Republic of the Philippines

Department of Education
REGION I
SCHOOLS DIVISION OF THE CITY OF BATAC

ACTIVITY SHEETS IN MATHEMATICS 9

QUARTER 4, WEEK 6

LAW OF SINES

Most Essential Learning Competency (MELC):

• Illustrates laws of sines and cosines. (M9GE-IVf-g-1)

Prepared by:

KAREN MAE A. MANGALLAY

Teacher I
 Background Information

LAW OF SINES

The ASA and SAA Cases

𝑎 𝑏 𝑐
For any triangle ∆𝐴𝐵𝐶, 𝑆𝑖𝑛 𝐴 = 𝑆𝑖𝑛 𝐵 = 𝑆𝑖𝑛 𝐶 , 𝑎 is the length of the side opposite ∡𝐴, 𝑏

is the length of the side opposite ∡𝐵 and 𝑐 is the length of the side opposite ∡𝐶.
B

c a

A C
b
The law of sines is used to solve for the measures of triangles when:

a. there are two angles and an included side (ASA Case);

b. there are two angles and a non-included side (SAA Case); or
c. there are two sides and an angle opposite one of the given sides (SSA
Case).

Examples:

1. Determine the measures of the missing parts of ∆𝐴𝐵𝐶 below.

B
Given: two angles and the included side (ASA case)
∠𝐴 = 67° 45º

∠𝐵 = 45° 8 cm b
c = 8 cm.
67º
A C
Solutions: a

Since the measurements of the two angles are known, the measure of the third
angle can be determined using the concept that the sum of the measures of the angles
of a triangle is 180°. Hence,

∠𝐴 + ∠𝐵 + ∠𝐶 = 180° ∠𝐶 = 180° − 112°

67° + 45° + ∠𝐶 = 180° ∠𝐶 = 𝟔𝟖°
112° + ∠𝐶 = 180°
 𝑎 𝑐
To solve for side 𝑎, we can use the formula 𝑠𝑖𝑛𝐴 = sin 𝐶.
𝑎 𝑐
= Formula to solve for 𝑎
𝑠𝑖𝑛 𝐴 sin 𝐶
𝑎 8
= Substitute the given values
𝑠𝑖𝑛 67° 𝑠𝑖𝑛 68°
𝑎 𝑠𝑖𝑛 68° = 8 𝑠𝑖𝑛 67° Cross multiply
0.9272𝑎 = 8(0.9205) Compute the values of 𝑠𝑖𝑛 68° and 𝑠𝑖𝑛 67° in four
decimal places using a scientific calculator.
0.9272𝑎 = 7.364 Simplify the resulting equation
7.364
𝑎 = 0.9272 Solve for 𝑎

𝑎 = 𝟕. 𝟗𝟒 𝒄𝒎 Round off final answer to two decimal places

𝑏 𝑐
To solve for side 𝑏, use the formula 𝑠𝑖𝑛𝐵 = sin 𝐶 , and follow the steps earlier.
𝑏 𝑐
= sin 𝐶
𝑠𝑖𝑛𝐵
𝑏 8
= 𝑠𝑖𝑛 68°
𝑠𝑖𝑛 45°

𝑏 𝑠𝑖𝑛 68° = 8 𝑠𝑖𝑛 45°

0.9272𝑏 = 8(0.7071)
0.9272𝑏 = 5.6568
5.6568
𝑏=
0.9272
𝑏 = 𝟔. 𝟏𝟎 𝒄𝒎

2. Find the missing parts of ∆𝐴𝐵𝐶 at the right. B

Given: two angles and the included side (SAA case)
∠𝐴 = 57°
c 9
∠𝐶 = 63°
57º 63º
A C
b
 Solve for c.
𝑎 𝑐
= Formula to solve for 𝑐
𝑠𝑖𝑛𝐴 sin 𝐶
9 𝑐
= Substitute the given values
𝑠𝑖𝑛 57° 𝑠𝑖𝑛 63°
c 𝑠𝑖𝑛 57° = 9 𝑠𝑖𝑛 63° Cross multiply
0.8387𝑐 = 9(0.8910) Compute the values of 𝑠𝑖𝑛 57° and 𝑠𝑖𝑛 63° in four
decimal places using a scientific calculator.
0.8387𝑐 = 8.019 Simplify the resulting equation
8.019
c= 0.8387 Solve for 𝑐

c= 9.56 𝒄𝒎
a = 9 cm.

Solve for ∠𝐵.

