the
Mathematics
Quarter 4-Module 7
Law of Sines and Its Applications
Week 6
Learning Code - M9GE-IVh-47.1.1
Mathematics – Grade 9
Alternative Delivery Mode
Mathematics – Grade 9
Alternative Delivery Mode
Quarter 4 – Module 7 – New Normal Math for G9
First Edition 2020
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Mathematics
Quarter 4-Module 7
Law of Sines and Its Applications
Week 6
Learning Code - M9GE-IVh-47.1.1
GRADE 9
Learning Module for Junior High School Mathematics
MODULE
7
LAW OF SINES AND ITS APPLICATIONS
In the previous modules, you have learned how to solve right triangle using
trigonometric ratios. Now you will learn to solve non-right triangles called oblique
triangles. Any triangle, right or oblique, may be solved using the Law of Sines and
the Law of Cosines. If any three of the six measures of a triangle are given, provided
at least one measure is a side, then the other three measures can be found. An
oblique triangle is a triangle that does not have a right angle. There are laws or
formulas that describe the relationships between the angles and the sides of an
oblique triangle. These are the Law of Sines and the Law of Cosines.
WHAT I NEED TO KNOW
PPREPREVIER!
LEARNING COMPETENCY
The learners will be able to:
• Illustrate law of sines. M9GE-IVf-g-1
WHAT I KNOW
PPREPREVIER
A. Find out how much you already know about the module. Write the letter of
!
the best answer to each question on a sheet of paper. Answer all items. After
taking and checking this short test, take note of the items that you were not
able to answer correctly and look for the right answer as you go through this
module.
1. Which of the following statements is the correct equation to solve for x.
a.
5.5
b.
𝑥
sin 95°
sin 95°
=
=
𝑥
sin 25°
5.5
sin 25°
c.
𝑥
d.
5.5
sin 25°
sin 95°
=
=
5.5
sin 95°
sin 25°
𝑥
2. Given 𝐴 = 40°, 𝐶 = 70°, and 𝑏 = 10, use the Law of Sines to solve the triangle.
a. B = 70°, a = 10, c = 6.84
c. B = 65°, a = 10, c = 6.84
b. B = 75°, a = 6.84, c = 10
d. B = 70°, a = 6.84, c = 10
3. Given A = 50°, B = 79°, and c = 8, solve the triangle for the value of a. Round
answer to two decimal places.
a. 7.89 units
b. 7.55 units
c. 7.77 units
d. 7.51 units
4. Use the Law of Sines to solve for x. Round answer to two decimal places.
a. 5.33 units
b. 5.44 units
1
c. 5.44 units
d. 5.66 units
GRADE 9
Learning Module for Junior High School Mathematics
5. In ∆𝐴𝐵𝐶, if 𝐴 = 35°, 𝐵 = 25°, and 𝑎 = 5 then 𝑏 = __________. Round answer to two
decimal places.
a. 3.68 units
b. 3.78 units
c. 3.88 units
d. 3.98 units
6. Given a triangle with a = 16, A = 13, and B = 44, determine c using the Law
of Sines.
a. c = 49.4 units
b. c =6.3 units
c. c = 59.65 units d. c = 5.1 units
B. In each ∆𝐴𝐵𝐶, sides 𝑎, 𝑏, and 𝑐 are opposite sides of angles 𝐴, 𝐵, and 𝐶,
respectively. Solve each triangle.
7. If 𝐴 = 42°, 𝐵 = 96°, 𝑎𝑛𝑑𝑏 = 12
9. If 𝐴 = 33°, 𝐶 = 128°, 𝑎𝑛𝑑𝑎 = 16
8. If 𝐴 = 42°, 𝐵 = 48°, 𝑎𝑛𝑑𝑐 = 12
10. If 𝐴 = 48°, 𝐵 = 37°, 𝑎𝑛𝑑𝑎 = 100
C. In each ∆𝐴𝐵𝐶, sides 𝑎, 𝑏, and 𝑐 are opposite sides of angles 𝐴, 𝐵, and 𝐶,
respectively. Determine whether ∆𝐴𝐵𝐶 has no solution, one solution, or two
solutions. Then solve each triangle.
