Download Math 10C Ch. 2 Lessons

Math 10C Ch. 2 Trigonometry Lessons
2.1 Tangent ratio
South-facing solar panels on a roof work best when the angle of inclination of the roof,
that is, the angle between the roof and the horizontal, is approximately equal to the
latitude of the house.
When an architect designs a house that will have solar panels on its roof, she has to
determine the width and height of the roof so that the panels work efficiently.
Is there an easier way to measure the angle than to use a protractor or some other
measuring device? There is! It is called Trigonometry. We are going to work with only
right triangles in this chapter. We name the sides of a right triangle in relation to one of
its acute angles. What is an acute angle? (an angle less than 90 degrees)
B
A
C
But first, let’s look at the letter names for the sides. Look at triangle ABC. Angles are
indicated by capital letters and the sides across from them are named by lowercase
letters.
B
c
a
A
C
b
We already know that the longest side in a right triangle is called the hypotenuse. The
other two sides are called opposite and adjacent. But the opposite and adjacent will
change depending on which acute angle you are using as a reference point. If we use
angle A, the side across from it is the opposite, the side beside it (that isn’t the
hypotenuse) is the adjacent and the side across from the 90 degree angle is the
hypotenuse.
B
hypotenuse
opposite
A
C
adjacent
There are three primary trigonometric ratios. The ratio that we are going to learn today
is the Tangent Ratio. It is the ratio of the opposite side over the adjacent side. We
represent it as:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑡𝑎𝑛𝐴 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
We often express it as a decimal that compares the lengths of the sides. For example,
if tanA = 1.5, that means the opposite side is 1.5 times as big as the adjacent side.
We will learn about the other two ratios in a different lesson.
Example #1:
Determine tanD and tanF.
F
3
D
Answer:
E
4
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑡𝑎𝑛𝐷 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑑
𝑓
𝑓
3
= 4 = 0.75
4
𝑡𝑎𝑛𝐹 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 𝑑 = 3 = 1.33
We can use the tangent ratio to find angles. It is more accurate than using a protractor.
There is a lovely button in our calculator that will help us find the angles. Someone
went to all the trouble to program all the tangent ratio answers for angles into the
calculator! Your calculator defaults to something called radians which we don’t want
until Grade 12. So the first thing you need to do is change your calculator to Degree
Mode. Press the Mode button on the calculator and choose Degrees with your arrow
buttons and then press enter.
So the thing to remember is that if you want an angle you need to use the 2 nd tan
button. It will show up on your calculator as 𝑡𝑎𝑛−1 .
Example #2:
Determine the measures of ∠𝐺 𝑎𝑛𝑑 ∠𝐽 to the nearest tenth of a degree.
H
4
5
J
G
Answer: In order to solve for ∠𝐺, we will use the tangent ratio.
4
𝑡𝑎𝑛𝐺 = 5
Since we want the angle, we need to use the 2nd tan button to find it.
4
𝑡𝑎𝑛−1 (5) = 38.7°
Now, if you remember triangles from junior high, you know that all the angles add to
180°. Since we will always be dealing with 90° triangles, the two acute angles add to
90°. Therefore, to find angle J, we can subtract our angle G from 90°. So,
90 − ∠𝐺 = 90 − 38.7 = 51.3°
Example #3:
The latitude of Fort Smith, NWT, is approximately 60°. Determine whether this design
for a solar panel is the best for Fort Smith.
(see diagram on p. 73)
Answer: The best angle of inclination for the solar panel is the same as the latitude,
60°. So let’s find the angle of inclination (angle from the horizontal). It’s always a great
idea to draw a triangle. The angle that we are looking for is C. Let’s set up the tangent
ratio.
A
9 ft
C
B
12 ft
9
𝑡𝑎𝑛𝐶 = 12
Since we want the angle, we use the 2nd tan button.
9
𝑡𝑎𝑛−1 (12) = 37°.
