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The Law
of SINESof SINES
The Law of SINES
a
sin A
b
sin B
c
sin C
For any triangle (right, acute or obtuse), you may use the following formula to solve for missing sides or angles:
Use Law of SINES when ...
AAS - 2 angles and 1 adjacent side ASA - 2 angles and their included side
SSA (this is an ambiguous case)
you have 3 dimensions of a triangle and you need to find the other 3 dimensions - they cannot be just ANY 3 dimensions though, or you won’t have enough info to solve the Law of Sines equation. Use the Law of Sines if you are given:
Example 1
You are given a triangle, ABC, with angle A = 70°, angle B = 80° and side a = 12 cm. Find the measures of angle C and sides b and c.
* In this section, angles are named with capital letters and the side opposite an angle is named with the same lower case letter .*
Example 1 (con’t)
C
The angles in a ∆ total 180°, so angle C = 30°.
Set up the Law of Sines to find side b:
12
sin 70
b
sin 80A
B
70°
a = 12c
b
80°
12 sin 80 b sin 70
.80
b12 sin
sin 7012 6cm
Example 1 (con’t)
Set up the Law of Sines to find side c:
12
sin 70
c
sin 3012 sin 30 csin70
c 12 sin 30
sin706.4cm
A C
B
70°
80°a = 12c
b = 12.630°
Example 1 (solution)
Angle C = 30°
Side b = 12.6 cm
Side c = 6.4 cm
A C
B
70°
80°a = 12
c =
6.4
b = 12.630°
Note:
We used the given values of A and a in both calculations. Your answer is more accurate if you do not use rounded values in calculations.
Example 2
You are given a triangle, ABC, with angle C = 115°, angle B = 30° and side a = 30 cm. Find the measures of angle A and sides b and c.
Example 2 (con’t)
AC
B
115°
30°
a = 30
c
b
To solve for the missing sides or angles, we must have an angle and opposite side to set up the first equation.
We MUST find angle A first because the only side given is side a.
The angles in a ∆ total 180°, so angle A = 35°.
Example 2 (con’t)
AC
B
115°
30°
a = 30
c
b35°
Set up the Law of Sines to find side b:
30
sin35
b
sin 30
30sin 30 bsin35
b30sin30
sin3526.2cm
Example 2 (con’t)
AC
B
115°
30°
a = 30
c
b = 26.235°
Set up the Law of Sines to find side c:
30
sin35
c
sin115
30 sin115 csin35
c 30 sin115
sin3547.4cm
Example 2 (solution)
AC
B
115°
30°
a = 30
c = 47.4
b = 26.235°
Angle A = 35°
Side b = 26.2 cm
Side c = 47.4 cm
Note: Use the Law of Sines whenever you are given 2
angles and one side!
The Ambiguous Case (SSA)
When given SSA (two sides and an angle that is NOT the included angle) , the situation is ambiguous. The dimensions may not form a triangle, or there may be 1 or 2 triangles with the given dimensions. We first go through a series of tests to determine how many (if any) solutions exist.
The Ambiguous Case (SSA)
In the following examples, the given angle will always be angle A and the given sides will be sides a and b. If you are given a different set of variables, feel free to change them to simulate the steps provided here.
‘a’ - we don’t know what angle C is so we can’t draw side ‘a’ in the right position
A B ?
b
C = ?
c = ?
The Ambiguous Case (SSA)
Situation I: Angle A is obtuse
If angle A is obtuse there are TWO possibilities
A B ?
ab
C = ?
c = ?
If a ≤ b, then a is too short to reach side c - a triangle with these dimensions is impossible.
A B ?
ab
C = ?
c = ?
If a > b, then there is ONE triangle with these dimensions.
The Ambiguous Case (SSA)
Situation I: Angle A is obtuse - EXAMPLE
Given a triangle with angle A = 120°, side a = 22 cm and side b = 15 cm, find the other dimensions.
Since a > b, these dimensions are possible. To find the missing dimensions, use the Law of Sines:
BA
a = 2215 = b
C
c 120°
22
sin120
15
sin B15sin120 22sin B
B sin 1 15sin12022
36.2
The Ambiguous Case (SSA)
Situation I: Angle A is obtuse - EXAMPLE
Angle C = 180° - 120° - 36.2° = 23.8°
Use Law of Sines to find side c:
B
36.2°
A
a = 2215 = b
C
c 120°
22
sin120
c
sin 23.8c sin120 22sin 23.8
c 22sin 23.8sin120
10.3cm
Solution: angle B = 36.2°, angle C = 23.8°, side c = 10.3 cm
The Ambiguous Case (SSA)
Situation II: Angle A is acute
If angle A is acute there are SEVERAL possibilities.
