Download Unit 5: Trigonometry

Grade 10 Essentials Math
Trigonometry
Unit 5: Trigonometry
Introduction:
This unit deal with similar triangles, Pythagorean Theorem, and the trigonometric ratios of sine,
cosine, and tangent. You will be using these to solve word problems as well. Trigonometry is
based on the relationship between the measure of the angles and the lengths of the sides of a right
angle triangle. These skills are necessary in occupations such as carpentry, aviation and
astronomy.
Assessment:
o
o
Learning Activity 5.1: Similar Triangles
Learning Activity 5.2: Similar Triangles Word Problems
o
o
o
o
o
Lesson 2 Assignment: Problem Solving using Pythagorean Theorem
Learning Activity 5.3: SINE Ratio
Learning Activity 5.4: COSINE Ratio
Learning Activity 5.5: TANGENT Ratio
o
o
o
Lesson 1 Assignment: Similar Triangles
Lesson 3 Assignment: SOH CAH TOA
Lesson 4 Assignment: Word Problems using Trigonometric Ratios
Lesson 5 Assignment: Angle of Elevation and Depression
Unit 5 Test: Trigonometry
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Grade 10 Essentials Math
Trigonometry
LESSON 1: SIMILAR TRIANGLES:
Similarity:
Similar triangles have the same shape, but they may be different in size (the lengths of the sides
are different). By definition, shapes are similar if they have the same angles and the sides are
proportional. To solve similar triangles we need to use proportions. It is important to label the
corresponding
Definitions:
1. Corresponding: sides have two angles that are the same size.
2. Proportional: means that it is a scaled up or a scaled down version of the other and all sides
of the triangle are scaled down (or up) by the same multiple.
x
a
2 cm
6 cm
6 cm
b
y
2 cm
18 cm
6 cm
c
z
These triangles are similar because their corresponding angles are the same and the sides are
proportional, to get the size of the smaller triangle sides we must divide the larger triangle sides
by 3.
2
Grade 10 Essentials Math
Trigonometry
Example:
Find the values for m and t.
A
X
6 mm
Z
m
Y
8mm
t
15mm
C
12mm
B
Solution:
Step 1: locate the corresponding sides on each triangle; to do this we match two pairs of angles
that are the same size. Angles that are the same size are indicated using the same symbols.
∠𝑋 = ∠𝐴
∠𝑌 = ∠𝐵
𝑆𝑜 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑠𝑖𝑑𝑒 𝑚 𝑚𝑎𝑡𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑑𝑒 15𝑚𝑚
∠𝑋 = ∠𝐴
∠𝑍 = ∠𝐶
𝑆𝑜 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑠𝑖𝑑𝑒 6𝑚𝑚 𝑚𝑎𝑡𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑑𝑒 𝑡
𝑆𝑜 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑠𝑖𝑑𝑒 8𝑚𝑚 𝑚𝑎𝑡𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑑𝑒 12𝑚𝑚
Step 2: Solve for the missing sides using proportions:
𝑚
8
=
15 12
8
𝑚 = ( ) (15) = 10𝑚𝑚
12
𝑡 12
=
6
8
𝑡=(
12
) (6) = 9𝑚𝑚
8
3
Grade 10 Essentials Math
Trigonometry
Example:
Sovle for a
A
18
D
10
5
C
a
E
B
Solution:
Step 1: find the corresponding sides: Remember, you must use the angles that are the same size
(same symbol) to find the corresponding sides.
∠𝐴 = ∠𝐸
∠𝐵 = ∠𝐷
𝑆𝑜 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑠𝑖𝑑𝑒 10 𝑚𝑎𝑡𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑑𝑒 5
∠𝐴 = ∠𝐶
∠𝐶 = ∠𝐷
𝑆𝑜 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑠𝑖𝑑𝑒 18 𝑚𝑎𝑡𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑑𝑒 𝑎
Step 2: Solve for the missing sides using proportions:
𝑎
5
=
18 10
𝑎=(
5
) (18) = 9
10
4
Grade 10 Essentials Math
Trigonometry
Learning Activity 5.1: Similar Triangles
The triangles in each pair are similar. Find the unknown side lengths.
