Transcript The Laws of SINES - McAllister's Math

Do Now

If the legs of the right triangle are 4
and 5 find the hypotenuse and all the
angles.
The Law
of SINES
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
Use Law of SINES when ...
you are given:
 AAS - 2 angles and 1 adjacent side
 ASA - 2 angles and their included side
 SSA (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 (continued)
B
The angles in a ∆ total 180°,
so angle C = 30°.
80°
a = 12
c
A 70°
b
Set up the Law of Sines to
find side b:
C
12
b

sin 70 sin 80
12sin 80  b sin 70
12sin80
b
12.6cm
sin 70
Example 1 (continued)
B
80°
c
A 70°
Set up the Law of Sines to
find side c:
a = 12
b = 12.6
30°
12
c

sin 70 sin 30
C
12sin 30  c  sin70
12sin 30
c
 6.4cm
sin70
Example 1 (solution)
A 70°
B
Angle C = 30°
80°
Side b = 12.6 cm
a = 12
b = 12.6
30°
Side c = 6.4 cm
Note:
C
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 (continued)
To solve for the missing sides or
angles, we must have an angle and
opposite side to set up the first
equation.
B
30°
c
a = 30
115°
C
b
We MUST find angle A first because
the only side given is side a.
A
The angles in a ∆ total 180°, so angle
A = 35°.
Example 2 (continued)
B
Set up the Law of Sines to find side b:
30
b

sin35 sin 30
30°
c
a = 30
115° 35°
C
b
A
30sin 30  b sin35
30sin30
b
 26.2cm
sin35
Example 2 (continued)
B
Set up the Law of Sines to find side c:
30°
c
a = 30
115° 35°
C
b = 26.2 A
30
c

sin35 sin115
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
a = 30
115° 35°
C
b = 26.2 A
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.
C=?
angle C is so we can’t draw
side ‘a’ in the right position
b
A
‘a’ - we don’t know what
c=?
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
C = ? impossible.
a
b
A
C=?
If a > b, then there is
ONE triangle with
these dimensions.
a
b
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.
C=?
b
A
a
c=?
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.
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
b
h
A
c
C
b
a
B
If we open side ‘a’ to the
outside of h, angle B is acute.
A
c
a 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.
C
b
a
h
A
c
B
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.
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=?
h  bsin A
a = 12
15 = b
h  15sin40  9.6
h
A
40°
c=?
B?
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
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
15sin40 
B  sin 1 
 53.5
 12 
C  180  40  53.5  86.5
c
12

sin86.5 sin 40
12sin86.5
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
40°
c
B
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
C
Angle B = 126.5°
Angle C = 180°- 40°- 126.5° = 13.5°
15 = b
a = 12
A
40° 126.5°
c
B
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 = 53.5°
Angle C = 86.5°
Side c = 18.6
Angle B = 126.5°
Angle C = 13.5°
Side c = 4.4
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=?
a = 12
10 = b
h
A
40°
c=?
B?
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:
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(Ex II-1)
one with angle B acute,
one with angle B obtuse
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)