Since the angle we found is not 60°, this design is not good for Fort Smith.
Example #4:
A 10 ft. ladder leans against the side of a building with its base 4 ft. from the wall. What
angle, to the nearest degree, does the ladder make with the ground?
Answer: Again, it is a good idea to draw a diagram.
P
10 ft.
Q
R
4 ft.
However, this time we don’t have the opposite and adjacent side so we can’t use the
tangent ratio. There are two other ratios, but we will learn them later.
BUT, we have our old friend Pythagorus to help us! We can find the missing side and
then use tangent.
102 − 42 = 𝑟 2
84 = 𝑟 2
√84 = 𝑟
It is a good idea to use the square root version; however, you can use the decimals as
long as you use all of them.
𝑡𝑎𝑛𝑅 =
√84
4
We need to use the 2nd tan button.
√84
)
4
𝑡𝑎𝑛−1 (
= 66°.
Homework: p. 75 #3 – 6 (last letter), 9, 10d, 11c, 13, 15, 18, 20
2.2 Tangent Ratio (often done on same day as 2.1)
In our last lesson we found angles. Now we are going to also learn how to find sides.
Example #1:
Determine the length of AB to the nearest tenth of a centimeter.
B
10 cm
C
30°
A
Answer: This time since we are looking for a side, we are going to use the regular tan
button.
𝑡𝑎𝑛30° =
𝐴𝐵
10
Since the unknown is on the top of the fraction, we need to cross multiply to find the
answer.
10𝑡𝑎𝑛30° = 𝐴𝐵
AB = 5.8 cm
Example #2:
Determine the length of d to the nearest tenth of a centimeter.
E
F
20°
3.5 cm
D
Answer: We are looking for side d which is adjacent to 20°. Set up the tangent ratio:
𝑡𝑎𝑛20° =
3.5
𝑑
This time the unknown is on the bottom of the equation so we would have to cross
multiply and then divide. We can just save some time and remember that we just have
to divide the side by the tangent with its angle.
3.5
𝑑 = 𝑡𝑎𝑛20° = 9.6 𝑐𝑚.
Example #3:
A searchlight beam shines vertically on a cloud. At a horizontal distance of 250 m from
the searchlight, the angle between the ground and the line of sight to the cloud is 75°.
Determine the height of the cloud to the nearest metre.
(see p. 81 for the diagram)
Answer:
C
S
75°
P
250 m
Let’s set up the tangent ratio to find our answer.
𝑝
𝑡𝑎𝑛75° = 250 = 250𝑡𝑎𝑛75° = 933 𝑚
Homework: p. 82 #3 – 5(last letter), 6, 7, 11, 12, 15
2.3 Measuring an Inaccessible Height
Tree farmers use a clinometer to measure the angle between a horizontal line and the
line of sight to the top of a tree. They measure the distance to the base of the tree.
How could they use the tangent ratio to calculate the height of the tree?
Do the clinometer experiment using the clinometers from the Math closet of fun. Have
the students use the clinometers to measure the angle to an object high up in the
classroom. Have them measure the horizontal ground distance as well. Show them
how to use the tangent ratio to solve.
x
A
y
15
1.5 m
10 m
We would use the tangent ratio to find the x value.
𝑥
𝑡𝑎𝑛15° = 10 = 10𝑡𝑎𝑛15° = 2.6 …
Then we need to add the height of the person holding the clinometer to get the actual
height (make sure you keep all the decimals)
2.6 … + 1.5 = 4.2 𝑚
Clinometers are very helpful to calculate distances that you can’t actually measure the
vertical height. Surveyors use something called a transit to measure heights as well.
Now, instead of assigning homework, we will do p. 86 #1 – 3 together and that counts
as homework.
1. Explain how the angle shown on the protractor of your clinometer is related to the
angle of inclination that the clinometer measures.
Answer: on a regular clinometer, you use whatever angle the clinometer gives you.
