# Triangle in circle?!

• Jan 23rd 2010, 10:27 PM
davidman
Triangle in circle?!
Triangle ABC has angle B 45 degrees, side AB 9 and side BC $6\sqrt2$.

The area of ABC is:

And since side AC is:

... the circle containing triangle ABC has the radius:

Would be great to get some insight on this. I'm completely lost as to what to do.
• Jan 23rd 2010, 11:20 PM
Prove It
Quote:

Originally Posted by davidman
Triangle ABC has angle B 45 degrees, side AB 9 and side BC $6\sqrt2$.

The area of ABC is:

And since side AC is:

... the circle containing triangle ABC has the radius:

Would be great to get some insight on this. I'm completely lost as to what to do.

Use the formula

$A = \frac{1}{2}ab\,\sin{C}$, where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them.
• Jan 23rd 2010, 11:31 PM
davidman
$A=\frac{1}{2}absinC=\frac{1}{2}\times9\times6\sqrt {2}sin45$

and looking at triangles with special angles 45 and 90 tells me that

$sin45=\frac{1}{\sqrt{2}}\:\:\therefore$

$27\sqrt{2}\times\frac{1}{\sqrt{2}}=27$

how to get side AC? I only know that one angle...
• Jan 23rd 2010, 11:44 PM
Prove It
Quote:

Originally Posted by davidman
$A=\frac{1}{2}absinC=\frac{1}{2}\times9\times6\sqrt {2}sin45$

and looking at triangles with special angles 45 and 90 tells me that

$sin45=\frac{1}{\sqrt{2}}\:\:\therefore$

$27\sqrt{2}\times\frac{1}{\sqrt{2}}=27$

how to get side AC? I only know that one angle...

Now to get side $c$, use the Cosine Rule.

$c^2 = a^2 + b^2 - 2ab\,\cos{C}$.

Also, since you found the area of the triangle before, you need to write $= 27\,\textrm{units}^2$
• Jan 24th 2010, 12:03 AM
davidman
$AC^2=AB^2+BC^2-2(AB\times{BC})cosB=9^2+(6\sqrt{2})^2-2(9\times{6}\sqrt{2})cos45=$ $81+36\times{2}-2\times{9}\times{6}\sqrt{2}\times{\frac{1}{\sqrt{2 }}}=45$

$AC=\sqrt{45}=\sqrt{9\times5}=3\sqrt{5}$

ok, managed with that somehow... about the circle though, is there a rule for that?
• Jan 24th 2010, 12:15 AM
Prove It
Quote:

Originally Posted by davidman
Triangle ABC has angle B 45 degrees, side AB 9 and side BC $6\sqrt2$.

The area of ABC is:

And since side AC is:

... the circle containing triangle ABC has the radius:

Would be great to get some insight on this. I'm completely lost as to what to do.

Are you saying that the vertices A, B, C need to lie on the circle?
• Jan 24th 2010, 12:22 AM
davidman

wasn't sure what it's called in English, but I guess circumscribed circle of a triangle.
• Jan 24th 2010, 12:34 AM
Prove It
OK I don't know if there's an easier way, but here goes:

I placed the length of 9 units along the $x$ axis of a set of cartestian axes, beginning at the origin.

So that means that two of the vertices lie at $(0,0)$ and $(9, 0)$.

There is also an angle of $45^\circ$ made with the $x$ axis, and a length of $6\sqrt{2}$ units.

Using some trigonometry, I know that the final vertex must have co-ordinate $(x, y) = (6\sqrt{2}\,\cos{45^\circ}, 6\sqrt{2}\,\sin{45^\circ})$

$(x, y) = (6, 6)$.

So now you have the three vertices as $(x, y)$ co-ordinates.

Substitute the three co-ordinates into the general equation for the circle

$(x - h)^2 + (y - k)^2 = r^2$.

You will end up with three equations in three unknowns that you will need to solve simultaneously. One of them ( $r$) is the radius of the circle.
• Jan 24th 2010, 01:52 AM
davidman
Quote:

Originally Posted by Wikipedia @ http://en.wikipedia.org/wiki/Law_of_sines#Relation_to_the_circumcircle
In the equation

the common value of the three fractions is actually the diameter of the triangle's circumcircle.

hence;

$\frac{a}{sin\:A}=2R$

or in my case;

$\frac{AC}{sin\:B}=\frac{3\sqrt{5}}{\frac{1}{\sqrt{ 2}}}=\frac{3\sqrt{10}}{1}$

$R=\frac{3\sqrt{10}}{2}$

so I guess wikipedia helped save the day this time.(Giggle)