm, andb = 39.2 cm.

7.2 Example 2 Solving the Ambiguous Case (Two Triangles)

Use the Law of Sines to find B.


There are two angles between 0° and 180° such that sin B ≈ .9695:

7.2 Example 2 Solving the Ambiguous Case (Two Triangles) (cont.)

so

is a valid possibility.

Solve separately for triangles


7.2 Example 3 Solving the Ambiguous Case (One Triangle)

Solve triangle ABC if B = 68.7°, b = 25.4 in., anda = 19.6 in.

Use the Law of Sines to find A.


There are two angles between 0° and 180° such that sin A ≈ .7189:

7.2 Example 3 Solving the Ambiguous Case (One Triangle) (cont.)

is not a valid possibility.

There is only one triangle


7.2 Example 3 Solving the Ambiguous Case (One Triangle) (cont.)


Without using the law of sines, explain why no triangle exists satisfying B = 93°, b = 42 cm, and c = 48 cm.

7.2 Example 4 Analyzing Data Involving an Obtuse Angle

Because B is an obtuse angle, it must be the largest angle of the triangle. Thus, b must be the longest side of the triangle.

We are given that c > b, so no such triangle exists.


The Law of Cosines7.3Derivation of the Law of Cosines ▪ Solving SAS and SSS Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a Triangle


Two boats leave a harbor at the same time, traveling on courses that make an angle of 82°20′ between them. When the slower boat has traveled 62.5 km, the faster one has traveled 79.4 km. At that time, what is the distance between the boats?

7.3 Example 1 Using the Law of Cosines in an Application (SAS)

The harbor is at C. The slower boat is at A, and the faster boat is at B.

We are seeking the length of AB.


Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle.

7.3 Example 1 Using the Law of Cosines in an Application (SAS)

The two boats are about 94.3 km apart.


Solve triangle ABC if B = 73.5°, a = 28.2 ft, and c = 46.7 ft.

7.3 Example 2 Using the Law of Cosines to Solve a Triangle (SAS)

Use the law of cosines to find b because we know the lengths of two sides of the triangle and the measure of the included angle.


7.3 Example 2 Using the Law of Cosines to Solve a Triangle (SAS) (cont.)

47.2 ft

Now use the law of sines to find the measure of another angle.


Solve triangle ABC if a = 25.4 cm, b = 42.8 cm, and c = 59.3 cm.

7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS)

Use the law of cosines to find the measure of the largest angle, C.


Use either the law of sines or the law of cosines to find the measure of angle B.

7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (cont.)

118.6°


Alternative:

7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (cont.)

118.6°


7.3 Example 4 Designing a Roof Truss (SSS)

Find the measure of angle C in the figure.

4-50

Harmonic Motion4.5Simple Harmonic Motion ▪ Damped Oscillatory Motion

4-51

4.5 Example 1 Modeling the Motion of a Spring

Suppose that an object is attached to a coiled spring. It is pulled down a distance of 16 cm from its equilibrium position and then released. The time for one complete oscillation is 6 seconds.

4-52

4.5 Example 1(a) Modeling the Motion of a Spring

(a) Give an equation that models the position of the object at time t.

When the object is released at t = 0, the object is at distance −16 cm from equilibrium.

Since the time needed to complete one oscillation is 6 sec, P = 6, and

4-53

4.5 Example 1(b, c) Modeling the Motion of a Spring

(b) Determine the position at t = 1.5 seconds.

At t = 1.5 seconds, the object is at the equilibrium position.

(c) Find the frequency.

The frequency is the reciprocal of the period.

4-54

4.5 Example 2 Analyzing Harmonic Motion

Suppose that an object oscillates according to the model s(t) = 2.5 sin 5t, where t is in seconds and s(t) is in meters. Analyze the motion.

a = 2.5, so the object oscillates 2.5 meters in either direction from the starting point.

and

The motion is harmonic because the model is of the form

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-1 Solving Right Triangles 2.4...

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