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Lesson 9.5
Apply the Law of Sines
Standard Accessed: Students will prove, apply, and
model trigonometric functions and ratios.
Warm-Up
Lesson Presentation
Lesson Quiz
Warm-Up
Solve Ξ” XYZ.
𝑋
17
𝟐𝟎. πŸπŸ•
𝑦
π‘Œ
πŸ“πŸ•°
π‘₯ 𝟏𝟏. πŸŽπŸ’
33°
βˆ π’ = ?
π‘Ίπ’Šπ’…π’† 𝒙 = ?
π‘Ίπ’Šπ’…π’† π’š = ?
𝑍
βˆ π™ = πŸ“πŸ•°
𝒙
tan πŸ‘πŸ‘ =
π‘Ίπ’Šπ’…π’† 𝒙 = 𝟏𝟏. πŸŽπŸ’
πŸπŸ•
𝟏𝟏. πŸŽπŸ’
π‘Ίπ’Šπ’…π’† π’š = 𝟐𝟎. πŸπŸ•
cos πŸ“πŸ• =
π’š
Warm-Up
Solve Ξ” ABC.
𝐴
𝑏 πŸ’πŸ—°
14. πŸ’πŸ‘πŸ‘
𝐢
βˆ π‘¨ = ?
π‘Ίπ’Šπ’…π’† 𝒂 = ?
π‘Ίπ’Šπ’…π’† 𝒃 = ?
22
41°
π‘Ž πŸπŸ”. πŸ”πŸŽπŸ’ 𝐡
βˆ π€ = πŸ’πŸ—°
𝒂
sin πŸ’πŸ— =
𝟐𝟐
𝒃
cos πŸ’πŸ— =
𝟐𝟐
π‘Ίπ’Šπ’…π’† 𝒂 = πŸπŸ”. πŸ”πŸŽπŸ’
π‘Ίπ’Šπ’…π’† 𝒃 = πŸπŸ’. πŸ’πŸ‘πŸ‘
Vocabulary
Can be used to solve triangles
with no right angle, when two
angles and the length of any side
are known (AAS or ASA cases), or
when the lengths of two sides and
an angle opposite one of the two
sides are known, (SSA).
Essential Understandings
When can the Law of sines be used to
solve a triangle?
 When two angles and the length of any
side are known (AAS or ASA cases), or
when the lengths of two sides and an
angle opposite one of the two sides are
known, (SSA). The law of sines is used to
solve triangles with no right angle.
EXAMPLE
1
Solve a triangle for the AAS or ASA case
Solve Ξ”ABC with 𝐴 = 94°, 𝐢 = 67°, and 𝑏 = 25.
𝐴
SOLUTION
1. Draw a picture 
94°
πŸ•πŸŽ. πŸ”πŸ–πŸ’ πŸπŸ—°
25
𝐢
67°
πŸ•πŸ”. πŸ”πŸŽπŸ
𝐡
2. Find the third angle, 180 βˆ’ 94 + 67 = πŸπŸ—°
sin 19 sin 94
3. Write the Law of Sine proportion
=
25
π‘Ž
sin 19 sin 67
4. Write the Law of Sine proportion
=
25
𝑐
∴In ABC, βˆ π‘© = πŸπŸ—°, 𝒂 β‰ˆ πŸ•πŸ”. πŸ”πŸŽπŸ, and cβ‰ˆ πŸ•πŸŽ. πŸ”πŸ–πŸ’.
EXAMPLE
1
Solve a triangle for the AAS or ASA case
Key Question
When you are given the measures of two angles and one side of
a triangle, why does it not matter whether the given side is the
one included between the two angles?
It does not matter since you are given the measures of any
two angles, you can always find the third angle measure.
Essential Understandings
What are the possible triangles for the SSA case?