However, when you use a homemade protractor clinometer, it actually measures the
angle at the top of the clinometer. So, if you want the angle at the bottom, remember
that the two acute angles in a triangle add to 90°. So if we want our bottom angle, just
subtract the protractor angle from 90°.
2. A tree farmer stood 10.0 m from the base of a tree. She used a clinometer to
sight the top of the tree. The angle shown on the protractor scale was 40°. The
tree farmer held the clinometer 1.6 m above the ground. Determine the height of
the tree to the nearest tenth of a metre. The diagram is not drawn to scale.
Answer:
40
x
y
A
1.6 m
10 m
If you take a look at the diagram, to get the bottom angle (angle of inclination), subtract
90 – 40 = 50. Then set up the tangent ratio:
𝑥
𝑡𝑎𝑛50° = 10 = 10𝑡𝑎𝑛50° = 11.9 …
Then we need to add the height of the person holding the clinometer to get the actual
height (make sure you keep all the decimals)
11.9 … + 1.6 = 13.5 𝑚
3. Use the information in the diagram to calculate the height of a totem pole
observed with a drinking-straw clinometer. Give the answer to the nearest metre.
The diagram is not to scale.
Answer:
12
x
y
A
1.5 m
5.0 m
If you take a look at the diagram, to get the bottom angle (angle of inclination), subtract
90 – 12 = 78. Then set up the tangent ratio:
𝑡𝑎𝑛78° =
𝑥
5.0
= 5.0𝑡𝑎𝑛78° = 23.5 …
Then we need to add the height of the person holding the clinometer to get the actual
height (make sure you keep all the decimals)
23.5 … + 1.5 = 25 𝑚
2.4 The Sine and Cosine ratios
Sometimes we don’t have the opposite and adjacent side, so we can’t use the tangent
ratio. Since there are three sides to a triangle, we actually have three ratios we can
use.
For example, the railroad track through the mountains between Field, BC and Hector,
BC, includes spiral tunnels. They were built in the early 1900s to reduce the angle of
inclination of the track between the two towns. You can see a long train passing under
itself after it comes out of a tunnel before it has finished going in.
Here is a diagram of the track (not to scale) before the tunnels were constructed:
Hector
6.6 km
297 m
Field
We would not be able to use the tangent function unless we used Pythagorus. So we
are going to use a different ratio.
The other two ratios are the sine ratio and the cosine ratio. Their formulas look like this:
𝑠𝑖𝑛𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑐𝑜𝑠𝐴 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
We can actually use the “word” SOHCAHTOA to help us remember the ratios:
S:
O:
H:
C:
A:
H:
T:
O:
A:
Sine
opposite
hypotenuse
Cosine
adjacent
hypotenuse
Tangent
opposite
adjacent
The tangent, sine and cosine ratios are called the primary trig ratios. The word
trigonometry comes from three Greek words “tri + gonia + metron” that together mean
“three angle measure”.
Example #1
a) In triangle DEF, identify the side opposite ∠𝐷 and the side adjacent to ∠𝐷.
Answer:
E
12 cm
D
5 cm
13 cm
F
The side opposite ∠𝐷 would be called either EF or d and is the 5 cm side.
The side adjacent ∠𝐷 would be called either DE or f and is the 12 cm side. The side
that is 13 cm is across from the 90 degree angle so it is the hypotenuse. It can’t be the
adjacent side even through it is beside ∠𝐷.
b) Determine sinD and cosD to the nearest hundredth.
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
5
Answer: 𝑠𝑖𝑛𝐷 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 13 = 0.38
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
12
𝑐𝑜𝑠𝐷 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 13 = 0.92
Example #2:
Determine the measures of ∠𝐺 and ∠𝐻 to the nearest tenth.
K
6
G
14
H
Answer: Let’s start by finding ∠𝐻. If we use ∠𝐻 as our reference point, 6 is the side
opposite of H and 14 is the hypotenuse. That means we will need to use the sine ratio.
