Download 4.3 Right Triangle Trigonometry

Section 4.3 Notes Page 1
4.3 Right Triangle Trigonometry
This is a very important section since we are giving definitions for the six trigonometric functions you be using
throughout the rest of this course and beyond. We need to first start with a drawing of a right triangle. The
following definitions only apply to RIGHT TRIANGLES.
sin  
opposite
hypotenuse
csc  
hypotenuse
opposite
cos  
adjacent
hypotenuse
sec  
hypotenuse
adjacent
tan  
opposite
adjacent
cot  
adjacent
opposite
EXAMPLE: Find the exact value of the 6 trig functions using the following figure:
5

12
First we need to find the missing side. In this case it is the hypotenuse. In order to find this we need to use the
formula a 2  b 2  c 2 . The side opposite the right angle is always side c. So we have 12 2  5 2  c 2 .
Simplifying we will get 144  25  c 2 , or 169  c 2 . We will get c  13 , however our answer is c = 13 since
we can’t have a negative side. Now we are ready to write our 6 trig functions. Here the hypotenuse is 13,
opposite is 5, and the adjacent side is 12. We will put these into the formulas and that will be our answers.
13

sin  
5
13
csc  
13
5
cos  
12
13
sec  
13
12
tan  
5
12
cot  
12
5
5
12
Section 4.3 Notes Page 2
EXAMPLE: Find the exact value of the 6 trig functions using the following figure:
7

4
First we need to find the missing side. In this case it is the hypotenuse. In order to find this we need to use the
formula a 2  b 2  c 2 . The side opposite the right angle is always side c, which is 7 in our figure. So we have
4 2  b 2  7 2 . Simplifying we will get 16  b 2  49 , or 33  c 2 . We will get c  33 . Now we are ready to
write our 6 trig functions. Here the hypotenuse is 7, opposite is 33 , and the adjacent side is 4. We will put
7
. This needs to be
these into the formulas and that will be our answers. The answer for cosecant is
33
7
33 7 33
rationalized by multiplying top and bottom by the square root of 33:


. You want to always
33
33 33
rationalize so that there is no square root in the denominator.
sin  
7
33
7
7
csc  
33
=
7 33
33
=
4 33
33
33
cos  

4
7
sec  
7
4
4
tan  
33
4
cot  
4
33
30 – 60 – 90 Triangle
60
2x

x
30
In this triangle the opposite side of 30 degrees is always half of the
hypotenuse. The adjacent side is always 3 times the opposite. From
this relationship we can get values for 30 and 60 degrees.

x 3
sin 30 
x 1

2x 2
sin 60 
x 3
3

2x
2
cos 30 
3
x 3

2x
2
cos 60 
x 1

2x 2
tan 60 
x 3
 3
x
tan 30 
x
x 3

1
3

3
3
Section 4.3 Notes Page 3
45 – 45 – 90 Triangle
45 
In this triangle the opposite and adjacent sides are the same. The
hypotenuse is always 2 times the opposite or adjacent. From
this relationship we can get values for 45 degrees.
x
x 2
45 
x
sin 45 
cos 45 
tan 45 
x
x 2

1

1
x
x 2
2

2
2

2
2
2
x
1
x
You will be referring to the following table A LOT, so please bookmark it. This summarizes what we just
covered in this section and the previous section. I am also including the 90 degree angle at this time.
Table of trigonometric values
 (degrees)  (radians) sin 
cos 
tan 
0
0
0
1
0
30

1
2
3
2
2
2
1
2
0
3
3
1
6

45
4

60
3

90
2
2
3
2
1
3
undefined
2
EXAMPLE: Find the exact value without using a calculator: 2 cos 2 30   sin 30 
This can be rewritten as: 2cos 30    sin 30  . Now substitute values by using the table.
2
2
 3
3
1
3 1
1
  . Now square everything inside the parenthesis: 2   . Reduce this fraction:  . After
2

2
2 2
4 2
 2 
subtracting we get 1.
Section 4.3 Notes Page 4
1  cos 60
.
sin 60 

EXAMPLE: Find the exact value without using a calculator:
1
2 . We can subtract on the top to get:
We can put in values right away from the table:
3
2
1
3
1 2

.
over the bottom fraction and multiply to get: 
which is
3
2 3
3
1
EXAMPLE: Find the exact value without using a calculator: cos

3
sec

3
 cot

3
1
2 . We can flip
3
2
.
We don’t have a cotangent or secant on our table, so we will need to first convert this into a sine by using the
1
1


 1

identity: sec 
and cot 
. From the table, we know that cos  and tan  3 .
3
3 cos 3
3 tan  3
3 2
We can put in these values into our expression:
3 3 3
1 1
1
1 2
3