11.𝐴 = 50°, 𝑎 = 15, 𝑎𝑛𝑑 𝑏 = 10
14. 𝐴 = 30°, 𝑎 = 5, 𝑎𝑛𝑑 𝑏 = 20
12.𝐴 = 136°, 𝑎 = 115, 𝑎𝑛𝑑 𝑏 = 99.6
15. 𝐴 = 150°, 𝑎 = 10, 𝑎𝑛𝑑 𝑏 = 30
13.𝐴 = 112°, 𝑎 = 84.2, 𝑎𝑛𝑑 𝑐 = 74
WHAT’S IN
PPREPREV
IER!RECALL
LET’S
A. What are the special angles?
B. Without using a scientific calculator, find the values of each of the following.
1. sin 45°, cos 45°
2. sin 30°, cos 30°
3. sin 60°, cos 60°
4. sin 90°, cos 90°
5. sin 0°, cos 0°
C. Using a scientific calculator, find the values of each of the following.
1. sin 20°
2. sin 50°
3. sin 125°
4. sin 162°
5. sin x° = 0.1115
6. sin x° = 0.2436
7. sin x° = 0.3219
8. sin x° = 0.5432
2
GRADE 9
Learning Module for Junior High School Mathematics
WHAT’S NEW
Look at the triangles below.
Questions:
1. What have you noticed about each of the given triangles?
2. Can you use trigonometric ratios to solve for the missing parts of these
triangles? Why?
Examine closely the triangles. Can you solve the missing parts of these triangles
using the previous concepts you have learned?
Consider two oblique triangles shown below.
The two triangles in (a) and (b) each have angles A, B, and C and corresponding
opposite sides a, b, and c.
Draw an altitude of length ℎ from vertex C of each of the triangles. Notice that two
right triangles are formed, and these are triangles ACD and BCD. Using the definition
of the sine of an angle, you have
𝑠𝑖𝑛 𝐵 =
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
ℎ
= ,
𝑠𝑖𝑛 𝐴 =
𝑎
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Note that the equations are true for triangles in (a) and (b).
Solving for ℎ in the two equations, you will get
ℎ = 𝑎 𝑠𝑖𝑛 𝐵
𝑎𝑛𝑑
ℎ = 𝑏 𝑠𝑖𝑛 𝐴
Therefore, equating the two expressions for ℎ gives you
𝑎 𝑠𝑖𝑛 𝐵 = 𝑏 𝑠𝑖𝑛 𝐴
Dividing both sides of the equation by 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛 𝐵 results to
𝑎 sin 𝐵
𝑏 sin 𝐴
=
sin 𝐴 sin 𝐵 sin 𝐴 sin 𝐵
Simplifying both sides of the equation gives you
3
ℎ
= ,
𝑏
GRADE 9
Learning Module for Junior High School Mathematics
𝑎
𝑏
=
sin 𝐴 sin 𝐵
That is, side 𝑎 divided by the sine of its opposite angle (sin 𝐴) is equal to side 𝑏 divided
by the sine of its opposite angle (sin 𝐵).
Repeating the process of drawing an altitude of length ℎ from vertex A of each
triangle, the following equation is obtained:
𝑏
𝑐
=
.
sin 𝐵 sin 𝐶
Combining the two equations, you get
𝑎
𝑏
𝑐
= sin 𝐵 = sin 𝐶 .
sin 𝐴
The above equation gives the 𝐿𝑎𝑤 𝑜𝑓 𝑆𝑖𝑛𝑒𝑠.
The Law of Sines
In any triangle, a side divided by the sine of its opposite angle is equal to any other
side divided by the sine of the corresponding opposite angle.
𝑎
𝑏
𝑐
= sin 𝐵 = sin 𝐶
sin 𝐴
This can also be expressed as
sin 𝐴
=
sin 𝐵
=
sin 𝐶
𝑎
𝑏
𝑐
Note that the derived formula for oblique triangles is also applicable to right triangles.
Communication, Critical
Thinking, and Collaboration
WHAT IS IT
The Law of Sines can be used in solving problems involving oblique triangles, given
the measures of two angles and one side. It can be used when the given are two
angles and the included side (ASA) or two angles and a nonincluded side (SAA).
Figure (a) shows triangle ABC given its two angles and the included side (ASA)
while (b) shows triangle DEF given its two angles and a nonincluded side (SAA).