Side ‘a’ may or may not be long enough to reach side ‘c’. We calculate the height of the altitude from angle C to side c to compare it with side a.
A B ?
b
C = ?
c = ?
a
The Ambiguous Case (SSA)
Situation II: Angle A is acute
First, use SOH-CAH-TOA to find h:
A B ?
b
C = ?
c = ?
ah
sin A h
bh bsin A
Then, compare ‘h’ to sides a and b . . .
The Ambiguous Case (SSA)
Situation II: Angle A is acute
A B ?
b
C = ?
c = ?
a
h
If a < h, then NO triangle exists with these dimensions.
The Ambiguous Case (SSA)
Situation II: Angle A is acute
cA B
b
C
ah
If h < a < b, then TWO triangles exist with these dimensions.
AB
b
C
c
a h
If we open side ‘a’ to the outside of h, angle B is acute.
If we open side ‘a’ to the inside of h, angle B is obtuse.
The Ambiguous Case (SSA)
Situation II: Angle A is acute
BcA
b
C
ah
If h < b < a, then ONE triangle exists with these dimensions.
Since side a is greater than side b, side a cannot open to the inside of h, it can only open to the outside, so there is only 1 triangle possible!
The Ambiguous Case (SSA)
Situation II: Angle A is acute
If h = a, then ONE triangle exists with these dimensions.
If a = h, then angle B must be a right angle and there is only one possible triangle with these dimensions.
C
AB
b
c
a = h
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
Given a triangle with angle A = 40°, side a = 12 cm and side b = 15 cm, find the other dimensions.
B ?
a = 12
A
15 = b
C = ?
c = ?
h40°
Find the height:
h bsin A
h 15sin40 9.6
Since a > h, but a< b, there are 2 solutions and we must find BOTH.
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
a = 12
A B
15 = b
C
c
h40°
FIRST SOLUTION: Angle B is acute - this is the solution you get when you use the Law of Sines!
12sin 40
15sin B
B sin 1 15sin4012
53.5
C 180 40 53.5 86.5c
sin86.5 12
sin 40
c 12sin86.5sin 40
18.6
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
SECOND SOLUTION: Angle B is obtuse - use the first solution to find this solution.
a = 12
AB
15 = b
C
c40°
1st ‘a’
1st ‘B’
In the second set of possible dimensions, angle B is obtuse, because side ‘a’ is the same in both solutions, the acute solution for angle B & the obtuse solution for angle B are supplementary.
Angle B = 180 - 53.5° = 126.5°
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
SECOND SOLUTION: Angle B is obtuse
a = 12
A B
15 = b
C
c40° 126.5°
Angle B = 126.5°
Angle C = 180°- 40°- 126.5° = 13.5°
c
sin13.5 12
sin 40
c 12sin13.5
sin 404.4
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EX. 1 (Summary)
Angle B = 126.5°Angle C = 13.5°Side c = 4.4
a = 12
A B
15 = b
C
c = 4.440° 126.5°
13.5°
Angle B = 53.5°Angle C = 86.5°Side c = 18.6
A B
15 = b
C
c = 18.6
a = 12
40° 53.5°
86.5°
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 2
Given a triangle with angle A = 40°, side a = 12 cm and side b = 10 cm, find the other dimensions.
A B ?
10 = b
C = ?
c = ?
a = 12
h40°
Since a > b, and h is less than a, we know this triangle has just ONE possible solution - side ‘a’opens to the outside of h.
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 2
Using the Law of Sines will give us the ONE possible solution:
BA
10 = b
C
c
a = 12
40°
12sin 40
10sin B
B sin 1 10sin 4012
32.4
C 180 40 32.4 107.6c
sin107.6 12
sin 40
c 12sin107.6sin 40
17.8
The Ambiguous Case - Summary
if angle A is acute
find the height,h = b*sinA
if angle A is obtuse
if a < b no solution
if a > b one solution
if a < h no solution
if h < a < b 2 solutions one with angle B acute, one with angle B obtuse
if a > b > h 1 solution
If a = h 1 solution angle B is right
(Ex I)
(Ex II-1)
(Ex II-2)
The Law of Sines
a
sin A
b
sin B
c
sin C
AAS ASA SSA (the
ambiguous case)
Use the Law of Sines to find the missing dimensions of a triangle when given any combination of these dimensions.