  
 1

.
3
3
2 12
3 2 1 3
Cofunction Identities
sin   cos90   
csc   sec90   
cos   sin 90   
sec   csc90   
tan   cot 90   
cot   tan 90   
EXAMPLE: Write the following as an equivalent cosine expression: sin 33 .
We can use a Cofunction Identity for this. We will use sin   cos90    . Here   33 . Substituting we get
sin 33  cos90  33 . So our answer is: sin 33  cos 57 . If you put both of these in your calculator you should
get the same decimal.
EXAMPLE: Simplify and find the EXACT value: tan 35  sec 55  cos 35
We need to use a Cofunction Identity to get rid of the 55 angle. The only formula we can use for secant is
sec   csc90    . We will get: sec 55  csc90  55 . We will get: sec 55  csc 35 . Now our problem
becomes: tan 35  csc 35  cos 35 . Let’s now change everything into sines and cosines:
sin 35
1
cos 35


. We can cross cancel the sines. Then the cosines also cancel and we get 1.
cos 35 sin 35
1
Section 4.3 Notes Page 5
EXAMPLE: Use trigonometric functions to find x and y:
58
x
30 
y
Let’s first start with x. We want to choose a trig function that relates the side I want to find with a side that is
given. In our drawing the opposite side is x and the hypotenuse is 58. You want to choose the trig function that
relates these two sides, which is sine. Now we can use the definition of sine to set up our equation:
sin 30  
x
58
58 sin 30   x
58 
1
 29
2
Now we can solve for x by using cross multiplication
From the table we know that sin 30  
1
.
2
So we know x = 29.
Now we can solve for y. I will solve this the same way as above. Note you could also use a 2  b 2  c 2 fo find
the missing side since we have a right triangle. This time y is the adjacent side and 58 is the hypotenuse. The
trig function that relates these two sides will be cosine. The formula is:
cos 30  
y
58
58 cos 30   y
58 
3
 29 3
2
More on next page…
Cross multiply
From the table we know that cos 30  
So we know that y  29 3 .
3
2
Section 4.3 Notes Page 6
EXAMPLE: A ladder is leaning against a building and forms an angle of 72 degrees with the ground. If the
foot of the ladder is 6 feet from the base of the building how far up the building does the ladder reach? How
long is the ladder?
First let’s draw a picture that describes what is happening. A ladder leaning against a house will give us a right
triangle:
This time we want to solve for y. We can do this the same as in the
previous problem. We want to find a trig function that relates y (opposite)
and 6 (adjacent). This would be tangent. Now we will set up an equation
and solve for y, which is the first thing they are asking us to find.
x
y
72 
6
tan 72  
y
6
Now cross multiply.
6 tan 72   y
This time we need to use our calculator. Make sure your calculator is in degree mode.
y = 18.47 feet
You can just round to 2 decimal places.
For the second question we need to solve for x, which will give us the length of the ladder. We need to use
cosine this time since we have an adjacent side and a hypotenuse.
cos 72  
6
x
x cos 72   6
x
6
 19.42
cos 72 
Cross multiply.
Divide both sides by cos 72 
So the length of the ladder is 19.42 feet.
Angle of elevation and depression
When you look up at something you have an angle of elevation, and when you look down on something you
have an angle of depression.
Section 4.3 Notes Page 7
EXAMPLE: While standing 4000 feet away from the base of the CN tower in Toronto, the angle of elevation
was measured to be 24.4 degrees. Find the height of the tower.
First we need a picture. The angle of elevation is 24.4 degrees, so this is as we look up to the top of the tower.
I will call the top of the tower x.
x
24.4
Now we need an equation that relate the opposite side (x) and the adjacent
side (4000). This is tangent. So we can set up an equation.

4000
tan 24.4  
x
4000
Cross multiply.
4000 tan 24.4   x
Make sure your calculator is in degree mode and solve for x.
x  1814.48 feet
EXAMPLE: An observer in a lighthouse is 66 feet above the surface of the water. The observer sees a ship and
finds the angle of depression to be 0.7  . Estimate the distance from the ship to the base of the lighthouse in
miles.
We first draw a picture. Notice that the 0.7 degrees is measured from the dotted line since it is an angle of
depression. Remember that angles of elevation and depression are always measured from the HORIZONTAL.
0.7 
66
0.7 
From geometry we know that if we have to parallel lines the
alternate interior angles are the same. So this is where the 0.7 
comes from that is inside the triangle. Now we can use tangent
again since it relates the opposite and adjacent sides.
x
tan 0.7  
66
x
x tan 0.7   66
x
66
 5401.09 feet
tan 0.7 
5401.09 ft 1 mile

 1.02 miles
1
5280 ft
Cross multiply.
Divide both sides by tan 0.7  .
This is in feet, so we need to convert it. We know 1 mile = 5280 feet.