LEARNING ACTIVITY SHEET

QUARTER 4-WEEK 5

Name:_________________________________________________Score:_______
Grade & Section ______________________________Subject: MATHEMATICS 9
Name of Teacher: __________________________________Date: _____________

I. Title: Law of Sine

II. Type of Activity: Concept notes with formative activities
LAS for summative assessment
( Written Work Performance Task)
III. MELC: Illustrates law of sines.
IV. Learning Objective/s:
• Illustrate the Law of Sines
• Determine the formula in finding the missing parts of an oblique triangle.
• Find the missing parts of oblique triangle using the law of sines

V. Reference/s:
Print Material/s:
• Advance Algebra and Trigonometry, Aurello P. Ramos Jr.pp.327-331.
• Learner’s Material for Mathematics Grade 9, Department of Education, 483-
488.
• Math Made Easy for Grade 9 ,Eldefonso Natividad, Jr.140-145

VI. Concept notes with formative activities

RECALL ON OBLIQUE TRIANGLES
Oblique triangle is a triangle which does not contain any right angle
Oblique triangles are classified into two:
1. Acute triangle – is a triangle whose angles are less than 90°.

280

710

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2. An obtuse triangle – a triangle in which one of the angles is more than 90°.

LAW OF SINE
𝒂 𝒃 𝒄
For any triangle ABC, 𝑺𝒊𝒏 𝑨 = 𝑺𝒊𝒏 𝑩 = , where a is the length of the side opposite
𝑺𝒊𝒏 𝑪
to ∠𝐴, b is the length side opposite to ∠𝐵, c is the length of opposite side to ∠𝐶.

The Law of Sines can also be written in the reciprocal form:

There are two cases to solve the Law of Sines

CASE I. Two angles and any side are known: SAA and ASA

CASE II. Two sides and an angle opposite of the given sides are known:
a. SSA Case – Part I
b. SSA Case – Part II

CASE I: Given Two Angles and One Side are known – (ASA )

Example 1.
For the triangle below C = 102, B = 29, and b = 28 feet. Find the remaining angle and
sides.

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Given: ∠𝑪 = 𝟏𝟎𝟐𝒐 ∠𝑩 = 𝟐𝟗𝒐 𝒃 = 𝟐𝟖
Solve for ∠𝑨

The third angle of the triangle is taken from ∠𝑨 + ∠𝑩 + ∠𝑪 = 180

∠𝐴 + 29𝑜 + 102𝑜 = 180 Substitute the given values then simplify
∠𝐴 + 131𝑜 = 180 Subtract both sides of the equation by 131
∠𝐴 + 131𝑜 − 131𝑜 = 180-131𝑜
∠A = 49.

Solve for a

By the Law of Sines, we have

Using b = 28 and ∠B = 29o produces
a b
=
Sin A Sin B
𝒂 𝟐𝟖
= substitute the given values
sin 49° sin 29°

a sin 29° = 28(sin 49°) cross multiply

a(. 4848) = 28(. 7547) Find the value of sin 29° and sin 49° using a scientific
calculator
𝑎(.4848) 21.1316
= divide both sides by .4848
.4848 .4848

𝒂 = 𝟒𝟑. 𝟓𝟗 𝒇𝒆𝒆𝒕 round answer to the nearest hundreths

Solve for c
c b
=
Sin C sin B
c 28
= substitute the given values
Sin 102° sin 29°
c 28
= find the values of sin 102 °and sin 29° using a scientific
.9781 .4848

calulator
.4848c = 28(.9781) cross multiply
.4848c = 27.3868
.4848𝑐 27.6388
= divide both sides by .4848
.𝟒𝟖𝟒𝟖 .4848

𝒄 = 𝟓𝟔. 𝟒𝟗 𝒇𝒆𝒆𝒕

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This is the triangle with the complete measures of its parts;

Example 2. ASA Case

Find the measures of the missing parts of ABC below.
Given: two angles and the included side
∠𝐴 = 60°
∠𝐵 = 40°
𝑐=7
B