6
𝑠𝑖𝑛𝐻 = 14
To find the angle, we need to use the 2nd sin button.
6
𝑠𝑖𝑛−1 (14) = 25.3 …
∠𝐻 = 25.3°
To find ∠𝐻, just remember that the two acute angles need to add to 90. So to find ∠𝐻,
simply do: 90 – 25.3 = 64.7°.
Example #3
A water bomber is flying at an altitude of 5000 ft. The plane’s radar shows that it is
8000 ft from the target site. What is the angle of elevation of the plane measured from
the target site, to the nearest degree?
Answer: it is actually helpful to draw a diagram to represent the situation.
A
8000 ft.
R
5000 ft.
X
We can always assume that the altitude is being measured vertically and that the
ground is perfectly horizontal. If we are looking for the angle of elevation, that is always
the angle up from the horizontal. Therefore, we are looking for ∠𝑅. Since we have the
5000 ft opposite ∠𝑅 and 8000 ft is the hypotenuse, we are going to use the sine ratio.
𝑠𝑖𝑛𝑅 =
5000
8000
To find the angle, we use the 2nd sine button.
𝑠𝑖𝑛−1 (
5000
) = 39°
8000
Therefore, our angle of elevation is 39°.
Homework: p. 95 #4 - 6 (last letter), 7 - 8 (odd letters), 9c, 10cd, 12 - 15
2.5 Sine and Cosine Ratios Part 2 (usually done the same day as 2.4)
We are going to continue with the sine and cosine ratios from the last lesson. However,
instead of finding angles, we are going to find side lengths.
Example #1:
Determine the length of BC to the nearest tenth of a centimetre.
B
50°
C
5.2 cm
A
Answer: Since we are looking for side BC (which is also called “a”), we need to pick an
angle as a reference point. Since we are never allowed to use the right angle, we need
to use one of the acute angles. We will use 50° because it is the one we are given.
From 50°, we have 5.2 cm which is the hypotenuse and we want BC which is adjacent.
Therefore, we are going to use the cosine ratio.
𝑎
𝑐𝑜𝑠50° = 5.2
Since we have the angle, we will use the regular cosine button. Since our unknown is
on the top, we will multiply.
5.2𝑐𝑜𝑠50° = a
3.3 cm = a
Example #2
Determine the length of DE (f) to the nearest tenth of a centimetre.
F
E
55°
6.8 cm
D
Answer: Again, we need to use an angle as a reference point. So we will use the 55°
angle. We are looking for side f, which is the hypotenuse. We are given 6.8 cm which
is the opposite. Therefore, we are going to use the sine ratio.
𝑠𝑖𝑛55° =
6.8
𝑓
Again, we are going to use the regular sine button since we have the angle. However,
since the unknown is in the denominator, we are going to divide in order to solve.
6.8
𝑓 = 𝑠𝑖𝑛55° = 8.3 𝑐𝑚
Example #3
A surveyor made the measurements shown in the diagram. How could the surveyor
determine the distance from the transit to the survey pole to the nearest hundredth of a
metre?
P
T
67.3°
20.86 m
S
Answer: we are given 67.3°, so we will use it as our reference point. We want the
hypotenuse and we have the adjacent. So we are going to use the cosine ratio.
𝑠
𝑐𝑜𝑠67.3° = 20.86
Since we have the angle, use the regular sine button. Since our unknown is in the
numerator, we multiply.
20.86𝑐𝑜𝑠67.3° = 𝑠
54.05 = s
Homework: p. 101 #3 - 5 (last letter), 6, 7, 9, 11, 12 ii, 14 (challenge)
2.6 Applying the Trig Ratios
By now we should be getting fairly comfortable with our trig ratios. We are just going to
add one extra concept in this section: solving triangles. To solve a triangle means to
find all the sides and angles. There are two types of triangles: given 2 sides or given 1
side and 1 angle. We will do one of each example.