A
a)
E
35
15
5
●
C
B
F
x
●
D
E
b)
C
♣
x
A
18
B
●
48
z
●
24
36
♣
D
c)
A
X
m
Y
35
C
Z
10
B
20
5
Grade 10 Essentials Math
Trigonometry
LESSON 2: WORD PROBLEMS WITH SIMILAR TRIANGLES:
Example:
One of the 7 natural wonders of the world is the Grand Canyon. From points along the canyon a
surveyor was able to take measurements and sketched the following diagram. What is the width
across the canyon?
D
40 km
20 km
E
A
°
C
°
200m
Width of
canyon
x
8 km
B
Solution:
Step 1: find the corresponding sides: Remember, you must use the angles that are the same size
(same symbol) to find the corresponding sides.
∠𝐷 = ∠𝐵
∠𝐸 = ∠𝐶
𝑆𝑜 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑠𝑖𝑑𝑒 20𝑘𝑚 𝑚𝑎𝑡𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑑𝑒 8𝑘𝑚
∠𝐷 = ∠𝐵
∠𝐴 = ∠𝐴
𝑆𝑜 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑠𝑖𝑑𝑒 40𝑘𝑚 𝑚𝑎𝑡𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑠𝑖𝑑𝑒 𝑥
Step 2: Solve for the missing sides using proportions:
𝑥
8
=
40 20
𝑎=(
8
) (40) = 16𝑘𝑚
20
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Grade 10 Essentials Math
Trigonometry
Example:
Two ladders are leaning against a wall such that they make the same angle with the ground. The
10 foot ladder reaches 8 feet up the wall, how high up the wall does the 18 foot ladder reach?
Solution:
x ft
18 ft
10 ft
8 ft
𝑥 18
=
8 10
18
𝑎 = ( ) (8) = 14.4 𝑓𝑡
10
Example:
At a certain time of the day, the shadow of a 5 foot tall boy is 8 feet long. The shadow of a tree at
the same time is 28 feet long. How tall is the tree?
Solution:
x ft
5 ft
8 ft
28 ft
𝑥 28
=
5
8
28
𝑎 = ( ) (5) = 17.5 𝑓𝑡
8
7
Grade 10 Essentials Math
Trigonometry
Learning Activity 5.2: Similar Triangles Word Problems
1. To find the distance PQ across the farm pond, Marty marks out points R and S. By
measuring, she finds that RS = 5.7 m, OP = 19.5 m, and OS = 4.2 m. What is the distance
PQ?
P
Q
O
R
S
2. Two trees cast shadows as shown, how tall is the evergreen? (diagram is not to scale)
x
14m
5m
8m
3. How far is it across the river?
7m
32m
12m
x
8
Grade 10 Essentials Math
Trigonometry
Curriculum Outcomes:
10E2.TG.2.Demonstrate an understanding of trigonometric by: applying similarity to right triangles, generalizing
patterns from similar right triangles, solving problems
Lesson 1 Assignment: Similar Triangles
See your teacher for Assignment
9
Grade 10 Essentials Math
Trigonometry
LESSON 3: PYTHAGOREAN THEOREM
The Pythagorean Theorem is for right angle triangle only.
Pythagorean Theorem:
c
a
Hypotenuse
𝑎2 + 𝑏 2 = 𝑐 2
b
The hypotenuse is the side that is always opposite (across from) the 90° angle. This is
always side C in the Pythagorean Theorem.
Example:
Determine the length of the missing for the following triangle.
Solution:
x
3
𝑥 2 = 32 + 42
𝑥 2 = 9 + 16
𝑥 2 = 25
𝑥 = √25 = 5
4
Example:
Determine the length of the missing side for each of the following triangle.
Solution:
102 = 𝑥 2 + 82
100 = 𝑥 2 + 64
100 − 64 = 𝑥 2
36 = 𝑥 2
𝑥 = √36 = 6
Example:
Determine the length of the missing side for each of the following triangle.
Solution:
152 = 𝑥 2 + 72
225 = 𝑥 2 + 49
225 − 49 = 𝑥 2
176 = 𝑥 2
𝑥 = √176 = 13.2
10
x
8
15
x
7
10
Grade 10 Essentials Math
Trigonometry
Word Problems using the Pythagorean Theorem
Example:
Scott wants to swim across a river that is 400m wide. He plans to swim directly across the river
but ends up 100m downstream because of the current. How far did he actually swim?
Solution:
Step 1: Draw a diagram for the right angle triangle:
400 m
x
100 m
Step 2: Label the sides of the triangle. Pay attention to where the hypotenuse is!!!