Solutions:
Since the measures of the two angles of the triangle are known, the measure of the
third angle can be determined using the concept that the sum of the measures of the angle
of a triangle is 180°.
Solve for ∠𝑪
∠𝑨 + ∠𝑩 + ∠𝑪 = 𝟏𝟖𝟎°
∠60° + ∠40° + ∠𝐶 = 180° Substitute the given values
100° + ∠𝐶 = 180°
100° − 100° + ∠𝐶 = 180° − 100° Subtract both sides by 100°
∠𝑪 = 𝟖𝟎°

To solve for side a, we can use the formula;

𝑆𝑖𝑛 𝐴 𝑆𝑖𝑛 𝐶
= formula to solve for a
𝑎 𝑐
𝑆𝑖𝑛 60° 𝑆𝑖𝑛 80°
= substitute the given values
𝑎 7

𝑎 𝑠𝑖𝑛 80° = 7 𝑠𝑖𝑛 60° cross multiply

0.9848 𝑎 = 7 (.8660) find the values of sin 80° and sin 60° using scientific
calculator
0.9848 𝑎 = 6.062 simplify
0.9848 𝑎 6.062
= 0.9848 divide both sides by 0.9848
0.9848
6.062
𝑎 = 0.9848 solve for a, round answer to the hundredths

𝒂 = 𝟔. 𝟏𝟔

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To solve for side b, use the formula;
𝑺𝒊𝒏 𝑩 𝑺𝒊𝒏 𝑪
=
𝒃 𝒄
𝑺𝒊𝒏 𝟒𝟎° 𝑺𝒊𝒏 𝟖𝟎°
= substitute the given values
𝒃 𝟕

𝒃 𝒔𝒊𝒏 𝟖𝟎° = 𝟕 𝒔𝒊𝒏 𝟒𝟎° cross multiply

0.9848 𝑏 = 7( 0. 6428) find the values of sin 80° and sin 40° using scientific
calculator
𝟎. 𝟗𝟖𝟒𝟖 𝒃 = 𝟒. 𝟒𝟗𝟗𝟔 simplify
𝟎.𝟗𝟖𝟒𝟖 𝒃 𝟒.𝟒𝟗𝟗𝟔
= divide both sides by 0.9848
𝟎.𝟗𝟖𝟒𝟖 𝟎.𝟗𝟖𝟒𝟖
4.4996
𝑏 = 0.9848 solve for b

𝐛 = 𝟒. 𝟓𝟕

This is the triangle with the complete

measures
of its parts

Example 3. SAA Case – two angles and a side

Example 1.
Find the missing parts of ABC at the
right.

Given: two angles and one side

∠𝐴 = 42°
∠𝐶 = 70°
a=6

Solutions:
Since, side b and ∠𝐵 are given, we can use the formula:
𝑺𝒊𝒏 𝑨 𝑺𝒊𝒏 𝑪
=
𝒂 𝒄
𝑺𝒊𝒏 𝟒𝟐° 𝑺𝒊𝒏 𝟕𝟎°
= substitute the given values
𝟔 𝒄
𝒄 𝒔𝒊𝒏 𝟒𝟐° = 𝟔 𝒔𝒊𝒏 𝟕𝟎° cross multiply
𝟎. 𝟔𝟔𝟗𝟏 𝒄 = 𝟔(𝟎. 𝟗𝟑𝟗𝟕) find the values of sin 42 °and sin 70° using a scientific
calculator
𝟎. 𝟔𝟔𝟗𝟏 𝒄 = 𝟓. 𝟔𝟑𝟖𝟐 simplify the resulting equation
𝟎.𝟔𝟔𝟗𝟏 𝒄 𝟓.𝟔𝟑𝟖𝟐
= 𝟎.𝟔𝟔𝟗𝟏 divide both sides of the equation by 0.6691
𝟎.𝟔𝟔𝟗𝟏

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 𝟓.𝟔𝟑𝟖𝟐
𝒄 = 𝟎.𝟔𝟔𝟗𝟏 solve for c then round your answer to the hundredths
c = 8.43

For∠𝑩, use the concept the sum of the measures of the three angles is 180°.
∠𝑨 + ∠𝑩 + ∠𝑪 = 𝟏𝟖𝟎° , so we have:
42° + ∠B + ∠70° = 180°
∠B + ∠112° = 180°
∠B = 180° − 112°
∠𝑩 = 𝟔𝟖°

