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Online Trigonometry Tutoring: Laws of sines & cosines

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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:

  1. To find the 3rd side of a triangle when we know the 2 sides and the angle between them (SAS)
  2. 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)


Example 1

In a triangle ABCa = 10, b = 5, and  ∠= 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

Example 2

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:

sinA = \frac{h}{b}

h = bsin A

Then, compare ‘h’ to sides a and b . . .

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

sinA = \frac{h}{b}

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 = 15 cm, find the other dimensions.

Using Law of Sines, \frac{a}{sinA} = \frac{b}{sinB}

\frac{10}{sin 30} = \frac{14}{sin B} \Rightarrow 10sinB = 14 sin30

sinB = \frac{14(0.5)}{10} = 0.7

B = sin^{-1} (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°.


Check Point

Problem 1. A = 40; B = 20; a = 2. Find side b.

Problem 2.  If  b = 5, c = 2, = 30, find a.

Problem 3.  Given ABC, a = 8, b = 5,c = 7,  find C using law of cosines.

Answer Key
  1. b = 3.76 approx
  2. a = 3.42 approx
  3. C = 60 degrees

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