Use the concept that the sum of the measures of the angles of triangle is 180°.
∠𝐴 + ∠𝐵 + ∠𝐶 = 180° ∠𝐵 = 180° − 120°
57° + ∠𝐵 + 63° = 180° ∠𝐵 = 𝟔𝟎°
120° + ∠𝐵 = 180°
Solve for b
𝑎 𝑏
Following the steps used earlier in solving for 𝑐, we now have 𝑠𝑖𝑛𝐴 = sin 𝐵
𝑎 𝑏
= sin 𝐵 0.8387𝑏 = 7.794
𝑠𝑖𝑛𝐴
9 𝑏 7.794
= 𝑠𝑖𝑛 60° 𝑏 = 0.8387
𝑠𝑖𝑛 57°

𝑏 𝑠𝑖𝑛 57° = 9 𝑠𝑖𝑛 60° 𝑏 = 𝟗. 𝟐𝟗 𝒄𝒎

0.8387𝑏 = 9(0.8660)
The SSA Case

When the measures of two sides and an angle opposite one of them are given, you
may be able to use the Law of Sines to find the measurement of the triangle’s another
angle. These may, however, lead to the following possibilities:

THE SSA POSSIBILITIES

If ∠𝐴 is an acute angle and 𝑎 ≥ 𝑏, If ∠𝐴 is an acute angle, 𝑎 < 𝑏, and 𝑎 >
then there is exactly one solution. 𝑏 sin 𝐴, then there are two solutions.
C
C 𝑏 sin 𝐴

b a b a
a

A B A
c B
B
(ambiguous case)

If ∠𝐴 is an acute angle, 𝑎 < 𝑏, and If ∠𝐴 is an acute angle, 𝑎 < 𝑏, and 𝑎 <

𝑎 = 𝑏 sin 𝐴, then there is exactly one 𝑏 sin 𝐴, then there are no solutions.
solution. C
a C
a
b 𝑎 = 𝑏 sin 𝐴 b
𝑏 sin 𝐴
A c B A c B

If ∠𝐴 is an obtuse or a right angle If ∠𝐴 is an obtuse or a right angle and 𝑎 ≤

and 𝑎 > 𝑏, then there is exactly one 𝑏, then there is exactly no solution.
solution. C
a C
C a
b
a b
b A A
c c
A c B

b a

A c B
Examples:
1. Solve for the missing parts of ∆𝐴𝐵𝐶 below.
Given: two sides and an angle opposite one of these sides.
B
𝑎 = 10
𝑐 = 19 c=19
a =10
∠𝐶 = 120°
120°
A C
b
Solutions:
∠𝐶 is an obtuse angle and 𝑐 > 𝑎, there is exactly one solution.
𝑎 𝑐
Since 𝑎, 𝑐 𝑎𝑛𝑑 ∠𝐶 are known, we can use the formula 𝑠𝑖𝑛𝐴 = sin 𝐶.
𝑎 𝑐
= sin 𝐶 Formula to solve for 𝑐
𝑠𝑖𝑛𝐴
10 19
= 𝑠𝑖𝑛 120° Substitute the given values
𝑠𝑖𝑛 𝐴

19𝑠𝑖𝑛 𝐴 = 10 𝑠𝑖𝑛 120° Cross multiply

19𝑠𝑖𝑛 𝐴 = 10(0.8660) Compute the values of 𝑠𝑖𝑛 120° using a scientific
calculator
19𝑠𝑖𝑛 𝐴 = 8.66 Simplify the resulting equation
8.66
𝑠𝑖𝑛𝐴 = = 0.4558 Solve for 𝐴
19

𝐴 = 𝟐𝟕. 𝟏𝟐°

Use the concept that the sum of the measures of the angles of a triangle is 180.
∠𝐴 + ∠𝐵 + ∠𝐶 = 180° ∠𝐵 = 180° − 147.12°
27.12° + ∠𝐵 + 120° = 180° ∠𝐵 = 𝟑𝟐. 𝟖𝟖°
147.12° + ∠𝐵 = 180°

𝑏 𝑐
Following the steps used earlier in solving for 𝑐, we now have 𝑠𝑖𝑛𝐵 = sin 𝐶
𝑏 𝑐
= sin 𝐶 0.8660𝑏 = 10.3151
𝑠𝑖𝑛𝐵
𝑏 19 10.3151
= 𝑠𝑖𝑛 120° 𝑏=
𝑠𝑖𝑛 32.88° 0.8660

𝑏 𝑠𝑖𝑛 120° = 19 𝑠𝑖𝑛 32.88° 𝑏 = 𝟏𝟏. 𝟗𝟏 𝒄𝒎

0.8660𝑏 = 19(0.5429)
2. Solve ∆𝐴𝐵𝐶 given that 𝑏 = 15, 𝑐 = 20, 𝑎𝑛𝑑 ∠𝐵 = 27°.
𝑏? 𝑐 sin 𝐵
15 ? 20𝑠𝑖𝑛27°
15 ? 20(0.4540)
15 > 9.08
Since ∠𝐵 is an acute angle and 𝑏 > 𝑐 sin 𝐵, this is an ambiguous case and there are
two solutions. A

c = 20 b = 15
b = 15

27°
B a C

Solution 1 Solution 2
A A

c = 20 b = 15

27° c = 20
B a C b = 15

27°
Solving for ∠𝐶, B
a C
𝑏 𝑐
= sin 𝐶
𝑠𝑖𝑛𝐵
15 20
=
𝑠𝑖𝑛 27° 𝑠𝑖𝑛 𝐶
Solving for ∠𝐶,
16 𝑠𝑖𝑛 𝐶 = 20 𝑠𝑖𝑛 27°
∠𝐶 = 180° − 34.58°
20 sin 27°
sin 𝐶 = ∠𝐶 = 𝟏𝟒𝟓. 𝟒𝟐°
16
20(0.4540)
sin 𝐶 = 16

