### Unit 2

```2.4
Solving Right Triangles
Significant Digits ▪ Solving Triangles ▪ Angles of Elevation or
Depression
2-1
2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (page 69)
Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm.
2-2
2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (cont.)
Three significant digits
2-3
2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (cont.)
Three significant digits
2-4
2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (cont.)
2-5
2.4
Example 2 Solving a Right Triangle Given Two Sides
(page 70)
Solve right triangle ABC, if a = 44.25 cm and
b = 55.87 cm.
Use the Pythagorean theorem to find c:
Four significant digits
2-6
2.4
Example 2 Solving a Right Triangle Given Two Sides
(cont.)
2-7
2.4
Example 2 Solving a Right Triangle Given Two Sides
(cont.)
2-8
2.4
Extra Example Finding a Length When the Angle of
Elevation is Known
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.
The measure of
equals the measure of the
angle of depression because the two angles are
alternate interior angles, so
2-9
2.4
Extra Example Finding a Length When the Angle of
Elevation is Known (cont.)
Three significant digits
The tree is about 27.3 m tall.
2-10
2.4
Example 3 Finding a Length When the Angle of
Elevation is Known (page 71)
The length of a shadow of
a flagpole 55.20 ft tall is
27.65 ft. Find the angle of
elevation of the sun.
Four significant digits
The angle of elevation of the sun is 63.39°.
2-11
2.5
Further Applications of Right Triangles
Bearing ▪ Further Applications
2-12
2.5
Example 4 Solving a Problem Involving Angles of
Elevation (page 85) – similar to Homework #28
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.
2-13
2.5
Example 4 Solving a Problem Involving Angles of
Elevation (cont.)
There are two unknowns, the
distance from the base of the
building, x, and the height of the
building, h.
In triangle ABC
In triangle BCD
2-14
2.5
Example 4 Solving a Problem Involving Angles of
Elevation (cont.)
Set the two expressions for h equal and solve for x.
Since h = x tan 74.2°, substitute the expression for x to
find h.
The building is about 69 feet tall.
2-15
2.5
Examples for homework
• See section 2.5 #15 and #25
• Help with sketch on #26
7-16
7
Applications of Trigonometry and Vectors
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)
7.5 Applications of Vectors (Unit 4)
7-17
7.1
Oblique Triangles and the Law
of Sines
Congruency and Oblique Triangles ▪ Derivation of the Law of
Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a
Triangle
7-18
7.1
Example 1 Using the Law of Sines to Solve a Triangle
(AAS)
Solve triangle ABC if A = 28.8°, C = 102.6°, and
c = 25.3 in.
7-19
7.1
Example 1 Using the Law of Sines to Solve a Triangle
(AAS) (cont.)
Use the Law of Sines to find
the lengths of the missing
sides.
7-20
7.1
Example 2 Using the Law of Sines in an Application
(ASA)
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.
7-21
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.
7-22
7.1
Example 3 Using the Law of Sines in an Application
(ASA)
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.
Let x = the distance to the
lighthouse at bearing N 52° W and
y = the distance to the lighthouse
at bearing N 23° W.
7-23
7.1
Example 3 Using the Law of Sines in an Application
(ASA) (cont.)
The lighthouse is located at Z, and
the ship is first located at Y and
then at X.
7-24
7.1
Example 3 Using the Law of Sines in an Application
(ASA) (cont.)
The distance between the ship and the
first location is about 4.7 km.
The distance between the ship and the
second location is about 9.4 km.
7-25
7.1
Example 4 Finding the Area of a Triangle (SAS)
Find the area of triangle DEF
in the figure.
7-26
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.
7-27
7.1
Example 5 Finding the Area of a Triangle (ASA) (cont.)
7-28
7.1
Example 5 Finding the Area of a Triangle (ASA) (cont.)
