Transcript Document

The Law of SINES
For any triangle (right, acute or obtuse), you may
use the following formula to solve for missing
sides or angles:
a
b
c


sin A sin B sin C
When to use the law of Sines
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:
•
•
•
AAS or SAA - 2 angles and 1 adjacent side
ASA - 2 angles and their included side
SSA or ASS (this is an ambiguous case)
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)
B
The angles in a ∆ total 180°, so angle C
= 30°.
80°
a = 12
c
A
Set up the Law of Sines to find side b:
70°
b
C
12
b

sin 70 sin 80
12sin 80  b sin 70
12sin80
b
12.6cm
sin 70
Example 1 (con’t)
B
Set up the Law of Sines to find side c:
12
c

sin 70 sin 30
80°
a = 12
c
A
70°
30°
b = 12.6
C
12sin 30  c  sin70
12sin 30
c
 6.4cm
sin70
Example 1 (solution)
B
Angle C = 30°
Side b = 12.6 cm
80°
Side c = 6.4 cm
a = 12
Note:
A
70°
30°
b = 12.6
C
We used the given values of A and a in
both calculations. Your answer is more
accurate if you do not used rounded values
in calculations.
LECTURE VIDEO
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)
To solve for the missing sides or angles, we must
have an angle and opposite side to set up the first
equation.
B
30°
We MUST find angle A first because the only side
given is side a.
c
a = 30
The angles in a ∆ total 180°, so angle A = 35°.
C
115°
b
A
Example 2 (con’t)
B
Set up the Law of Sines to find side b:
30°
30
b

sin35 sin 30
c
a = 30
C
115°
35°
b
A
30sin 30  b sin35
30sin30
b
 26.2cm
sin35
Example 2 (con’t)
B
Set up the Law of Sines to find side c:
30
c

sin35 sin115
30°
c
a = 30
C
115°
35°
b = 26.2
A
30sin115  c  sin35
30sin115
c
 47.4cm
sin35
Example 2 (solution)
B
Angle A = 35°
30°
Side b = 26.2 cm
c = 47.4
Side c = 47.4 cm
a = 30
C
115°
35°
b = 26.2
A
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.
C=?
b
A
c=?
‘a’ - we
don’t know what
angle C is so we can’t draw
side ‘a’ in the right position
B?
The Ambiguous Case (SSA)
Situation I: Angle A is obtuse
If angle A is obtuse there are TWO possibilities
If a ≤ b, then a is too short to reach
side c - a triangle with these
dimensions is impossible.
C=?
C=?
If a > b, then there is ONE
triangle with these
dimensions.
a
a
b
b
A
c=?
B?
A
c=?
B?
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:
C
22
15

sin120 sin B
15sin120  22sin B
a = 22
15 = b
A
120°
c
B
15sin120 
B  sin 1 
 36.2
 22

The Ambiguous Case (SSA)
Situation I: Angle A is obtuse - EXAMPLE
Angle C = 180° - 120° - 36.2° = 23.8°
C
Use Law of Sines to find side c:
a = 22
15 = b
A
120°
c
B
36.2°
22
c

sin120 sin 23.8
c sin120  22sin 23.8
22sin 23.8
c
 10.3cm
sin120
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.
C=?
b
A
a
c=?
B?
The Ambiguous Case (SSA)
Situation II: Angle A is acute
First, use SOH-CAH-TOA to find h:
C=?
b
a
h
A
c=?
B?
h
sin A 
b
h  bsin A
Then, compare ‘h’ to sides a and b . . .
The Ambiguous Case (SSA)
Situation II: Angle A is acute
If a < h, then NO triangle exists with these dimensions.
C=?
a
b
h
A
c=?
B?
The Ambiguous Case (SSA)
Situation II: Angle A is acute
If h < a < b, then TWO triangles exist with these dimensions.
C
C
b
h
A
c
b
a
a
B
If we open side ‘a’ to the
outside of h, angle B is acute.
A
c
h
B
If we open side ‘a’ to the
inside of h, angle B is obtuse.
The Ambiguous Case (SSA)
Situation II: Angle A is acute
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!
C
b
a
h
A
c
B
The Ambiguous Case (SSA)
Situation II: Angle A is acute
If h = a, then ONE triangle exists with these dimensions.
C
b
A
a=h
c
B
If a = h, then angle B must be a right
angle and there is only one possible
triangle with these dimensions.
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.
Find the height:
C=?
a = 12
15 = b
h
A
40°
c=?
B?
 but
b sin
Athere are 2
Since a h
> h,
a< b,
solutions and we must find BOTH.
h  15sin 40  9.6
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
FIRST SOLUTION: Angle B is acute - this is the solution you get when you
use the Law of Sines!
C
a = 12
15 = b
h
A
40°
c
B
12
15

sin 40 sin B
 15sin 40 
B  sin 1 
  53.5
12


C  180  40  53.5  86.5
c
12

sin 86.5 sin 40
12sin 86.5
c
 18.6
sin 40
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.
C
1st ‘a’
15 = b
A
a = 12
Angle B = 180 - 53.5° = 126.5°
40°
c
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.
B
1st ‘B’
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
SECOND SOLUTION: Angle B is obtuse
C
15 = b
a = 12
A
40° 126.5° B
c
c
12

sin13.5 sin 40
12sin13.5
c
 4.4
sin 40
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
Angle B = 53.5°
Angle C = 86.5°
Side c = 18.6
13.5°
C
15 = b
A
40°
86.5°
15 = b
a = 12
53.5° B
c = 18.6
C
a = 12
A
40° 126.5° B
c = 4.4
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.
C=?
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.
a = 12
10 = b
h
A
40°
c=?
B?
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 2
Using the Law of Sines will give us the
ONE possible solution:
C
a = 12
10 = b
A
40°
c
B
12
10

sin 40 sin B
1 10sin 40
B  sin 
 32.4

12
C  180  40  32.4  107.6
c
12

sin107.6 sin 40
12sin107.6
c
 17.8
sin 40
The Ambiguous Case - Summary
if angle A is
obtuse
if a < b  no solution
if a > b  one solution
(Ex I)
if a < h  no solution
if angle A is acute
find the height,
h = b*sinA
if h < a < b  2 solutions
one with angle B acute,
one with angle B obtuse
(Ex II-1)
if a > b > h  1 solution
(Ex II-2)
If a = h  1 solution
angle B is right
The Law of Sines
a
b
c


sin A sin B sin C
Use the Law of Sines to find the
missing dimensions of a triangle
when given any combination of these
dimensions.

AAS
 ASA
 SSA (the
ambiguous case)