Online Trigonometry Tutoring: Laws of sines & cosines
Relationship between the angles and the side lengths of triangle
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Learn the laws of sines and cosines from certified online trigonometry tutor
✓ It states that “When we divide side a by the sine of angle A, it is equal to side b divided by the
sine of angle B, and also equal to side c divided by the sine of angle C”.
✓ It can be used to –
Calculate the unknown sides (Triangulation)
✓ Angles of a Triangle Law can be applied if
- SSA – two sides and angle not included between them are given
- ASA – two angles and side between them are given
- SAS – two angles and one side that is not included in the angles
What is the Law of Cosines
Relationship between the side lengths and the angles of a triangle
The Law of Cosines can be given as
Other two versions of Law of Cosines are:
a2 = b2 + c2 – 2 bc cos A
b2 = a2 + c2 – 2ac cos B
What are the Uses of Laws of Cosines?
The Laws of Cosines is used:
- To find the 3rd side of a triangle when we know the 2 sides and the angle between them (SAS)
- To find the angles of a triangle when we know all the 3 sides (SSS) of triangle.
When should we use the Laws of Sines?
1.If you are given: Angle-Side-Angle (ASA) or Angle-Angle-Side (AAS). OR
2.If you are given: Hypotenuse-Leg (HL), you have a right triangle. OR
3.If you are given: Side-Side-Angle (SSA– in that order!!), then you have the AMBIGUOUS CASE ( we’ll discuss this case after the examples)
In a triangle ABC, a = 10, b = 5, and ∠A = 45°. Find the value of ∠B.
a/sin A = b/sin B
10/sin 45 = 5/sin B
sin B = 1/2√2 or √2/4 (because sin 45 = 1/√2)
sin B = 0.3535
B = sin-1(0.3535) = 20.7
In triangle ABC, side b = 5 cm, c = 10 cm, and the angle at A is 60°. Find side a.
According to law of cosines, a2 = b2 + c2 – 2bc cos 60
a2 = 52 + 102 – 2 x 5 x 10 cos 60
a2 = 125 – 2 x 5 x 10 x ½ (cos 60 = ½)
= 125 – 50
a2 = 75
a = √75
Ambiguous Case – Law of Sines
There are 5 situations when we need to use the Ambigious Case of the Law of Sines:
Case I: Angle is acute.
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.
Using Case I: First, use SOH-CAH-TOA to find h:
h = bsin A
Then, compare ‘h’ to sides a and b . . .
Case I: If a < h, then NO triangle exists with these dimensions.
h = bsin A
Then, compare ‘h’ to sides a and b . .
Case II: If h < a < b, then TWO triangles exist with these dimensions.
If we open side ‘a’ to the outside of h, If we open side ‘a’ to the inside of h, angle B is obtuse.
angle B is acute.
Case III: 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!
Case IV: 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.
Given a triangle with angle A = 30°, side a = 14 cm and side b = 15 cm, find the other dimensions.
Using Law of Sines, =
= 10sinB = 14 sin30
sinB = = 0.7
B = (0.7) = 44.42 ≅ 44
Angles could be 30°, 44°, and 106°: sum 180°.
The angle from Quadrant II could create angles 30°, 14°, and 136°: sum 180°.
Problem 1. A = 40; B = 20; a = 2. Find side b.
Problem 2. If b = 5, c = 2, A = 30, find a.
Problem 3. Given ABC, a = 8, b = 5,c = 7, find C using law of cosines.
- b = 3.76 approx
- a = 3.42 approx
- C = 60 degrees
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