4.2 THE LAW OF SINES
Solving an Oblique Triangle (meaning finding all sides and
angles) depending on which sides and angles are given:
S
S
A
A
Case 1: ASA
1 Side and 2 Angles
known.
A
S
Case 2: SSA
2 Sides and the Angle
opposite one of them
known.
S
S
S
A
S
S
Case 3: SAS
2 Sides and the
included Angle known.
Case 4: SSS
3 Sides known.
Solved using
Law of Cosines
Solved using
Law of Sines
Law of Sines
For a triangle with sides a,b,c and
angles α,β,γ as in the picture on the
right, the following holds true:
sin α sin β sin γ
=
=
a
b
c
Once two angles are known, the third
is easily calculated using the fact that
the sum of the angles of any triangle
equals 180 degrees.
α + β + γ = 180°
c
a
β
γ
α
b
{ Proof done in class }
Example 1 (SAA)
Solve the triangle: α=40°, β=60°, a = 4
γ = ?, b = ?, c=?
Using the fact that the angles add up 180°, we can solve for γ
α + β + γ = 180°
40° + 60° + γ = 180°
γ = 180° - 40° - 60° = 80°
Now let’s solve for sides b and c.
sin α sin β sin γ
=
=
a
b
c
sin 40° sin 60° sin 80°
=
=
4
b
c
(4 sin 60°) ≈ 5.39
b=
sin 40°
c=
Now you do #1
(4 sin 80°) ≈ 6.13
sin 40°
Example 3 Solving an SSA Triangle
Solve the triangle: a = 3, b = 2, α=40°
sin α sin β sin γ
=
=
a
b
c
sin 40° sin β sin γ
=
=
3
2
c
Solve for β first :
2 sin 40°
≈ 0.43
3
Use inverse sin to solve for β :
sin β =
β = sin -1 0.43 ≈ 25.4°
However, this only gives the Quadrant I answer.
We know from the Supplement ary Angle theorem that
sin(180 - β ) = sin β , so another possible answer is 180 ° - 25.4 ° = 154 .6°
We can check which angle will work for our triangle by testing it with the
fact that the sums of the angles should add up to 180°
The second possible answer, β = 154.6°, makes the sum α + β + γ > 180°,
so that won’t work. Therefore, β = 25.4°
Now that we know two angles,
γ = 180° - 40° - 25.4° = 114.6°
Now we can solve for c.
sin 40° sin 114.6°
=
3
c
3 sin 114.6°
c=
≈ 4.24
sin 40°
NOW YOU DO #17
Examples 6, 7 done in class
HOMEWORK
4.2 on p.285-285
#1,5,15, 17,23,25, 29, 31, 33
A
c = a + b − 2ab cos γ
2
2
2
S
S
Case 3: SSA
2 Sides and the
included Angle known.
b 2 = a 2 + c 2 − 2ac cos β
a 2 = b 2 + c 2 − 2bc cos α
x/a = cos γ
x =acos γ
y/a = sin γ
y =asin γ
S
S
S
4.3 LAW OF COSINES
Case 4: SSS
3 Sides known.
Solved using
Law of Cosines
B(x,y) =(acos γ, asin γ)
β
a
γ
c
y
α
x
A = (b,0)
b
We can use the distance formula for the distance from point A to B to calculate the length of side c.