To solve for side b, the formula to be used is

𝑺𝒊𝒏 𝑨 𝑺𝒊𝒏 𝑩
=
𝒂 𝒃
𝑆𝑖𝑛 42° 𝑆𝑖𝑛 68°
= 𝑏 substitute the given values
6
𝑏 𝑆𝑖𝑛 42° = 6 𝑠𝑖𝑛 68° cross multiply
0.6691 𝑏 = 6(0.9272) find the values of Sin 42° and sin 68° using scientific
calculator
0.6691 𝑏 = 5.5632
0.6691 𝑏 5.5632
= divide both sides of the equation by 0.6691
0.6691 0,6691
5.5632
𝑏 = simplify
0.6691
𝒃 = 𝟖. 𝟑𝟏 the length of b

This is the triangle with the complete

measures of its parts;

Case II. SSA

THE AMBIGOUOS CASE
If two sides and an angle opposite one of them are given three possibilities can
occur
1. No such triangle exists.
2. Two different one triangle exists.
3. Exactly one triangle exists.

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 SUMMARY

Condition
Angle A Triangles possible
(h = b sin A)
Acute a<𝒉 none
Acute 𝒂=𝒉 one
Acute 𝒉 < 𝒂 <𝒃 two
Acute 𝒂≥𝒃 one
Obtuse or right 𝒂≤𝒃 none
Obtuse or right a>𝒃 one

NOTE: h is the altitude or height of the triangle.

∠𝑨 𝒊𝒔 𝒂𝒄𝒖𝒕𝒆 ∠𝑨 𝒊𝒔 𝒐𝒃𝒕𝒖𝒔𝒆
Examples :

1. Let a = 8
b = 12
∠𝐴 =60°
Solution:
Use h = b sin A
Calculate h: h = 12 sin 60° Find the value of sin 60°
= 12 (0.8660) = 10.39
Since a < 𝒉 where 8 < 12, no is formed, Hence, there is no possible triangle to
be formed.
• This is sometimes described as an ambiguous case because it is possible
that the solution to this case is not unique solution, or there are no solution
at all.

2. Let a = 3
b=6
∠𝐴 = 30°
A

7
Solution:
Use h = b sin A
Calculate h: h = 6 sin 30° Find the value of sin 30°
= 6 (0.5) = 3
Since a = h where 3 = 3, one is formed. Thus, there is one solution.

3.Let b = 8
c = 7.5
C= 50°

Solution:
Use h = b sin C
Calculate h: h = 8 sin 50° Find the value of sin 50°
= 8 (.7660) = 6.13
Since h< 𝑐 < 𝑏, where 6.13< 7.5 < 8 then two s are formed. There are two
solutions.

4. Let a = 6
b = 15
A= 1100
Solution:
Use h = b sin A
= 15 (sin 1100)
= 15 (.9397) = 14.1
Since a < ℎ where 6 < 14.1 then there is no formed, hence, there is no solution.

5. Solve for the missing parts of ABC below

Given: two sides and an angle opposite one
of these sides
a = 10
c = 19
∠𝐂 = 𝟏𝟐𝟎°

8
Solutions:
∠C is obtuse angle and c > a, thus there is exactly one solution
Since a, c, and ∠C are known, we can use the formula;
𝐒𝐢𝐧 𝐀 𝐒𝐢𝐧 𝐂
=
𝐚 𝐜
𝐒𝐢𝐧 𝐀 𝐒𝐢𝐧 𝟏𝟐𝟎°
= 𝟏𝟗 substitute the given values
𝟏𝟎
19 sin A = 10 Sin 120 cross multiply
19 sin A = 10 (0.8660) find the values of sin 120° using a scientific calculator
19(sin A) = 8.66 simplify
8.66
sin A = solve for A
19
8.66
sin A =
19
sin A = .4558
A = sin-1 0.4558 Press shift sin 0.4558 then press equal sign (=)
A = 27.12°