400 m
x HYP
100 m
𝑥 2 = 4002 + 1002
𝑥 2 = 160000 + 10000
𝑥 2 = 170000
𝑥 = √170000 = 412.3𝑚
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Grade 10 Essentials Math
Trigonometry
Example:
To get from point A to point B you must avoid walking through a building. To avoid the
building, you walk 14m south and 25m east. How many metres would you have saved had the
building not been there?
Solution:
Must walk 14 + 25 = 39𝑚
x
14 m
142 + 252 = 𝑥 2
196 + 625 = 𝑥 2
821 = 𝑥 2
𝑥 = √821 = 28.65𝑚
25 m
If was able to walk through the building would only have to walk 28.65 m. So would save 39 −
28.65 = 10.35𝑚
Example:
The foot of a 6m ladder is placed 2m from the base of a building. How far up the building does
that ladder reach?
Solution:
6m
x
𝑥 2 + 22 = 62
𝑥 2 + 4 = 36
𝑥 2 = 36 − 4
𝑥 2 = 32
𝑥 = √32 = 5.7𝑚
2m
12
Grade 10 Essentials Math
Trigonometry
Example:
The diagram below shows a baseball diamond. How far must the catcher throw the ball from
home plate for the ball to reach second base
Solution:
902 + 902 = 𝑥 2
8100 + 8100 = 𝑥 2
16 200 = 𝑥 2
𝑥 = √16200 = 127.3 𝑓𝑡
The catcher must make a throw of 127.3 feet for a ball to travel from home
plate to second base
13
Grade 10 Essentials Math
Trigonometry
Curriculum Outcomes:
10E2.TG.1. Solve problems involving right triangles using the Pythagorean Theorem
10E2.TG.3. Solve problems that require the manipulation & application of the Pythagorean Theorem
Lesson 2 Assignment: Problem Solving using Pythagorean Theorem
See your teacher for Assignment
14
Grade 10 Essentials Math
Trigonometry
LESSON 4: TRIGONOMETRY
Lesson Introduction:
We will now move on to the trigonometric functions.
The three trig functions are SIN, COS, and TAN
SOH CAH TOA
Stands for:
𝑠𝑖𝑛  =
𝑜𝑝𝑝
ℎ𝑦𝑝
𝑐𝑜𝑠  =
𝑎𝑑𝑗
ℎ𝑦𝑝
𝑡𝑎𝑛  =
𝑜𝑝𝑝
𝑎𝑑𝑗
When we are using the trigonometric functions, we need to label the sides of the given right
angle triangle so that we know what number goes where in our formulas.
The three sides are:
1. Hypotenuse: is the side of the triangle that is always opposite the 90° angle
2. Opposite: is the side that is opposite (across from) the given angle,
we never use the 90° angle to find the opposite side
3. Adjacent: is the side that is adjacent (next to) the given angle
we never use the 90° angle to find the adjacent side
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Grade 10 Essentials Math
Trigonometry
SINE RATIO
𝑜𝑝𝑝
The sine ratio 𝑠𝑖𝑛𝜃 = ℎ𝑦𝑝 requires three values, the acute angle, the opposite side and the
hypotenuse. We label the opposite side by locating the angle and going to the side directly
opposite from it. The hypotenuse is located by finding the 90 angle and going to the side across
from it. You would need to know any two of the values and then you can substitute these two
known values into the formula to find the unknown value.
When you have or are looking for values for the opposite side and the hypotenuse, you use the
sine ratio.
Notice that the x-side is opposite the given angle and the side with a value of 16 is on the
hypotenuse, the side across from the right angle.
Example:
Solve for the missing side in the following triangle.
Solution:
Before solving for an unknown, it is helpful to mark the diagram and to list the values as follows:
the side opposite the designated angle is x cm and is OPP and the hypotenuse across from the
90 angle and is 16 cm
𝑜𝑝𝑝
ℎ𝑦𝑝
𝑥
𝑠𝑖𝑛50 =
16
𝑥
0.766 =
16
𝑠𝑖𝑛𝜃 =
(0.766)(16) = 𝑥
12.3 = 𝑥
16
Grade 10 Essentials Math
Trigonometry
Example:
Solve for x.
