# 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:

*a*^{2} = *b*^{2} + *c*^{2} – 2 *bc* cos *A*

*b*^{2} = *a*^{2} + *c*^{2} – 2*ac* cos *B*

### What are the Uses of Laws of Cosines?

The Laws of Cosines is used:

- To find the 3
^{rd}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)

**Example 1**

**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

**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,* a*^{2} = *b*^{2} + *c*^{2} – 2*bc* cos 60

* * a^{2} = 5^{2} + 10^{2} – 2 x 5 x 10 cos 60

* a*^{2} = 125 – 2 x 5 x 10 x ½ (cos 60 = ½)

= 125 – 50

* a*^{2} = *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 =

h = bsin A

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

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

sinA =

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°.

**Check Point**

**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.

**Answers**

*b*= 3.76 approx*a*= 3.42 approx*C*= 60 degrees

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