Determining the area of a triangle

[diredragon]

Homework Statement

http://i.stack.imgur.com/KzCIl.png
From the given picture the known quantities are:
$r = 7*3^{1/2}$
$BC = 13$
angle opposite to AC is
$120$ degrees

Homework Equations

3. The Attempt at a Solution [/B]
I figured i could use a sine rule to get the side $AC$
$\frac{AC}{sinb} = 2r$
I got $AC = 21$
I don't know where to go from this now. cosine rule to get side $AB$ gives me weird numbers. What should be done now?

 

[JeremyG]

angle opposite to AC is
120 degrees

Are you talking about angle B? Because if you are, that angle is clearly acute

 

[diredragon]

Are you talking about angle B? Because if you are, that angle is clearly acute

Yes angle B, in the picture it is shown as acute but in a given problem its not. The picture represents the general diagram.

 

[cnh1995]

cosine rule to get side AB gives me weird numbers.

If you get AB, you can use Heron's formula to calculate the area of the triangle.

 

[cnh1995]

yes angle B, in the picture it is shown as acute but in a given problem its not. The picture represents the general diagram.

From your diagram, it is clear that the circle is circumcircle of the triangle ABC. However, since it is an obtuse angled triangle, the circumcenter should be outside the triangle. This will affect the calculations. I think you should draw and refer the exact diagram.

 

[diredragon]

http://postimg.org/image/xspgzewtt/ here is a more specific diagram. How would the center affect the calculations?

 

[Samy_A]

There may be a shorter method, but the sine rule can give you the angle in A.
Then, having the three angles and two sides, finding the area should be easy.

 

[diredragon]

There may be a shorter method, but the sine rule can give you the angle in A.
Then, having the three angles and two sides, finding the area should be easy.

What would be the easier method? I don't really want to deal with the cosine rule, there has to be simpler approach.

 

[Samy_A]

What would be the easier method? I don't really want to deal with the cosine rule, there has to be simpler approach.

Once you have the three angles, you don't need the cosine rule.
Remember the basic formula for the area of a triangle? Area=base*height/2.

 

[diredragon]

Once you have the three angles, you don't need the cosine rule.
Remember the basic formula for the area of a triangle? Area=base*height/2.

I don't see a way of getting all three angles. I have $a$ $b$ and an angle $ABC$ . I tried to calculate the $c$ side by using the cosine rule
$b^2 = a^2 + c^2 - 2accosB$
from this i get
$c^2 + 13c - 272 = 0$ but the numbers don't come out right

 

[cnh1995]

I have a b and an angle ABC

ABC is not the angle between a and b right? You can get all the three sides and angles using sine rule here.

 

[diredragon]

ABC is not the angle between a and b right? You can get all the three sides and angles using sine rule here.

I need to make a correction. I have written that $AB = 13$, it actually equals $9$. That then gives me fine numbers. Using the sine rule:
$\frac{b}{sinb}=2r$ i get $b = 21$
Then using the cosine rule
$b^2=a^2 + c^2 - 2accosb$ i get $a=16$ and $c = 9$ is known.
Now the height:
$h_a=csin60 = \frac{9*3^{1/2}}{2}$
Finally $A=\frac{(h_a)*a}{2} = 36\sqrt{3}$