Solution:
𝒐𝒑𝒑
𝒉𝒚𝒑
10
𝑠𝑖𝑛35 =
𝑥
𝐬𝐢𝐧  =
10
0.574 =
𝑥
10
𝑥=
0.574
𝑥 = 17.43
Example:
Solve for x.
Solution:
𝑜𝑝𝑝
ℎ𝑦𝑝
16
𝑠𝑖𝑛30 =
𝑥
sin  =
16
0.5 =
𝑥
16
𝑥=
0.5
𝑥 = 32
17
Grade 10 Essentials Math
Trigonometry
Example:
Find the missing angle.
Solution:
𝑜𝑝𝑝
ℎ𝑦𝑝
12
𝑠𝑖𝑛 =
18
sin  =
sin  = 0.67
 = sin−1(0.67)
 = 𝟒𝟏. 𝟖
Example:
Find the missing angle.
Solution:
𝑜𝑝𝑝
ℎ𝑦𝑝
22
𝑠𝑖𝑛 =
40
sin  =
sin  = 0.55
 = sin−1(0.55)
 = 𝟑𝟑. 𝟒
18
Grade 10 Essentials Math
Trigonometry
Learning Activity 5.3: SINE Ratio
1. Use the SINE ratio to solve for the missing sides in the following triangles.
a.
b.
x
5
x
15
420
340
c.
d.
x
x
170
430
14
8
e.
x
f.
25
32
x
250
100
g.
h.
6
x
16
x
540
0
22
19
Grade 10 Essentials Math
Trigonometry
2. For the following triangles, determine the size of the angle indicated using the sine ratio
(round to the nearest degree).
a.
b.
12
17
22
8
c.
d.
9
7
17
29
8
e.
15
20
Grade 10 Essentials Math
Trigonometry
COSINE RATIO
𝑎𝑑𝑗
The cosine ratio 𝑐𝑜𝑠𝜃 = ℎ𝑦𝑝 requires three values, the acute angle, the adjacent side and the
hypotenuse. We label the adjacent side by locating the angle and going to the side directly next
(beside it) from it. The hypotenuse is located by finding the 90 angle and going to the side
across from it. You would need to know any two of the values and then you can substitute these
two known values into the formula to find the unknown value.
When you have or are looking for values for the adjacent side and the hypotenuse, you use the
cosine ratio.
Notice that the x-side is adjacent (beside) the given angle and the side with the value of 18 is
across from the 90 angle so it is the hypotenuse. We have the adjacent and hypotenuse sides
therefore we will use the cosine ratio.
Example:
Solve for x.
Solution:
𝑎𝑑𝑗
ℎ𝑦𝑝
𝑥
cos 50  =
18
𝑐𝑜𝑠𝜃 =
0.643 =
𝑥
18
(0.643)(18) = 𝑥
11.57 = 𝑥
21
Grade 10 Essentials Math
Trigonometry
Example:
Solve for x.
Solution:
𝑎𝑑𝑗
ℎ𝑦𝑝
𝑥
cos 55  =
16
𝑐𝑜𝑠𝜃 =
0.574 =
𝑥
16
(0.574)(16) = 𝑥
𝑥 = 9.18
Example:
Solve for x.
Solution:
𝑎𝑑𝑗
ℎ𝑦𝑝
14
cos 62 =
𝑥
𝑐𝑜𝑠𝜃 =
14
0.469 =
𝑥
14
𝑥=
0.469
𝑥 = 29.85
22
Grade 10 Essentials Math
Trigonometry
Example:
Solve for the missing angle.
Solution:
𝑐𝑜𝑠𝜃 =
𝑐𝑜𝑠 =
12
18
𝑎𝑑𝑗
ℎ𝑦𝑝
𝑐𝑜𝑠  = 0.67
 = 𝑐𝑜𝑠 −1 (0.67)
 = 𝟒𝟖. 𝟐
Example:
Solve for the missing angle.
Solution:
𝑎𝑑𝑗
ℎ𝑦𝑝
25
𝑐𝑜𝑠 =
40
𝑐𝑜𝑠𝜃 =
𝑐𝑜𝑠  = 0.625
 = 𝑐𝑜𝑠 −1 (0.625)
 = 𝟓𝟏. 𝟑
23
Grade 10 Essentials Math
Trigonometry
Learning Activity 5.4: COSINE Ratio
Show all of your work.
1. Use the COSINE ratio to solve for the missing sides in the following triangles.
a.
b.