Example 1:
Solve ∆𝐴𝐵𝐶 𝑔𝑖𝑣𝑒𝑛 𝐴 = 60°, 𝐵 = 45°, 𝑎𝑛𝑑 𝑎 = 20𝑐𝑚.
4
GRADE 9
Learning Module for Junior High School Mathematics
Solution:
You are given two angles and a
nonincluded side (SAA)
First, find the measure of angle 𝐶. Since
the sum of the measures of the interior
angles of any triangle equals 180°, that is,
𝐴 + 𝐵 + 𝐶 = 180°
60° + 45° + 𝐶 = 180°
𝐶 = 180° − (60° + 45°)
𝐶 = 180° − 105°
𝐶 = 75°
Use the Law of Sines to find b.
𝑎
𝑏
=
sin 𝐴 sin 𝐵
You can also use the Law of Sines to find c.
𝑎
𝑐
=
sin 𝑎 sin 𝑐
20
𝑏
=
sin 60° sin 45°
20
𝑐
=
sin 60° sin 75°
𝑏=
20 sin 45°
sin 60°
𝑐=
𝑏 = 16.33 𝑐𝑚.
20 sin 75°
sin 60°
𝑐 = 22.31 𝑐𝑚
Thus, 𝐶 = 75°, 𝑏 = 16.33 𝑐𝑚, 𝑎𝑛𝑑 𝑐 = 22.31 𝑐𝑚.
Example 2:
Solve ∆𝐴𝐵𝐶 𝑔𝑖𝑣𝑒𝑛 𝐴 = 78°, 𝐵 = 52°, 𝑎𝑛𝑑 𝑐 = 35 𝑐𝑚.
Solution:
You are given two angles and an
included side (ASA)
First, find the measure of angle 𝐶. Since the sum
of the measures of the interior angles of any
triangle equals 180°, that is,
𝐴 + 𝐵 + 𝐶 = 180°
78° + 52° + 𝐶 = 180°
𝐶 = 180° − (78° + 52°)
𝐶 = 180° − 130°
𝐶 = 50°
Use the Law of Sines to find 𝑎.
𝑎
𝑐
=
sin 𝐴 sin 𝐶
𝑎
35
=
sin 78° sin 50°
35 sin 78°
sin 50°
𝑎 = 44.69 𝑐𝑚.
𝑎=
5
GRADE 9
Learning Module for Junior High School Mathematics
You can also use the Law of Sines to find b.
𝑏
𝑎
=
sin 𝐵 sin 𝐴
𝑏
44.69
=
sin 52° sin 78°
44.69 sin 52°
sin 78°
𝑏 = 36 𝑐𝑚.
𝑏=
Thus,
𝐶 = 50°, 𝑎 = 44.69 𝑐𝑚, 𝑎𝑛𝑑 𝑏 = 36 𝑐𝑚
The Law of Sines can also be used when two sides and a nonincluded angle
aregiven (SSA). In this case, there may be no triangle having the given
measurements or there may be one or two triangles that satisfy the given
conditions. This case is often referred to as the 𝑎𝑚𝑏𝑖𝑔𝑢𝑜𝑢𝑠 𝑐𝑎𝑠𝑒.
Now, suppose that in ∆𝐴𝐵𝐶, angle 𝐴 and sides 𝑎 and 𝑏 are given (SSA). Based
on the Law of Sines,
𝑏
𝑎
𝑏 sin 𝐴
=
sin 𝐵 =
sin 𝐵
sin 𝐴
𝑎
Consider the following cases:
Case 1. 0° < 𝐴 < 90°
a. If 𝑎 < 𝑏 sin 𝐴 𝑡ℎ𝑒𝑛 sin 𝐵 > 1. This means
that no angle 𝐵 is determined; hence, no
triangle is formed and there is no
solution.
b. If 𝑎 = 𝑏 𝑠𝑖𝑛 𝐴 then 𝑠𝑖𝑛 𝐵 = 1. This
means 𝐵 = 90° and a right triangle is
determined.
c. If 𝑎 > 𝑏 sin 𝐴 𝑎𝑛𝑑 𝑎 < 𝑏, then two angles
are formed: an acute angle B in triangle
ABC and an obtuse angle B' in triangle
AB'C. Hence, there are two solutions.
d. If 𝑎 > 𝑏 sin 𝐴 and 𝑎 ≥ 𝑏 then there
is exactly one angle determined.
Hence, there is only one solution.
