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Math-3
Lesson 5-3 (non honors)The Law of Sines
The Law of Sines is for Triangles that are NOT right triangles.
AA BB
aabb
CC
vocabulary
Solve a triangle: find the unknown lengths of sides and measures of angles of a triangle.
The problem will given you some of the sides or angles.
Sin, cos, and tan ratios are for solving Right triangles!
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What do we do if the triangle is not a right triangle?
In this lesson we learn how to solve triangles that are NOT right triangles.
The standard method of labeling triangles is:
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The length of the side opposite Angle A is lower case a.
The standard method of labeling triangles is:
bb
The length of the side opposite Angle A is lower case a.
The standard method of labeling triangles is:
bb
The length of the side opposite Angle C is lower case c.
AA BB
CC
aabbhh
b
hAsin
a
hB sin
Solving for ‘h’ then setting the Solving for ‘h’ then setting the equations equal to each other.equations equal to each other.
AbhBa sinsin Eliminating ‘h’, dividing by ‘a’ Eliminating ‘h’, dividing by ‘a’ and ‘b’.and ‘b’.
b
B
a
A sinsin
We could repeat this for any We could repeat this for any combination of sides and angle.combination of sides and angle.
Law of Sines
cC
bB
aA
cbaCBA
sinsinsin
: trueisequation following thely,respective
, and , , sides and and , , angles with ABC In
By the By the Transitive PropertyTransitive Property this means this means eacheach of the expressions are equal to each other.of the expressions are equal to each other.
We could also write it this way (using sequential property of equality steps):
C
c
B
b
A
a
sinsinsin
Law of Sines
b
B
a
A sinsin
Which one we use depends upon whether we Which one we use depends upon whether we need to find the measure of an need to find the measure of an unknown angle unknown angle or or an an unknown sideunknown side..
B
b
A
a
sinsin
Pick the version that puts the unknown variable in the numerator!
When a problem is given, they either
107107ºº
2525ºº
1515
(1) Give you the drawing and provide some of the measured sides or angles. Then ask you to solve for the other sides.
In this example we’ll solve for ‘c’.
120120ºº
2525ºº
1515
B
b
C
c
sinsin
25sin
15107sinc 25sin
15
107sin
c
Pick the version that puts the unknown variable in the numerator!
b
B
a
A sinsin
B
b
A
a
sinsin
Replace letters in the Replace letters in the formula with numbers formula with numbers from the triangle.from the triangle.
““Solve” for ‘c’ (isolate Solve” for ‘c’ (isolate the variable ‘c’ on one the variable ‘c’ on one side of the equal sign..side of the equal sign..
9.33
Another way a problem is given, is that they just give you some measurements.
A = 35A = 35ºº C = 120C = 120ººa = 10a = 10
Draw the general triangle that has capital letters for angles and lower case letters for the lengths of the sides opposite the angles.
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A= 35A= 35ºº C = 120C = 120ººa = 10a = 10
bb 120120ºº
3535ºº
2. Label the triangle with 2. Label the triangle with values given in the problem.values given in the problem.
1010
bb
bb 120120ºº
3535ºº
1010
Pick the version that puts the unknown variable in the numerator!
b
B
a
A sinsin B
b
A
a
sinsin
bb 120120ºº
3535ºº
1010
Pick the version that puts the unknown variable in the numerator!
b
B
a
A sinsin B
b
A
a
sinsin
B
b
C
c
sinsin
35sin
10
120sin
c
35sin
10120sinc 1.15
Solve for the measure of angle C, given the following triangle.
3131ºº
1010
1919Pick the version that puts the unknown variable in the numerator!
b
B
c
C sinsin B
b
C
c
sinsin
b
B
c
C sinsin
10
31sin
19
)(
CSin
Replace letters in the Replace letters in the formula with numbers formula with numbers from the triangle.from the triangle.
Solve for the measure of angle C, given the following triangle.
3131ºº
1010
1919
““Solve” for ‘C’ (isolate the variable ‘C’ on one side of Solve” for ‘C’ (isolate the variable ‘C’ on one side of the equal sign). First we will have to isolate sin(C)the equal sign). First we will have to isolate sin(C)
10
31sin
19
)(
CSin
10
31sin19)(
CSin
97857.0)( CSin
Sine (angle) = ratio
3131ºº
1010
1919
““Inverse sine (ratio) = angleInverse sine (ratio) = angle
)97857.0(sin 1C
97857.0)( CSin
078C
What if they given you two angles but not the two that you need?
107107ºº
2525ºº
Using the Using the Triangle Sum TheoremTriangle Sum Theorem (angle is a triangle (angle is a triangle always add up to 180always add up to 180º): º): 48)25107(180 Am
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B = 25B = 25ºº C = 107C = 107ºº c = 15c = 15
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Find “little” ‘a’Find “little” ‘a’
Now solve using the Law of Sines.
Your Turn:
solve the triangle.solve the triangle.
1.1. Draw and label the triangle.Draw and label the triangle.
2. A = ?2. A = ?
2020
7575ººbb3. a = ?3. a = ?
2020ºº
C = 75C = 75º B = 20º c = 20º B = 20º c = 20
4. b = ?4. b = ?
Your turn: Solve for a:Your turn: Solve for a:
5555
2828ºº 6464ºº
A = 28A = 28ºº B = 64B = 64ºº c = 55c = 55
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