15
300
x
21
170
x
c.
d.
x
x
240
450
18
12
e.
9
f.
420
x
x
330
7
g.
10
h.
100
x
8
x
140
24
Grade 10 Essentials Math
Trigonometry
2. Use the cosine ratio to solve for the missing angle in each of the following triangles (round to
the nearest degree).
a.
b.
9
15
30
8
c.
d.
17
15
27
10
25
Grade 10 Essentials Math
Trigonometry
TANGENT RATIO
When you use the ratio 𝑡𝑎𝑛𝜃 =
𝑜𝑝𝑝
𝑎𝑑𝑗
there are three values, , opposite and adjacent that need to
be considered. You would need to know any two of the values and then you can substitute these
two known values into the formula to find the unknown value.
𝑡𝑎𝑛𝜃 =
𝑜𝑝𝑝
𝑎𝑑𝑗
Example:
Solve for the length of the side opposite the given angle.
Solution:
𝑡𝑎𝑛𝜃 =
𝑜𝑝𝑝
𝑎𝑑𝑗
tan 50 =
1.19 =
𝑥
8
𝑥
8
𝑥 = (1.19)(8) = 9.5 𝑐𝑚
26
Grade 10 Essentials Math
Trigonometry
Example:
Find the side adjacent to the given angle.
Solution:
𝑡𝑎𝑛𝜃 =
𝑜𝑝𝑝
𝑎𝑑𝑗
tan 35 =
0.700 =
𝑥=
10
𝑥
10
𝑥
10
= 14.3
0.700
Example:
Find the measure of .
Solution:
𝑜𝑝𝑝
𝑎𝑑𝑗
12
𝑡𝑎𝑛  =
8
𝑡𝑎𝑛𝜃 =
𝑡𝑎𝑛  = 1.5
 = 𝑡𝑎𝑛−1 (1.5)
 = 𝟓𝟔. 𝟑
27
Grade 10 Essentials Math
Trigonometry
Learning Activity: TANGENT Ratio
Show all of your work.
1. Use the TANGENT ratio to solve for the missing sides in the following triangles.
a.
b.
x
21
300
15
150
x
c.
d.
x
16
220
170
19
x
x
e.
f.
400
11
8
x
320
x
g.
h.
x
140
34
15
510
28
Grade 10 Essentials Math
Trigonometry
2. For each of the following triangles use the tangent ratio to determine the angle (round to the
nearest degree).
a.
b.
15
10
20
14
c.
d.
15
17
12
8
29
Grade 10 Essentials Math
Trigonometry
Curriculum Outcomes:
10E2.TG.3. Solve problems that require the manipulation and application of primary trigonometric ratios
10E2.TG.2.Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent)
Lesson 3 Assignment: SOH CAH TOA
See your teacher for Assignment
30
Grade 10 Essentials Math
Trigonometry
LESSON 4: WORD PROBLEMS USING TRIG RATIOS
Lesson Introduction
In this lesson, you apply your understanding of trigonometric ratios to solve problems.
Example:
Skiers ski down a slope that is inclined at a 35 angle to the horizontal at ground level. If the
skiers reach the ground level after travelling a distance of 325.6 m, how high is the ski slope
where they began their run?
Solution:
𝑠𝑖𝑛 35 =
0.574 =
ℎ
325.6
ℎ
325.6
ℎ = (0.574)(325.6) = 18.9𝑚
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Grade 10 Essentials Math
Trigonometry
Example:
A ship is nearing the rocks. The captain can see the top of the cliff at an angle of 28° with the
horizon. His charts tell him that the cliff is 175 m above sea level. How far is the ship from
hitting the rocks?
Solution:
𝑡𝑎𝑛 28 =
175
𝑥
175
𝑥
175
𝑥=
0.532
0.532 =
𝑥 = 328.9 𝑚
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Grade 10 Essentials Math
Trigonometry
Example:
A tree is 15.6 rn tall and has fallen against your cottage, which is 7 m tall. Find the angle at
which the tree meets the cottage.