Example #1: (given 2 sides)
Solve triangle XYZ. Give the measures to the nearest tenth.
Y
6.0 cm
10.0 cm
X
Z
Answer: we are missing ∠𝑋, ∠𝑍 𝑎𝑛𝑑 𝑠𝑖𝑑𝑒 𝑦. We are going to start by solving for side y.
We can use another useful formula called Pythagorus’ theorem….remember him?
Since we are looking for the hypotenuse, we would use:
62 + 102 = 𝑦 2
36 + 100 = 𝑦 2
136 = 𝑦 2
𝑦 = √136 = 11.7 𝑐𝑚
You can choose which angle you want to find. We will find ∠𝑍. Using ∠𝑍 as a
reference point, we have 6.0 which is opposite and 10.0 which is adjacent. That means
we will use tangent.
6.0
𝑡𝑎𝑛𝑍 = 10.0
Since we want the angle, we use the 2nd tan button.
6.0
𝑡𝑎𝑛−1 (10.0) = 30.9 … . = 31.0°
To find ∠𝑋, simply subtract 90 – 31.0 = 59.0°. Therefore, we found our three missing
quantities: y = 11.7 cm, ∠𝑍 = 31.0° 𝑎𝑛𝑑 ∠𝑋 = 59.0°.
Example #2: (given a side and an angle)
Solve this triangle. Give the measures to the nearest tenth where necessary.
D
5.0 cm
F
25°
E
We need to find ∠𝐷 and side f and side d. It is easy to find ∠𝐷 because ∠𝐷 𝑎𝑛𝑑 ∠𝐸 add
to 90°. So, 90 – 25 = 65°. To find side f, we will use 25° as a reference point. Side f is
the hypotenuse and we have 5.0 which is the opposite. We will use the sine ratio.
𝑠𝑖𝑛25 =
5.0
𝑓
Since we have the angle, we will use the regular sine button. Since the unknown is on
the bottom, we will divide.
5.0
𝑓 = 𝑠𝑖𝑛25 = 11.8 𝑐𝑚
To find side d, use 25° as a reference point again. This time we have 5.0 which is the
opposite side, and we want d which is the adjacent side. We will use the tangent ratio.
𝑡𝑎𝑛25 =
5.0
5.0
𝑑
𝑑 = 𝑡𝑎𝑛25 = 10.7 𝑐𝑚
Example #3:
A small table has the shape of a regular octagon. The distance from one vertex to the
opposite vertex, measured through the centre of the table, is approximately 30 cm.
There is a strip of wood veneer around the edge of the table. What is the length of this
veneer to the nearest centimetre?
30 cm
Answer: To determine the length of veneer, calculate the perimeter of the surface of the
table. Since the surface of the table is a regular octagon, congruent isosceles triangles
are formed by drawing line segments from the centre of the surface to each vertex.
15 cm
Take a look at the isosceles triangle. The top angle would be
45°
15 cm
15 cm
360
8
= 45.
If we draw a line down the centre of the triangle, we can find a 90°. Since it is isosceles,
the top angle splits into 22.5°.
22.5°
15 cm
x
To find side x, we will use 22.5° as the reference point. We are looking for side x which
is opposite. We have 15 cm which is hypotenuse. We will use the sine ratio.
𝑥
𝑠𝑖𝑛22.5 = 15
15𝑠𝑖𝑛22.5 = 𝑥
𝑥 = 5.7 …
Now if we double the answer, we will get the bottom of our original isosceles triangle.
Then if we multiply by 8 we will get the perimeter of the entire octagon.
𝑥 = 5.7 … (2)(8) = 91.8 … = 92 𝑐𝑚
Homework: p. 111 #3 – 6 (last letter), 7, 9, 11, 13, 15
2.7 Solving Problems Involving More than one right triangle
We are going to continue to use trigonometry, but we are going to look at shapes that
have more than one right triangle.
