Download Chapter 7 Sections 1 and 2

Chapter 7 Sections 1, 2, and 3
Application problems with trigonometric functions and areas.
Use your calculator to do your homework. Test problems will be only “famous” angles.
Right triangle applications
Suppose theta is an acute angle.
Let’s discuss angle of elevation and angle of depression:
An eight foot ladder leans against a 15 foot wall. The angle of elevation of the ladder is
60. How far up the wall is the ladder?
1
Given ABC is a right angle with sides 3, 4, and hypotenuse 5, what are the measures of
the angles?
Onnie is 5 feet tall. When looking a the top of a 120 foot building she makes an angle of
elevation of 45. How far is she from the building?
2
Given an isosceles right triangle with hypotenuse length 9 2 , what is the side length?
Given an isosceles right triangle with hypotenuse length 5, what is the side length?
What is the area of an dodecagon with a length of 2 from the center to a vertex?
3
Given an isosceles triangle with side length 3 and base length 5, what are the base angle
measures?
hint: in the homework, you’ll need a calculator. On the test: famous angles or inverse
function form is the way to do.
Given a right triangle with the hypotenuse length 12 and an angle of 35 what is the leg
length opposite the angle?
4
Trigonometric area
A = ½ ab sin x
Given triangle ABC with AB = 6cm and BC = 18 cm and measure of angle B = 120.
What is the area? Note: this is NOT a right triangle!
What are the famous angles that you’ll need to know
30
45
60
90
120
135
150
5
ATTENTION: homework change: do 39 instead of 37! F’06
Section 3:
Law of Sines and Law of Cosines
All the trigonometry information was based on having a right triangle as the starting
point. It turns out that some trigonometric facts can be extended to arbitrary triangles.
Let’s look at a triangle:
C
AC = 3.03 cm
47
23
A
B
This is called an oblique triangle, because it is not a right triangle. What can we find out
about this triangle – well, the area is ½ AC(CB)sin 110. True, but we only know one
side.
Actually, I contend that we know all the sides and all the angles. Let’s look at how I can
claim this.
C
AC = 3.03 cm
47
A
23
D
B
If I construct the altitude from C to side AB, I can turn the triangle into two right
triangles. I know that
sin 47 
CD
3.03cm
sin 23 
CD
CB
6
Solve both of these for CD and equate them:
Now look, one equation and one unknown: CB.
Which turns out to be:
CB 
3.03(sin 47)
. It’s not unknown at all, it’s about
sin 23
C
sin23
3.03 cm
47
A
3.03sin47
= 5.67
23
D
B
7
To find AB, I’ll use a different altitude and solve in a similar way
E
C
sin 47 
70
3.03
5.67
47
A
EB
EB
and sin 70 
3.03(sin 47)
AB
sin 23
23
D
B
I got the 70 by taking 180  110 and technically I’m not supposed to use the rounded off
quantity 5.67 in a further calculation…it introduces propagating round off error.
Now solve each side for EB and equate:
Now solve for AB
AB 
3.03 sin 70
which is approximately 7.29 cm.
sin 23
8
Let’s get the big picture view here:
I’m using the sines and the sides…note the two s’s…There’s a fundamental relationship
here:
C
A
B
It’s simple and I’ll expect you to know how to use it.
9
Here’s a problem  completely describe this triangle in exact measures, no rounding.
Note what you know is AAS
C
6 cm
45
A
30
B
There’s only one wrinkle…if you know SSA you can’t be sure of your answer because
the sine function has the same values for it’s Q1 and Q2 angles.
10
Let’s look at an example of this:
If we know the length of angle D, side DB, and the length of the side across from angle
D…which is the scenario for SSA.
mBDC' = 32.73
DB = 5.74 cm
BC' = 3.55 cm
BC = 3.55 cm
B
mDCB = 60.80
sin mDCB = 0.87
D
C'
mDC'B = 119.19
sin mDC'B = 0.87
C
DBsin mBDC'
sin mDCB
= 3.55 cm
DBsin mBDC'
sin mDC'B
= 3.55 cm
mDCB+mDC'B = 180.00
The key to the relationship is right there on the left – these two angles add to 180. When
you use arcsin in your calculator, you will only get the Quadrant 1 angle (why?), you
have to supply the Quadrant 2 angle on your own.
11
Here’s another problem:
Given triangle ABC
Find the measure of angle A, given that B = 32, a = 42 and b = 30. “a” is the side across
from A. Is this SSA?
How many answers might there be?
(hint using a calculator sin 1 (.7419)  48 )
12
Now on to the law of cosines…another formula series to know by heart.
It turns out that if you know 3 sides of a triangle you can get all the angle measures…and
if you know SAS you can get the third side. You use this formula:
a 2  b 2  c 2  2bc cos 
B
c
A
a

b
C
There is no ambiguity when using this formula the answer from your calculator is the
only answer.
13
Suppose we have the following scenario. What is BC?
Give your answer in exact numbers.
B
15 cm
110
A
10 cm
C
Or we have
What is the measure of angle A. Leave your
answer in exact terms, please.
B
40 ft
32 ft
C
A
20 ft
14