In the triangle below, 1) Find the length of BC2) Find the are area of Δ ABC

1. Since we have two sides (AC and AB) and the included angle (60°), we are going to use the law of cosines to find the length of BC:
[tex]BC= \sqrt{AB^2+AC^2-2(AB)(AC)Cos \alpha } [/tex]
[tex]BC= \sqrt{6^2+4^2-2(6)(4)Cos(60)} [/tex]
[tex]BC= \sqrt{36+16-48Cos(60)} [/tex]
[tex]BC= \sqrt{52-48Cos(60)} [/tex]
[tex]BC=2 \sqrt{7} [/tex]

We can conclude that the length of the side BC is [tex]2 \sqrt{7} [/tex].

2. To find the area of triangle ABC we are going to use Heron's formula: [tex]A= \sqrt{S(S-A)(S-B)(S-C)} [/tex] 
where
[tex]A[/tex] is the area of the triangle 
[tex]S[/tex] is the semi-perimeter of the triangle 

But first, we are going to find the semi-perimeter of our triangle using the formula: [tex]S= \frac{A+B+C}{2} [/tex]
We can infer for our triangle and from our previous calculation that [tex]A=2 \sqrt{7} [/tex], [tex]B=4[/tex], and [tex]C=6[/tex]. Lets replace those values to find the semi-perimeter of our triangle:
[tex]S= \frac{A+B+C}{2} [/tex]
[tex]S= \frac{2 \sqrt{7+4+6} }{2} [/tex]
[tex]S=7.65[/tex]

Finally, we can use Heron's formula to find the area of our triangle:
[tex]A= \sqrt{7.65(7.65-2 \sqrt{7})(7.65-4)(7.65-6)}[/tex]
[tex]A=10.42[/tex]

We can conclude that the area of our triangle is 10.42 square units.