𝑠𝑖𝑛 𝐶 = 0.5675
∠𝐶 = 𝟑𝟒. 𝟒𝟖°
Solution 1 cont’d Solution 2 cont’d
Solving for ∠𝐴, *Note:
∠𝐴 + ∠𝐵 + ∠𝐶 = 180° If ∠𝐴 and ∠𝐵 are supplementary angles,
∠𝐴 + 27° + 34.58° = 180° then sin 𝐴 = sin 𝐵.
∠𝐴 + 61.58° = 180° Solving for ∠𝐴,
∠𝐴 = 180° − 61.58° ∠𝐴 + ∠𝐵 + ∠𝐶 = 180°
∠𝐴 = 𝟏𝟏𝟖. 𝟒𝟐° ∠𝐴 + 27° + 145.42° = 180°
∠𝐴 + 172.42° = 180°
∠𝐴 = 180° − 172.42°
∠𝐴 = 𝟕. 𝟓𝟖°
Solving for 𝑎, Solving for 𝑎,
𝑎 𝑏 𝑎 𝑏
= sin 𝐵 = sin 𝐵
𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐴
𝑎 15 𝑎 15
= 𝑠𝑖𝑛 27° = 𝑠𝑖𝑛 27°
𝑠𝑖𝑛 118.42° 𝑠𝑖𝑛 7.58°

𝑎 𝑠𝑖𝑛 27° = 15 𝑠𝑖𝑛 118.42° 𝑎 𝑠𝑖𝑛 27° = 15 𝑠𝑖𝑛 7.58°

0.4540𝑎 = 15(0.8795) 0.4540𝑎 = 15(0.1319)
0.4540𝑎 = 13.19 0.4540𝑎 = 1.9785
13.19 1.9785
𝑎 = 0.4540 𝑎 = 0.4540
𝑎 = 𝟐𝟗. 𝟎𝟓 𝑎 = 𝟒. 𝟑𝟔
Name:___________________________________________ Date:____________
Grade/Section: ____________________________________Score:____________

Activity 1: 0, 1 or 2!
Directions: Identify whether there is one, two solutions, or no
solution at all for the given conditions. Write your answers on
the space provided.

Answer
1. 𝑎 = 9, 𝑐 = 7, ∡𝐴 = 95° ____________________________

2. 𝑎 = 5, 𝑏 = 6, ∡𝐴 = 45° ____________________________

3. 𝑏 = 4, 𝑐 = 6, ∡𝐶 = 30° ____________________________

4. 𝑎 = 75, 𝑏 = 100, ∡𝐴 = 50° ____________________________

5. 𝑎 = 10, 𝑏 = 20, ∡𝐴 = 10° ____________________________

6. 𝑎 = 8, 𝑐 = 7, ∡𝐶 = 133° ____________________________

7. 𝑎 = 28, 𝑏 = 15, ∡𝐴 = 110° ____________________________

8. 𝑏 = 8, 𝑐 = 10, ∡𝐶 = 65° ____________________________

9. 𝑎 = 350, 𝑏 = 351, ∡𝐴 = 85° ____________________________

10. 𝑎 = 24, 𝑏 = 6, ∡𝐵 = 175° ____________________________

Name:___________________________________________ Date:____________
Grade/Section: ____________________________________Score:____________

Activity 2: Picture Me!

Directions: Draw your own triangles using the indicated
measures and solve for the other parts using the Law of Sines.
Round off all decimals to the nearest tenth.

1. ∠𝐴 = 73°, 𝑎 = 18, 𝑏 = 11

∠𝐵 = ________ ∠𝐶 = ________ 𝑐 = ________

2. ∠𝐴 = 26°, ∠𝐶 = 35°, 𝑏 = 13

∠𝐵 = ________ 𝑎 = ________ 𝑐 = ________

Name:___________________________________________ Date:____________
Grade/Section: ____________________________________Score:____________

Activity 3: Complete me!

Directions: Solve for the measurements of the following
triangles and round off your answers to the nearest tenth.
After which, write the corresponding letter on the blanks to
decode a message.

_____ _____ _____ _____ _____ _____ _____ _____ _____

14.1 15° 24.2 14.1 21.9 60° 19.6 24.2 14.7

_____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____
105° 120° 15° 15° 14 20 30° 55° 150° 45° 19.3 105°

_____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____
105° 14 38.6 120° 25° 14 105° 65 150° 19.3 5.4

G=?
T=?

10 o=?
c=?
i=?

45° 30° 120°

30°
a=?
14
W=?
20 n=?

15° L=?
m=?

H=?
45°
f=? 18
k=?
s= ?
60° 75°
16 15°
E=? r= ?
References:

Bryant, Merden L. et al. Mathematics Learner’s Material for Grade 9 Module 7:

Triangle Trigonometry. Pasig City: Department of Education, 2014

Natividad, Eldefonso Jr. B. et al. Math Made Easy for Grade 9. Makati City:
Salinlahi Publishing House, Inc., 2017