Case 2: 90° < 𝐴 < 180°
a. If 𝑎 ≤ 𝑏 then it can be seen from the
figure below that there is no solution.
b. If 𝑎 > 𝑏 then there is exactly one
triangle formed. Hence, there is
exactly one solution.
6
GRADE 9
Learning Module for Junior High School Mathematics
Example 3:
Solve ∆𝐴𝐵𝐶 given 𝐴 = 60°, 𝑎 = 8 𝑐𝑚, 𝑎𝑛𝑑 𝑏 = 53 𝑐𝑚.
Solution:
To solve for 𝐵, use the Law of Sines.
𝑎
𝑏
=
sin 𝐴 sin 𝐵
8
53
=
sin 60° sin 𝐵
53 sin 60°
sin 𝐵 =
8
sin 𝐵 = 5.7374
Notice that sin 𝐵 = 5.7374, which is
greater than 1. This is not possible
since −1 ≤ sin 𝐵 ≤ 1 for any angle 𝐵.
This means that there is no solution,
that is, there is no triangle having the
given measurements. This is an
example of 𝑎𝑚𝑏𝑖𝑔𝑢𝑜𝑢𝑠 𝑐𝑎𝑠𝑒.
Example 4:
Solve ∆𝐴𝐵𝐶 given 𝐵 = 28°, 𝑏 = 16 𝑐𝑚, 𝑎𝑛𝑑 𝑐 = 20 𝑐𝑚.
Solution:
𝐴 =?
𝐵 = 28°
𝐶 =?
Compare the values of 𝑏 𝑎𝑛𝑑 𝑐 sin 𝐵.
𝑏 ? 𝑐 sin 𝐵
16 > 20 sin 28°
So, 𝑏 > 𝑐 sin 𝐵.
Since 𝐵 is an acute angle and 𝑐 > 𝑏, then
this is Case 1.c.
This is an ambiguous case and there are
two possible solutions.
𝑎 =?
𝑏 = 16
𝑐 = 20
The two possible solutions are shown below.
Solution No. 1:
Solving for 𝐴
𝐴 = 180° − (28° + 36°)
𝐴 = 116°
Solving for 𝑎
𝑏
𝑎
=
sin 𝐵 sin 𝐴
16
𝑎
=
sin 28° sin 116°
Solving for 𝐶
𝑏
sin 𝐵
16
sin 28°
𝑐
=
sin 𝐶
20
= sin 𝐶
sin 𝐶 =
𝑎=
20 sin 28°
16
16 sin 116°
sin 28°
𝑎 = 31
Thus, 𝐴 = 116°, 𝐶 = 36° , 𝑎𝑛𝑑 𝑎 = 31 𝑐𝑚.
sin 𝐶 = .5868
𝐶 = 36°
7
GRADE 9
Learning Module for Junior High School Mathematics
Solution No. 2:
Solving for 𝐴,
𝐴 = 180° − (28° + 144°)
𝐴 = 8°.
Solving for 𝑎,
𝑏
𝑎
=
sin 𝐵 sin 𝐴
Solving for the other value of 𝐶,
𝐶 = 180° − 36°
𝐶 = 144°.
Remember:
For 0° < 𝐶 < 180°, there are two angles
with a sine value of
20 sin 28°
16
16
𝑎
=
sin 28° sin 8°
𝑎=
∶
one acute angle and one obtuse angle.
16 sin 8°
sin 28°
𝑎=5
Thus, 𝐴 = 8°, 𝐶 = 144° , 𝑎𝑛𝑑 𝑎 = 5 𝑐𝑚.
WHAT’S MORE
A. Determine whether each statement is true or false.
1. It is possible to solve a triangle if the only given information consists of
the measures of the three angles of the triangle.
2. In general, it is not possible to use the Law of Sines to solve a triangle for
which the given are the lengths of all the sides.
3. Given ∆𝐴𝐵𝐶 with 𝐴 = 30°, 𝑐 = 3 𝑐𝑚, . 𝑎𝑛𝑑 𝑎 = 2.5 𝑐𝑚. There can be more than
one triangle that can be drawn with the given dimensions.
4. In a scalene triangle, the largest angle is always opposite the longest side
and the smallest angle is always opposite the shortest side.
