The Law

The LawThe Law

of SINESof SINES

The Law of SINESThe 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:

A C

B

ac

b

Proof of the Law of SINESProof of the Law of SINES

A B

C

a b h

c

Prove that sin A = a sin B b

In triangle CDA,sin A = h / b

In triangle CDB,sin B = h/a

So,sin A = h/bsin B h/a

Since h/b = a / b h/a

Therefore sin A = a sin B b

D

a

sin A

b

sin B

c

sin C

Use Law of SINES when ...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 .

Use the Law of Sines if you are given:

Example 1Example 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.

Example 1 (con’t)Example 1 (con’t)

A C

B

70°

80°a = 12c

b

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 8012 sin 80 b sin 70

b 12 sin80

sin 7012.6cm


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)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 used rounded values in calculations.

Example 2Example 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.

AC

B

115°

30°

a = 30

c

b


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) 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

30 sin 30 b sin35

b 30 sin30

sin3526.2cm


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)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)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.


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 = ?


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.


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:

A B

a = 2215 = b

C

c 120°

22

sin120

15

sin B15sin120 22sin B

B sin 1 15sin12022

36.2


Situation I: Angle A is obtuse - EXAMPLE

Angle C = 180° - 120° - 36.2° = 23.8°

Use Law of Sines to find side c:

A B

a = 2215 = b

C

c 120°

36.2°

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


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



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 . . .



A B ?

b

C = ?

c = ?

a

h

If a < h, then NO triangle exists with these dimensions.



A B

b

C

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.



A B

b

C

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!



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.

AB

b

C

c

a = h


Situation II: Angle A is acute - EXAMPLE 1


A B ?

15 = b

C = ?

c = ?

a = 12

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.



A B

15 = b

C

c

a = 12

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



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°



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


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°




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.



Using the Law of Sines will give us the ONE possible solution:

A B

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 The Ambiguous Case - - SummarySummary

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 SinesThe 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.

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