DISTANCE FORMULA
The distance between two points P1 = (x1,y1) and P2 = (x2,y2) is
d(P1,P2) =
( x2 − x1 ) 2 + ( y2 − y1 ) 2
Squaring the distance formula for points B=(acos γ, asin γ) and A(b,0) gives:
c 2 = (b − a cos γ ) 2 + (0 − a sin γ ) 2
c 2 = b 2 − 2ab cos γ + a 2 cos 2 γ + a 2 sin 2 γ
c 2 = b 2 − 2ab cos γ + a 2 (cos 2 γ + sin 2 γ )
=1
c 2 = b 2 + a 2 − 2ab cos γ
Example 1
Solve the triangle: a=2, b=3, γ = 60°
Solve for c first: 2
2
2
c = b + a − 2ab cos γ
c 2 = 32 + 2 2 − 2(2)(3) cos(60°)
c 2 = 9 + 4 − 12( 1 )
2
c 2 = 13 − 6 = 7
c= 7
Now we need to find α and β
The Law of Cosines formulas with an α in it is:
a 2 = b 2 + c 2 − 2bc cos α
Solve this equation for cos α and then solve for α:
2bc cos α = b 2 + c 2 − a 2
( )
2
2 7
b 2 + c 2 − a 2 32 + 7 − 2 2 9 + 7 − 4 12
=
=
=
=
cos α =
2bc
7
2(3)( 7 )
6 7
6 7
⎛2 7 ⎞
⎟ ≈ 40.9°
⎟
⎝ 7 ⎠
α = cos −1 ⎜⎜
Use a similar process to solve for β:
b 2 = a 2 + c 2 − 2ac cos β
2
7
a 2 + c 2 − b 2 2 2 + ( 7 ) 2 − 32 4 + 7 − 9
=
=
=
=
cos β =
2ac
2(2)( 7 )
4 7
4 7 14
⎛ 7⎞
⎟ ≈ 79.1°
⎟
14
⎝
⎠
β = cos −1 ⎜⎜
We could have just found β using β=180°-α-γ=180°-60°-40.9°=79.1°
Example 1
Solve the triangle: a=2, b=3, γ = 60°
Solve for c first: 2
2
2
c = b + a − 2ab cos γ
c 2 = 32 + 2 2 − 2(2)(3) cos(60°)
c 2 = 9 + 4 − 12( 1 )
2
c 2 = 13 − 6 = 7
c= 7
Now we need to find α and β
The Law of Cosines formulas with an α in it is:
a 2 = b 2 + c 2 − 2bc cos α
Solve this equation for cos α and then solve for α:
2bc cos α = b 2 + c 2 − a 2
( )
2
2 7
b 2 + c 2 − a 2 32 + 7 − 2 2 9 + 7 − 4 12
=
=
=
=
cos α =
2bc
7
2(3)( 7 )
6 7
6 7
⎛2 7 ⎞
⎟ ≈ 40.9°
⎟
⎝ 7 ⎠
α = cos −1 ⎜⎜
Use a similar process to solve for β:
b 2 = a 2 + c 2 − 2ac cos β
2
7
a 2 + c 2 − b 2 2 2 + ( 7 ) 2 − 32 4 + 7 − 9
=
=
=
=
cos β =
2ac
2(2)( 7 )
4 7
4 7 14
⎛ 7⎞
⎟ ≈ 79.1°
⎟
14
⎝
⎠
β = cos −1 ⎜⎜
We could have just found β using β=180°- α – γ = 180°-60°-40.9° = 79.1°
Example 2
Solve the triangle: a=4, b=3, c=6
Find α the same way we did before:
cos α =
b 2 + c 2 − a 2 32 + 6 2 − 4 2 9 + 36 − 16 29
=
=
=
2bc
2(3)(6)
36
36
⎛ 29 ⎞
⎟ ≈ 36.3
⎝ 36 ⎠
a 2 + c 2 − b 2 4 2 + 6 2 − 32 16 + 36 − 9 43
cos β =
=
=
=
2ac
2(4)(6)
48
48
α = cos −1 ⎜
⎛ 43 ⎞
⎟ ≈ 26.4°
48
⎝ ⎠
β = cos −1 ⎜
And the “shortcut” to finding γ is:
γ =180°- α - β=180°-60°-40.9° = 79.1°
NOW YOU DO #7
Example 3
Let a=150, b=15mph*4hrs = 60 miles, γ = 20 °.
θ
Naples
20°
150 miles
15mph*4hours
=60miles
Key West
c) How much time has been added to the trip?
With no crosswinds, it should have taken 150
miles/15mph = 10 hours.
Now it will take 4 hours + 96 miles/15mph = 10.4
hours
Therefore, the extra time is 10.4-10=.4 hours =
.4*60 minutes = 24 minutes
a) How far is the boat from Key West after 4 hours?
That is the third side of the triangle. We know one side is
150, and one side is 60, and one angle is 20 °.
2
2
c =a + b2 – 2abcos γ
= 1502 + 602 – 2(150)(60)cos 20°
= 9186
c = 95.8
b) Through what angle should the boat turn to correct its
course?
The angle the boat must turn is supplementary to the
angle, α, in the triangle. That is, θ = 180 – α
Find α :
cosα =
b 2 + c 2 − a 2 602 + 962 − 1502
=
≈ −.8406
2bc
2(60)(968)
α = cos−1 (−.8406) ≈ 147.2°
REMEMBER,THIS IS NOT THE ANSWER.THE ANSWER IS
θ = 180° - 147.2° = 32.8°
* This problem assumes the crosswinds stopped after 4 hours.
HOMEWORK
4.3
P.292
#1, 7, 9, 11, 17, 25, 27, 31