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# Law of Sine and Law of Cosine

# Law of Sine

# What is the relation among the sides and angles in an obtuse triangle?

# The Law of Sine

# Apply the law of Sine to solve problems

# Law of Cosine

# The law of Cosine

The figure below show an acute triangle ABC, its three vertex A, B and C opposite three sides a, b and c respectively. What is the relation among the sides and the angles in the triangle? (Note: a capital letter represents the vertex and a small letter represents the length of a side in the triangle. )

Now we draw a line that pass through the pint C and perpendicular to AB at the point D. (D is the perpendicular point.)

- In right triangle ACD, use the definition of Sine, sin A = CD/AC = CD/b
- so CD = b sin A
- In right triangle BCD, sin B = CD/BC = CD/a
- so CD = a sin B
- Then b sin A = a sin B
- (Both sides of the equation divide by sin B sin A)
- so b/sin B = a/sin A
- The formula above tell us, in triangle ABC, the ratio of b to sin B equals the ratio of a to sin A (Note: in triangle ABC, side b opposite the angle B and the side a opposite the angle A).

**Now we will find the relation between the side c and the angle C.**

- angle C = angle 1 + angle 2
- sin C = sin(angle1 + angle2) = sin(angle1) cos(angle2) + cos(angle1) sin(angle2)
- sin(angle1) = BD/BC = BD/a
- cos(angle2) = CD/AC = CD/b
- cos(angle1) = CD/BC = CD/a
- sin(angle2) = AD/AC = AD/b
- Then sin C = BD/a × CD/b + CD/a × AD/b = [CD/ab] × (BD + AD)
- Note: BD + AD = AB = c
- Then sin C = [CD/ab] × c
- Note: sin B = CD/a
- So sin C = [sin B/b] × c
- Both sides divide by the side c
- So sin C/c = sin B/b
- So c/sin C = b/sin B
- Therefore, we get the law of Sine:
- a/sin A = b/sin B = c/sin C

The law of Sine describe the relation between three sides and the Sine of their corresponding angles. If we know two angles and a side, we can find other angle in the triangle from the sum of a triangle theorem. Then apply the law of Sine to calculate the other two sides. If we know any two sides and the angle opposite one of the sides, apply the law of Sine, we can find the value of Sine of the angle of other side opposite, from here to find other sides and angles.

The figure below show an obtuse triang (90^{o} < angle C < 180^{o}). Now we prove the law of Sine.

Draw a line passing through the point C and perpendicular to AB at point D (D is the perpendicular point.)

- In the figure above, in the right triangle BCD, sin B = CD/BC = CD/a,
- so CD = a sin B
- In the right triangle ACD, sin A = CD/AC = CD/b,
- so CD = b sin A
- So a sin B = b sin A,
- Both sides of the equation divide by sin A sin B
- so a/sin A = b/sin B

**Now we will find the relation between sin C and c**

Extended the BC to the left and draw a line passing through the point A and perpendicular to the extended BC line at the point E (E is the perpendicular point.)

- In the figure above, since BE is a straight line, so the angle BCE = 180
^{o} - angle C = 180
^{o}- angle 1 - sin C = sin(180
^{o}- angle1) = sin180^{o}cos(angle1) - cos180^{o}sin(angle1) - Since sin180
^{o}= 0, cos180^{o}= -1 - So sin C = sin(angle1)
- In right triangle ACE, sin(angle1) = AE/AC = AE/b,
- so AE = b sin(angle1) = b sin C
- In right triangle ABE, sin B = AE/AB = AE/c,
- so AE = c sin B
- So b sin C = c sin B
- Both sides of the equation divide by sin B sin C
- So b/sin B = c/sin C
- Therefore, we get,
- a/sin A = b/sin B = c/sin C

Thus we prove that the law of Sine satisfy the obtuse triangle.

**Example 1**

- Question:
- In triangle ABC, a = 49, B = 75
^{o}and C = 45^{o}, what is the value of c?

- Solution:
- A = 180
^{o}- (B + C) = 180^{o}- (75^{o}+ 45^{o}) = 180^{o}- 120^{o}= 60^{o} - The degree measure of the angle A is 60
^{o} - Apply the law of Sine
- The value of c is 40.

**Example 2**

- Question:
- In triangle ABC, c = 120, C = 105
^{o}and b = 88, find a, A and B.

- Solution:
- sin C = sin105
^{o} - = sin(180
^{o}- 75^{o}) - = sin75
^{o} - = sin(45
^{o}+ 30^{o}) - = sin45
^{o}cos30^{o}+ cos45^{o}sin30^{o} - = 0.9659
- Apply the law of Sine
- Because sin B = 0.7083, so B = 45
^{o} - A = 180
^{o}- (B + C) - = 180
^{o}- (45^{o}+ 105^{o}) - = 180
^{o}- 150^{o} - = 30
^{0} - Apply the law of Sine
- Thus, we get the triangle,
- In triangle ABC, because C > B > A, so c > b > a

If we know two sides and the angle between them, how to find the third side in the triangle? For example, if we know the sides b, c and the angle A, how to find the length of the side a?

In xy-plane, let the point A be the origin and the side AB lies on the a-axis. In the figure below, the coordinate of the point A is (0, 0), the coordinate of the point B is (x_{1}, 0),
[note: the y-coordinate of the point B is 0, because the point B lies on the x-axis], the coordinate of the point C is (x_{2}, y_{2}). Now we find the length of the side a.

- Note: cos A = x
_{2}/square root of (x_{2}^{2}+ y_{2}^{2}) - Thus, we get the law of Cosine, a
^{2}= b^{2}+ c^{2}- 2bc cosA

- If we know three sides in a triangle, how to find the angles in this triangle? From the law of Cosine, we get the following formula.

**Example 3**

- Question:
- In triangle ABC, the length of the side a = 36, B = 45
^{o}and the length of the side c = 40, find the triangle.

- Solution:
- Find the length of the side b
- b
^{2}= a^{2}+ c^{2}- 2ac cos B - = 36
^{2}+ 40^{2}- 2 × 36 × 40 cos45^{o} - = 1296 + 1600 - 2880 × (square root of 2)/2
- = 860
- b = square root of 860 = 29.32
- Find the angle B
- Find the angle C
- The triangle shown below