Example #1:
Calculate the length of CD to the nearest tenth of a centimetre.
C
B
47°
26°
A
4.2 cm
D
Answer: We don’t have any sides in triangle BCD, so we need to find a side in ABD
that we can use to help us.
Since they both share BD, we could find that side and solve for CD. Using ∠𝐵 (47°) as
our reference point, we have 4.2 cm which is opposite and we want BD which is the
hypotenuse. We will therefore use the sine ratio.
4.2
𝑠𝑖𝑛47 = 𝐵𝐷
Since we have the angle, we will use the regular sine button. Since the unknown is on
the bottom, we will divide.
4.2
𝐵𝐷 = 𝑠𝑖𝑛47 = 5.7 …
Make sure that you keep all the decimals in your next calculation. Now, we will use 26°
as our reference point. We want CD which is the adjacent side and we just found 5.7…
which is the hypotenuse. We will use the cosine ratio.
𝐶𝐷
𝑐𝑜𝑠26 = 5.7… = 5.7 … 𝑐𝑜𝑠26 = 5.2 𝑐𝑚
Example #2:
From the top of a 20m high building, a surveyor measured the angle of elevation of the
top of another building and the angle of depression of the base of that building. The
surveyor sketched this plan of her measurements. Determine the height of the taller
building to the nearest tenth of a metre. (diagram is not to scale)
P
x
S
30°
15°
Q
20m
R
T
Answer: First, what is an angle of depression? It is the angle DOWN from the
horizontal. In this diagram, 15° is the angle of depression and 30° is the angle of
elevation (since it the angle UP from the horizontal).
In order to find the height of the taller building, we need to find the value of x. However,
we don’t have any sides in that right triangle. So, we need to find a side in the bottom
right triangle. Both triangles share side QS. To find side QS, we will use the 15° as our
reference point. We have RS, which is the opposite side and we want QS which is the
adjacent side. We are going to use the tangent ratio.
20
𝑡𝑎𝑛15 = 𝑄𝑆
20
𝑄𝑆 = 𝑡𝑎𝑛15 = 74.6 …
Now, let’s use 30° as our reference point. We want x which is the opposite and we
have 74.6… which is the adjacent. We will use tangent again.
𝑥
𝑡𝑎𝑛30 = 74.6… = 74.6 … 𝑡𝑎𝑛30 = 43.0 …
To find the total height of the building, we need to add 20 + 43.0… = 63.0…= 63.1 m.
Example #3
From the top of a 90-ft observation tower, a fire ranger observes one fire due west of
the tower at an angle of depression of 5°, and another fire due south of the tower at an
angle of depression of 2°. How far apart are the fires to the nearest foot? The diagram
is not drawn to scale.
5°
2°
90 m
Fire #2
Fire #1
Answer: the angles of depression are not useful to us outside of the triangles.
However, if you remember the transversals and parallel lines from junior high, you know
that that bottom angle in the triangle is the same as the angle of depression.
5°
2°
90 m
5°
2°
Fire #1
x
Fire #2
Let’s look at the triangle with Fire #1.
90 m
2°
Fire #1
x
Since we are using 2° as our reference point, we have 90m which is opposite and we
want x which is adjacent. We will use the tangent ratio.
𝑡𝑎𝑛2 =
90
𝑥
90
𝑥 = 𝑡𝑎𝑛2 = 2577.2 …
In the second triangle, we will do the same ratio, except now we are using 5°.
𝑡𝑎𝑛5 =
90
𝑥
90
𝑥 = 𝑡𝑎𝑛5 = 1028.7 …
To find the distance between the fires, we will use pythagorus since the distance is the
third side of a right triangle! (we are finding x in the diagram)
2577.2 … 2 + 1028.7 …2 = x 2
√2577.2 … 2 + 1028.7 …2 = x
x = 2775 ft.
Homework: p. 118 #3 – 5 (last letter), 6, 9, 11, 12, 14, 16, 17, 20