# In triangle ABC, AC = 6, CB = 8 and angle ACB = 30 degrees. What is the length of side AB

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Given `Delta ABC ` . Let A,B,C be the vertices and a,b,c be the side lengths where a=BC,b=AC and c=AB.

We are given a=8,b=6 and `m/_ ACB=30^@ ` . We are asked to find c=AB.

We know two sides and the included angle of the triangle, so we use the Law of Cosines.

Here `c^2=a^2+b^2-2abcosC ` :

`c^2=64+36-2(8)(6)cos30^@ `

`c^2=100-96(sqrt(3)/2)`

` ``c^2~~16.86156` ==> `c~~4.10628 `

**Using the law of cosines, we find that the length of side `bar(AB) ` is approximately 4.1**

AC = 6;BC = 8;angle ACB = 30° {given}

Applying sine rule to the triangle ABC,

AC/sinABC = AB/sinACB = BC/sinBAC

Let angle ABC be x,then angle BAC = 180-30-x = 150-x

AC/sinABC = BC/sinBAC

6/sinx = 8/sin(150-x)

Expanding sin(150-x) = sin150cosx-cos150sinx

8sinx = 6×(sin150cosx-cos150sinx)

8sinx = 3cosx+5.2sinx

2.8sinx = 3cosx

tanx = 1.071

x = 46.97°

AC/sinABC = AB/sinACB

6/sin46.97 = AB/sin30

AB = 4.1

Since we are given an angle measure and 2 sides, we can use The Law of Cosines to solve for the third side. The Law of Cosines works with any type of triangle and all you have to do is input your known values into the formula:

`c^(2) = a^2 + b^2 - 2ab(cos(C))`.

You are given the values AC = 6, CB = 8 and `/_` ACB = 30 degrees.

`c^(2) = 64 + 36 - 2(6)(8)(cos(30))`

cos of 30, according to the unit circle, would equal `(sqrt(3))/(2)`.

`c^(2) = 100 - 96(cos((sqrt(3))/(2)))`

`c^(2) = 16.86`

**C would equal about 4.1**