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

Solve for b;
𝑺𝒊𝒏 𝑩 𝑺𝒊𝒏 𝑪
=
𝒃 𝒄
𝑺𝒊𝒏 𝟑𝟐.𝟖𝟖° 𝒔𝒊𝒏 𝟏𝟐𝟎°
= 𝟏𝟗 substitute the given values
𝒃
𝑏 𝑠𝑖𝑛 120° = 19 𝑠𝑖𝑛 32.88° cross multiply
𝑏(0.8660) = 19 (0.5429) find the values of sin 120° 𝑎𝑛𝑑 𝑠𝑖𝑛 32.88° using a
scientific
calculator
𝒃(𝟎.𝟖𝟔𝟔𝟎)= 𝟏𝟎.𝟑𝟏𝟓𝟏
𝟎.𝟖𝟔𝟔𝟎 𝟎.𝟖𝟔𝟔𝟎
𝒃 = 𝟏𝟏. 𝟗𝟏

9
 SUMMARY ON THE APPLICATION OF THE LAW OF SINE

NOTE: Given parts of a triangle are in a box.

1. Two angles and the included side ( ASA)
∠𝐶 = 180 − (∠𝐴 + ∠𝐶 )
𝑆𝑖𝑛 𝐴 𝑆𝑖𝑛 𝐶
𝑎 =
𝑎 𝑐
𝑆𝑖𝑛 𝐵 𝑆𝑖𝑛 𝐶
𝑏 =
𝑏 𝑐

2. Two angles and a side ( AAS or SAA)

∠𝐶 = 180 − (∠𝐴 + ∠𝐵 )
𝑆𝑖𝑛 𝐵 𝑆𝑖𝑛 𝐴
𝑏 =
𝑏 𝑎
𝑆𝑖𝑛 𝐶 𝑆𝑖𝑛 𝐴
𝑐 =
𝑐 𝑎

3. Two sides and an angle ( SSA)

𝑆𝑖𝑛 𝐴 𝑆𝑖𝑛 𝐵
∡𝐴 =
𝑎 𝑏
∡𝐶 = 180 − ( ∠𝐴 + ∠𝐵)

𝑆𝑖𝑛 𝐶 𝑆𝑖𝑛 𝐵
𝑐 =
𝑐 𝑏

10
 LEARNING ACTIVITY SHEET
QUARTER 4-WEEK 5

Name:_________________________________________________Score:_______
Grade & Section ______________________________Subject: MATHEMATICS 9
Name of Teacher: __________________________________Date: _____________

I. Title: Law of Sine

II. Type of Activity: Concept notes with formative activities
LAS for summative assessment
( Written Work Performance Task)
III. MELC: Illustrates law of sines.
IV. Learning Objective/s:
• Illustrate the Law of Sines
• Determine the formula in finding the missing parts of an oblique triangle.
• Find the missing parts of oblique triangle using the law of sines

V. Reference/s:
Print Material/s:
• Advance Algebra and Trigonometry, Aurello P. Ramos Jr.pp.327-331.
• Learner’s Material for Mathematics Grade 9, Department of Education, 483-
488.
• Math Made Easy for Grade 9 ,Eldefonso Natividad, Jr.140-145

VI. Summative Test.

I. Identify the side/angle opposite the given side/ angles of the following. Refer to the
figure at the right side.
1. ∠𝑇 = __________
2.4.77 = _______
3. ∠𝐼 = ____________
4. 6.11 = _______
5. ∠𝑁 = __________

11
II. Multiple Choice. Solve the missing values using the Law of Sines. Write the
letter only.

6 ∠𝐴 = 45°, ∠𝐵 = 65° , 𝑎 = 10, 𝑏 =?

a.13 b.12.49 c.12. 94 d. 14

7. ∠𝐴 = 30°, ∠𝐶 = 45° , 𝑏 = 10, 𝑐 =?

a.7.29 b.8 c.7.92 d. 7.09

8. ∠𝐵 = 30°, ∠𝐴 = 135° , 𝑏 = 10, 𝑎 =?

a. 14.23 b.14.2 c.15 d. 14.26

9. ∠𝐵 = 122°, 𝑏 = 10° , 𝑐 = 8, ∠𝐶 = ?

a.41.33 b.43 c.42.20 d. 42.22

10. 𝑎 = 16, ∠𝐴 = 35° , ∠𝐵 = 65° 𝑏 =?

a.23.55 b.25.44 c.24.55 d.23.44

III. Use the figure below and answer the following. Show your complete solution.
11- 15 . ∠𝑅 = __________

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