5. Given ∆𝐴𝐵𝐶 with 𝐴 = 57°, 𝑎 = 15 𝑚, and 𝑐 = 20 𝑚. There is no triangle that
can be formed for these values of 𝐴, 𝑎, and 𝑐.
B. Write an equation that will solve x.
C. For each triangle given in Exercise B, solve for the value of 𝑥. Round
answer to the nearest tenth.
8
GRADE 9
Learning Module for Junior High School Mathematics
D. Find the length of AB. Round answer to the nearest tenth.
E. Find measure of angle𝐴. Round answer to the nearest tenth.
WHAT I HAVE LEARNED
Given Triangle ABC, with angles 𝐴, 𝐵, and 𝐶 and corresponding opposite sides
𝑎, 𝑏, and 𝑐.
The Law of Sines states that
𝒂
𝑺𝒊𝒏𝑨
𝒃
𝒄
𝑺𝒊𝒏 𝑨
𝑺𝒊𝒏 𝑩
𝑺𝒊𝒏 𝑪
=
=
or
=
=
𝑺𝒊𝒏𝑩
𝑺𝒊𝒏 𝑪
𝒂
𝒃
𝒄
To solve an oblique triangle using the Law of Sines, you need to know the
measure of one side and the measures of two other parts of the triangle:—two
angles, or one angle and another side.
This breaks down into the following cases:
1. Two angles and any side (AAS or ASA)
2. Two sides and an angle opposite one of them (SSA or Angle Side Side)
WHAT I CAN DO
A. Determine the unknown part of the triangle marked with”?”. Round answers to
the nearest tenth.
9
GRADE 9
Learning Module for Junior High School Mathematics
B. Solve each ∆𝐴𝐵𝐶 given that 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 are opposite 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶, respectively.
6. SSA
A = 73°
B = ____
C = ____
7. SSA
A = _____
B = _____
C = 27°
8. ASA
A = 26°
B = _____
C = 35°
9. AAS
A = _____
B = 50°
C = 45°
a = 18 cm
b = 11 cm
c = _____
a = 17 cm
b = ______
c = 13 cm
a = _____
b = 13 cm
c = _____
a = ______
b = 14 cm
c = ______
ASSESSMENT
A. Read the questions carefully. Write the letter of the correct answer on a sheet of
paper.
1. Two sides and an angle opposite one of the given sides of a triangle are known.
To find the angle opposite the other given side , you should use:
a. Law of Sines b. Heron’s Formula
c. Law of Cosines d. All of the above
2. Which of the following is the correct equation to solve for x?
a.
b.
25
sin 100°
𝑥
sin 50°
=
=
𝑥
sin 50°
25
sin 100°
𝑥
c.
d.
=
sin 100°
25
sin 50°
=
25
sin 50°
sin 100°
𝑥
3. In the given triangle, A is 81° and B is 67° . If side a is 34 units long,
approximately how long is side b?
a. 32 units
b. 30 units
c. 22 units
d. 20 units
4. In the given triangle, A is 137° and B is 28° . If side b is 71 units long,
approximately how long is side a?
a. 77 units
5.
b. 98 units
c. 84 units
d. 103 units
In the given triangle, A is 98° degrees and B is 12° . If side a is 84 units long,
approximately how long is side b?
a. 20 units
b. 18 units
c. 16 units
d. 14 units
6. In the given triangle, 𝑄 is= 98°, 𝑞 = 17.5 𝑚, 𝑎𝑛𝑑 𝑟 = 15 𝑚. Find 𝑅.
a. 78°
10
b. 68°
c. 58°
d. 44°
GRADE 9
Learning Module for Junior High School Mathematics
B. In each ∆𝐴𝐵𝐶, sides 𝑎, 𝑏, and 𝑐 are opposite sides of angles 𝐴, 𝐵, and 𝐶,
respectively. Solve each triangle.
7. 𝐴 = 49°, 𝐵 = 57°, 𝑎𝑛𝑑 𝑎 = 8
9.
8. 𝐶 = 110°, 𝑐 = 18, 𝑎𝑛𝑑 𝑎 = 13
10. 𝐴 = 105°, 𝑎 = 18, 𝑎𝑛𝑑 𝑏 = 14
𝐵 = 49°, 𝑎 = 11.6, 𝑎𝑛𝑑 𝑏 = 14.9
C. In each ∆𝐴𝐵𝐶 sides 𝑎, 𝑏, and 𝑐 are opposite sides of angles 𝐴, 𝐵, and 𝐶,
respectively. Determine whether ∆𝐴𝐵𝐶 has no solution, one solution, or two
solutions. Then solve each triangle.
