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
A
A
S
A A
Case 1: ASA or SAA
1 Side and 2 Angles known.
S
A
S
Case 2: SSA
2 Sides and the Angle opposite one of them known.
S
A
S
Case 3: SAS
2 Sides and the included Angle known.
S
S
Case 4: SSS
3 Sides known.
S
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
(alpha, beta, gamma) as in the picture on the right, the following holds true: sin
a
= sin b
= sin c
c
α
β b
γ a
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°
{ Proof done in class }
Solving for a, b, & c we get a b c
=
=
= b sin sin
β
α a sin sin
α
β a sin sin
α
γ
= c sin
α sin
γ
=
= c sin sin
γ
β b sin sin
β
γ
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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° c
40 ° + 60 ° +
γ
= 180°
γ
= 180° - 40 ° - 60 ° = 80°
40 ° b
60 ° 4
Now let’s solve for sides b and c.
sin
α a
= sin
β b
= sin
γ c b
= c
= a sin a sin sin
sin
=
=
(
4 sin 60
°
)
(
4 sin sin sin
40
°
80
°
)
40
°
≈
≈
5 .
39
6 .
13
Now you do #9
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Example 2 (ASA)
Solve the triangle:
α
=35 ° ,
β
=15 ° , c = 5
γ
= ?, a = ?, b=?
Using the fact that the angles add up 180°, we can solve for
γ
α
+
β
+
γ
= 180°
35 ° + 15 ° +
γ
= 180°
50 ° +
γ
= 180°
γ
= 180°- 50° = 130°
Now let’s solve for sides a and b.
a
= b
= c sin
sin
c sin sin
=
=
(
5 sin 35
°
(
5 sin sin
130
°
15
°
)
) sin 130
°
≈
3 .
74
≈
1 .
69
35 °
5
5 b
15 ° a
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Example 3 Solving an SSA Triangle
Solve the triangle: a = 3, b = 2,
α
=40 ° sin
α
= sin
β
= sin a b sin 40
°
= sin
3
Solve for
β
2
β first
=
: c
γ sin c
γ sin
β = bsin a
Use inverse
α
=
2 sin
3
40
°
≈
0 .
43 sin to solve for
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 c
=
3
40
3
° sin
= sin sin
40
°
114
114 .
6
° c
≈
.
6
°
4 .
24
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Example 4 (SSA with two solutions)
Solve the triangle: a = 6, b=8 ,
α
=35 ° c =?
β
=?,
γ
= ?
Solve for one of the angles first.
sin β = bsin
α a
=
8 sin
6
35
°
≈
0 .
76
Use inverse sin to solve for
β
:
β = sin
1
0 .
76
≈
49 .
9
°
Supplement ary Angle Theorem says sin(180
°
so
β also
=
180
°
49.9
° =
130.1
°
β
)
= sin
β
Are both these answers possible?
α
+
β
+
γ
= 180° and so far
α
+
β
=35°+49.9° =84.9°
OR
α
+
β
=35°+130.1°=165.1°
Which are both less than 180°.
So the third angle,
γ
, could be
γ
= 180° - 84.9° = 95.1°
OR
γ
= 180° - 165.1° = 14.9° c
Use the Law of Sines to find c
=
OR a sin sin
α
γ c
= a sin sin
α
γ
=
=
6
6 sin sin sin sin
95
35
.
1
°
14 .
9
°
35
°
°
≈
≈
10 .
42
2 .
69
8
γ
=
95.1°
6
α
=35° β
=49.9°
10.42
γ
=
14.9°
8
α
=35°
2.69
6
β
=130.1°
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Example 5 SSA (no solution ) a = 2, c = 1,
γ
=50 ° sin a
α sin
α
= sin
β
= b asin
γ c
=
= sin
γ c
2 sin
1
50
°
≈
1 .
53
Impossible! sin values can only be between -1 and 1!
So there is NO SOLUTION!
NOW YOU DO #25
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Examples 6, 7 done in class
HOMEWORK
4.2 on p.302-306
#11,13,27,31,33,37,43,51
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