Solution:
𝑐𝑜𝑠  =
7
15.6
cos  = 0.449
 = cos −1(0.449) = 63.3
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Grade 10 Essentials Math
Trigonometry
Curriculum Outcomes:
10E2.TG.3. Solve problems that require the manipulation and application of primary trigonometric ratios
10E2.TG.2.Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent)
Lesson 4 Assignment: Word Problems using Trigonometric Ratios
See your teacher for Assignment
34
Grade 10 Essentials Math
Trigonometry
LESSON 5: ANGLE OF ELEVATION AND DEPRESSION
Angle of Elevation: is the angle formed up from the horizontal
Angle of Depression: is the angle formed down from the horizontal
Example:
From a point 8m from the base of a tree, you measure the angle up to the top of the tree from eye
level and find that it is 50°. If you are 1.2m tall, how tall is the tree?
Solution:
tan 50 =
x
1.19 =
y
50
𝑥
8
𝑥
8
(1.19)(8) = 𝑥
8m
1.2m
𝑥 = 9.5𝑚
𝑆𝑜 𝑡ℎ𝑒 ℎ𝑖𝑒𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑖𝑠 9.5 + 1.2 = 10.7 𝑚 𝑡𝑎𝑙𝑙
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Grade 10 Essentials Math
Trigonometry
Example:
The highest point on a cliff is 90m above the shore. From the top of the cliff, a surveyor
measures the angle of depression to a boat in the lake to be 42°. How far away from shore is the
boat?
Solution:
x
tan 42 =
42
90m
0.9 =
𝑥=
x
90
𝑥
90
𝑥
90
0.9
𝑥 = 100𝑚
Example:
From a point 5m from the base of a tree, you measure the angle of elevation using a 1.5m tall
instrument to be 39°. How tall is the tree?
Solution:
tan 39 =
0.81 =
y
X
𝑦
5
𝑦
5
𝑦 = (5)(0.81) = 4.05𝑚
39
1.5m
5m
𝑆𝑜 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑖𝑠 𝑋 = 4.05 + 1.5 = 5.55𝑚 𝑡𝑎𝑙𝑙.
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Grade 10 Essentials Math
Trigonometry
Example:
A 5m tall lighthouse sits at the top of a 30m cliff above sea level. The angle of depression from
the top of the lighthouse to a fishing boat is 24°. The angle of depression to a second boat is 16°.
How far apart are the two boats?
y
Solution:
z
24
16
35m
35m
35m
x
𝑡𝑜 𝑓𝑖𝑛𝑑 ℎ𝑜𝑤 𝑓𝑎𝑟 𝑎𝑝𝑎𝑟𝑡 𝑡ℎ𝑒 𝑏𝑜𝑎𝑡𝑠 𝑎𝑟𝑒 𝑓𝑟𝑜𝑚 𝑒𝑎𝑐ℎ𝑜𝑡ℎ𝑒𝑟 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑙𝑎𝑐𝑢𝑙𝑎𝑡𝑒: 𝑋 = 𝑌 − 𝑍
𝐹𝑖𝑟𝑠𝑡 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑘𝑛𝑜𝑤 ℎ𝑜𝑤 𝑓𝑎𝑟 𝑎𝑤𝑎𝑦 𝑏𝑜𝑎𝑡 1 𝑖𝑠 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑙𝑖𝑓𝑓
35
tan 24 =
𝑧
0.445 =
𝑧=
35
𝑧
35
= 78.7𝑚
0.445
𝑵𝒆𝒙𝒕, 𝒘𝒆 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒃𝒐𝒂𝒕 𝟐 𝒊𝒔 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒄𝒍𝒊𝒇𝒇
𝒚=
𝐭𝐚𝐧 𝟏𝟔 =
𝟑𝟓
𝒚
𝟎. 𝟐𝟖𝟕 =
𝟑𝟓
𝒚
𝟑𝟓
= 𝟏𝟐𝟏. 𝟗𝒎
𝟎. 𝟐𝟖𝟕
𝑿 = 𝟏𝟐𝟏. 𝟗 − 𝟕𝟖. 𝟕 = 𝟒𝟑. 𝟐𝒎 𝒔𝒐 𝒕𝒉𝒆 𝒃𝒐𝒂𝒕𝒔 𝒂𝒓𝒆 𝟒𝟑. 𝟐𝒎 𝒂𝒑𝒂𝒓𝒕
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Grade 10 Essentials Math
Trigonometry
Curriculum Outcomes:
10E2.TG.3. Solve problems that require the manipulation and application of primary trigonometric ratios
10E2.TG.2.Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent)
Lesson 5 Assignment: Angles of Elevation and Depression
See your teacher for Assignment
38