 ∠𝐴 is obtuse & a > b, one triangle
 ∠𝐴 is acute & a = height, one triangle
 ∠𝐴 is acute & a β‰₯ 𝑏, one triangle
 ∠𝐴 is acute & b > a > h, Two triangles
Note: ∠𝐴 is always across from side a
Note: To find height = 𝑏 sin 𝐴
EXAMPLE
2
Solve the SSA case with one solution
Solve Ξ”ABC with 𝐴 = 127°, π‘Ž = 63, and 𝑏 = 42.
𝐴
SOLUTION
2. Draw a picture 
πŸπŸ–. πŸŽπŸ“πŸ
𝐡
42
127°
𝐢
𝟐𝟎. πŸ–πŸ‘πŸ°
63
πŸ‘πŸ. πŸπŸ”πŸ—°
1. Evaluate SSA case, a > b & ∠𝐴 is obtuse, ∴ One triangle
3. Write the Law of Sine proportion
sin 127 sin 𝐡
=
63
42
4. Find the third angle, 180 βˆ’ 127 + 32.169 = 𝟐𝟎. πŸ–πŸ‘πŸ°
sin 127 sin 20.831
5. Write the Law of Sine proportion
=
63
𝑐
∴In ABC, βˆ π‘© β‰ˆ πŸ‘πŸ. πŸπŸ”πŸ—°, π’‚βˆ π‘ͺ β‰ˆ 𝟐𝟎. πŸ–πŸ‘πŸ, and cβ‰ˆ πŸπŸ–. πŸŽπŸ“πŸ.
EXAMPLE
3
Examine the SSA case with no solution
Solve Ξ”ABC with B = 52°, π‘Ž = 6.5, and 𝑏 = 4.7
𝐴
4.7
SOLUTION
52°
2. Draw a picture 
𝐡
1. Evaluate SSA case, a < b
β„Ž = 6.5 sin 52 = 5.122, a < h
6.5
∴No solution (The triangle is not possible.)
What will happen if you try to solve the triangle by applying
the law of sines and using your calculator?
sin 52 sin 𝐴
Write the Law of Sine proportion
=
4.7
6.5
Trying to find sinβˆ’πŸ 𝟏. πŸŽπŸ— will give an ERROR message, since
the value of the sine function is never greater than 1.
EXAMPLE
4
Solve the SSA case with two solutions
Solve Ξ”ABC with 𝐴 = 62°, π‘Ž = 32, and 𝑏 = 34.
SOLUTION
69. πŸ•πŸ’°
πŸπŸ•. πŸŽπŸ’πŸ‘
2. Draw a picture 
𝐴
𝐡
32
πŸ’πŸ–. πŸπŸ”°
62°
34
𝟏𝟏𝟎. πŸπŸ”°
32
𝐡
𝐢 πŸ’. πŸ–πŸ–πŸ 62°
34
𝐴
1. Evaluate SSA case, a < b β„Ž = 34 sin 62 = 30.02, a > h
∴ Two possible triangles
3. Write the Law of Sine proportion
sin 62 sin 𝐡
=
32
34
4. Find the third angle, 180 βˆ’ 62 + 69.74 = πŸ’πŸ–. πŸπŸ”°
4. Find the third angle, 180 βˆ’ 62 + 110.26 = πŸ•. πŸ•πŸ’°
5. Write the Law of Sine proportion
5. Write the Law of Sine proportion
sin 62 sin 48.26
=
32
𝑐
sin 62 sin 7.74
=
32
𝑐
πŸ•. πŸ•πŸ’°
𝐢
Vocabulary
EXAMPLE
4
Write and Solve a real – Use trig to find area of triangle
KIS Vegetable Garden Esther and Frederick are setting up
a triangular vegetable garden at KIS. They lay off lengths of
15ft and 11ft for two sides of the garden with a 62° angle
between these two sides, what will be the area of the
garden?
SOLUTION
1. Draw a picture 
1
2
2. Write an equation, π‘Žπ‘ sin 62
∴The area of Esther’s and Frederick’s
vegetable garden is β‰ˆ72.843π’‡π’•πŸ .
Lesson 9.5 Homework:
Practice B
Practice C β€œHonors”