11. 𝐴 = 20°, 𝑎 = 15, 𝑎𝑛𝑑 𝑏 = 20
14. 𝐴 = 90°, 𝑎 = 12, 𝑎𝑛𝑑 𝑏 = 14
12. 𝐴 = 37°, 𝑎 = 24, 𝑎𝑛𝑑 𝑏 = 32.2
15. 𝐴 = 25°, 𝑎 = 125, 𝑎𝑛𝑑 𝑏 = 150
13. 𝐴 = 31°, 𝑎 = 9, 𝑎𝑛𝑑 𝑏 = 20
ADDITIONAL ACTIVITIES
Communication, Critical Thinking,
Creativity
Challenge Problems:
1. Find 𝐴 to the nearest degree.
2. Find angle 𝐷𝐺𝐹 to the nearest degree.
PROBLEM – BASED LEARNING WORKSHEET
Problem:
Two fire-lookout stations are 15 miles apart, with station A directly east of station
B. Both stations spot a fire at position C. The angular direction of the fire from
station B is N52°E and the angular direction of the fire from station A is N36°W.
How far is the fire from station A?
C
11
GRADE 9
Learning Module for Junior High School Mathematics
Draw the triangle and use 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶 as angles and 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 as the opposite
sides, respectively.
Questions:
1. How do you find 𝐴?
2. How about 𝐵?
3. What is 𝐶?
4. What law are you going to use to find the distance from 𝐹𝑖𝑟𝑒 𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝐴 to
point 𝐶 where the fire is 𝑜𝑟 what law are you going to use to solve for 𝑏?
5. Write the equation to solve for 𝑏.
6. How far is the fire from station 𝐴?
E-Search
To further explore the concept learned today and if it possible to connect the
internet, you may visit the following links:
www.onlinemath learning.com>law of sine 2
www.math is fun .com>algebra>trig-sine-law
www.teacherspayteachers .com>Browse>Search law of sines
www.mathworksheetsgo.com
study.com>academy>lesson>law of sine lesson-plan
mathopenref.com/lawofcosinesproof.html
https://cdn.kutasoftware.com>
https://www.buffaloschool.org>
REFERENCES
Gladys C. Nivera, Minie Rose C. Lapinid(2013) Grade 9 Mathematics Patterns and
Practicalities Salesiana BOOKS by Don Bosco Press, Inc. Chino Roces Avenues,
Makati City, Philippines
Soledad Jose Dilao, Ed D. Fernando B. Orines, Julieta G. Bernabe (2013)Advances
Algebra Trigonometry and Statistics JTW Corporation Araneta Ave. Quezon City
Philippines
Orlando A. Oronce, Marilyn O. Mendoza (2013) e-math Advanced Algebra and
Trigonometry, Published by Rex Bookstore. C. M. Recto Avenue, Manila, Philippines
Regina MacarangalTresvalles, Wilson Cordova (2010)Math Ideas and Applications
Series Advanced Algebra, Trigonometry, and Statistics Aviba Publishing House,
Inc. Abiva Bldg., 851 G. Araneta Ave., Quezon City, Manila, Philippines
Orlando A. Oronce, Marilyn O. Mendoza (2010) Worktext in Mathematics e-math
Advanced Algebra and Trigonometry Rex Book Store 856 Nicanor Reyes, Sr. Street
1977 C.M.RectoAvenue, Manila, Philippines
Merden L. Bryant, Leonides E. Bulalayao, Melvin M. Callanta, Jerry D. Cruz,
Richard F. De Vera, Gilda T. Garcia, Sonia E. Javier, Roselle A. Lazaro, Bernadette
J. Mesterio, and Rommel Hero A. Saladino (2014)MATHEMATICS, Learner's
Material 9, Vibal Group, Inc. 5th Floor Mabini Building , Deped Complex Meralco
Avenue, Pasig City, Philippines
12
13
Grade 9_ Quarter 4_Module 7 (Answer Key)
WHAT I KNOW
1. b
2. d
3. a
4. d
7. 𝐶 = 42°, 𝑎 = 8.07, 𝑎𝑛𝑑 𝑐 = 8.07
8. 𝐶 = 90°, 𝑎 = 8.03 𝑎𝑛𝑑 𝑏 = 8.92
9. 𝐵 = 19°, 𝑏 = 9.56, 𝑎𝑛𝑑 𝑐 = 23.15
10. 𝐶 = 95°, 𝑏 = 80.98, 𝑎𝑛𝑑 𝑐 = 134.05
11. one solution: 𝐵 = 30.71°, 𝐶 = 99.29°, 𝑎𝑛𝑑 𝑐 = 19.32
12. one solution: 𝐵 = 37°, 𝐶 = 7°, 𝑎𝑛𝑑 𝑐 = 20.18
13. one solution: 𝐵 = 13.43°, 𝐶 = 54.57°, 𝑎𝑛𝑑 𝑏 = 21.1
14. (No solution)