48°20′
37.0 cm
7-29
7.2
The Ambiguous Case of the Law
of Sines
Description of the Ambiguous Case ▪ Solving SSA Triangles
(Case 2) ▪ Analyzing Data for Possible Number of Triangles
7-30
7.2
Example 1 Solving the Ambiguous Case (No Such
Triangle)
Solve triangle ABC if a = 17.9 cm, c = 13.2 cm, and
C = 75°30′.
Use the Law of Sines to find A.
Since sin A > 1 is impossible, no such triangle exists.
7-31
7.2
Example 1 Solving the Ambiguous Case (No Such
Triangle) (cont.)
An attempt to sketch the triangle leads to this figure.
7-32
7.2
Example 2 Solving the Ambiguous Case (Two Triangles)
Solve triangle ABC if A = 61.4°, a = 35.5 cm, and
b = 39.2 cm.
Use the Law of Sines to find B.
7-33
7.2
Example 2 Solving the Ambiguous Case (Two Triangles)
(cont.)
There are two angles between 0° and 180° such that
sin B ≈ .9695:
so
is a valid possibility.
Solve separately for triangles
7-34
7.2
Example 2 Solving the Ambiguous Case (Two Triangles)
(cont.)
7-35
7.2
Example 2 Solving the Ambiguous Case (Two Triangles)
(cont.)
7-36
7.2
Example 3 Solving the Ambiguous Case (One Triangle)
Solve triangle ABC if B = 68.7°, b = 25.4 in., and
a = 19.6 in.
Use the Law of Sines to find A.
7-37
7.2
Example 3 Solving the Ambiguous Case (One Triangle)
(cont.)
There are two angles between 0° and 180° such that
sin A ≈ .7189:
is not a valid possibility.
There is only one triangle
7-38
7.2
Example 3 Solving the Ambiguous Case (One Triangle)
(cont.)
7-39
7.2
Example 4 Analyzing Data Involving an Obtuse Angle
Without using the law of sines, explain why no
triangle exists satisfying B = 93°, b = 42 cm, and
c = 48 cm.
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.
7-40
7.3 The Law of Cosines
Derivation of the Law of Cosines ▪ Solving SAS and SSS
Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a
Triangle
7-41
7.3
Example 1 Using the Law of Cosines in an Application
(SAS)
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?
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.
7-42
7.3
Example 1 Using the Law of Cosines in an Application
(SAS)
Use the law of cosines
because we know the lengths
of two sides of the triangle and
the measure of the included
angle.
The two boats are about 94.3 km apart.
7-43
7.3
Example 2 Using the Law of Cosines to Solve a Triangle
(SAS)
Solve triangle ABC if B = 73.5°,
a = 28.2 ft, and c = 46.7 ft.
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.
7-44
7.3
Example 2 Using the Law of Cosines to Solve a Triangle
(SAS) (cont.)
Now use the law of sines to find the
measure of another angle.
47.2 ft
7-45
7.3
Example 3 Using the Law of Cosines to Solve a Triangle
(SSS)
Solve triangle ABC if
a = 25.4 cm, b = 42.8 cm, and
c = 59.3 cm.
Use the law of cosines to find
the measure of the largest
angle, C.
7-46
7.3
Example 3 Using the Law of Cosines to Solve a Triangle
(SSS) (cont.)
Use either the law of sines
or the law of cosines to find
the measure of angle B.
118.6°
7-47
7.3
Example 3 Using the Law of Cosines to Solve a Triangle
(SSS) (cont.)
Alternative:
118.6°
7-48
7.3
Example 4 Designing a Roof Truss (SSS)
Find the measure of angle C
in the figure.
7-49
4.5
Harmonic Motion
Simple Harmonic Motion ▪ Damped Oscillatory Motion
4-50
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.
4-51
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
4-52
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.
4-53
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.
The motion is harmonic because the model is of the
form
a = 2.5, so the object oscillates 2.5 meters in either
direction from the starting point.
and
4-54
```