15. no solution
5. a
6. c
WHAT'S IN
A. The specials angle are 30°, 45°, 60°, 𝑎𝑛𝑑 90°angles.
ξ2 ξ2
B. 1. ,
2
2
C. 1. 0.3420
2. 0.7660
3. 0.8192
4. 0.3090
WHAT’S MORE
2.
5.
6.
7.
8.
1 ξ3
,
2 2
6.4°
14.1°
18.8°
32.9
A. 1. False 2. False (True)
𝑥
10
B. 1.
=
sin 36°
2.
5
sin 𝑥
=
sin 41°
20
sin 125°
𝑥
3. ,
ξ3 1
,
2 2
3. True
37
4.
=
sin 𝑥
5.
5.True
4. True
5. 0,1
4. 1, 0
30
sin 30°
𝑥
sin 100°
=
25
sin 50°
5.5
3.
=
sin 95°
sin 25°
C. 1. 𝑥 = 9
2. 𝑥 = 11.8°
3. 𝑥 = 13
D. a. 𝐴𝐵 = 22.2
E. a. 𝑚∠𝐴 = 36.4
4. 𝑥 = 38.1°
5. 𝑥 = 32.5 (𝑥 = 32.1)
b. 𝐴𝐵 = 18
b. 𝑚∠𝐴 = 38.7
WHAT I CAN DO
A.1. 7.9
4. 33.04
2. 35°
5. 18.66(18.7)
3. 22.1
B. 6. 𝐵 = 35.8°, 𝐶 = 71.2°, 𝑎𝑛𝑑 𝑐 = 17.8 𝑐𝑚
7. 𝐴 = 36.4°, 𝐵 = 116.6°, 𝑎𝑛𝑑 𝑏 = 25.6 𝑐𝑚.
8. 𝐵 = 119°, 𝑎 = 6.5 𝑐𝑚, 𝑎𝑛𝑑 𝑐 = 8.5 𝑐𝑚.
9. 𝐴 = 85°, 𝑎 = 18.2 𝑐𝑚, 𝑎𝑛𝑑 𝑐 = 12.9 𝑐𝑚.
Learning Module for Junior High School Mathematics
GRADE 9
14
ASSESSMENT
A. 1. A
4. D
2. C
5. B
3. A
6. C
B.7. 𝐶 = 74°, 𝑏 = 8.9, 𝑎𝑛𝑑 𝑐 = 10.2
8. 𝐴 = 43°, 𝐵 = 27°, 𝑎𝑛𝑑 𝑏 = 8.7
9. 𝐴 = 36°, 𝐶 = 95°, 𝑎𝑛𝑑 𝑐 = 19.7
10. 𝐵 = 48.7°, 𝐶 = 26.3°, 𝑎𝑛𝑑 𝑐 = 8.3
11. one solution: (Two Solutions)
(1) 𝐵 = 27.1°, 𝐶 = 132.9°, 𝑎𝑛𝑑 𝑐 = 32.1
(2) 𝐵 = 152.9°, 𝐶 = 7.1°, 𝑎𝑛𝑑 𝑐 = 5.4
12. one solution:(Two Solutions)
(1) 𝐵 = 53.8°, 𝐶 = 89.2°, 𝑎𝑛𝑑 𝑐 = 39.9
13. No solution
14. No Solution
15. One Solution:(Two Solutions)
(1) 𝐵 = 30.5°, 𝐶 = 124.5, 𝑐 = 243.8
Learning Module for Junior High School Mathematics
GRADE 9