Trigonometric Ratio Problems

• Apr 26th 2010, 07:03 AM
Dovud
Trigonometric Ratio Problems
10) Three circles with centres A, B and C are mutually tangent and no circle lies inside another circle. The circle with centre A has a radius of 3 cm and the circle with centre B has a radius of 5 cm.
a) If angle BAC= 60 degrees, determine the radius of the third circle.
b) Find the area of triangle ABC

I drew a picture and went out on a wing and tried to use the cosine law:

$\displaystyle (3+x)^2=8^2+(5+x)^2-2(8)(5+x)cos 60$
$\displaystyle 9+6x+x^2=64+25+10x+x^2-80-x$
$\displaystyle 4x=64+25+40-9$
$\displaystyle x=10$

Now that's all fine and dandy, but the answer at the back of the book says the answer is 2 cm! I don't know what to do!

12In any triangle ABC, prove:
a)$\displaystyle a^2-b^2=c(a cos B - b cos A)$
b)$\displaystyle c= a cos B + b cos A$

I don't even know where to start...

Any help would be greatly appreciated!
• Apr 26th 2010, 07:43 AM
Soroban
Hello, Dovud!

Check your diagram . . . You used the wrong angle.

Quote:

Three circles with centres A, B and C are externally mutually tangent.
The circle with centre A has a radius of 3 cm and the circle with centre B has a radius of 5 cm.

(a) If angle BAC= 60°, determine the radius of the third circle.

(b) Find the area of triangle ABC.

Code:

          3          5     A * * * * * o * * * * * * * B       * 60°                *       3 *                *         *              * 5           o          *           *        o           x *    *             *  * x               *               C

Law of Cosines: .$\displaystyle (x+5)^2 \;=\;8^2 + (x+3)^2 - 2(8)(x+3)\cos60^o$

. . . . . . . . .$\displaystyle x^2 + 10x + 25 \;=\;x^2 + 6x + 9 + 64 - 8x - 24$

And we have: .$\displaystyle 12x \:=\:24 \quad\Rightarrow \quad \boxed{x \:=\:2\text{ cm}}$

The triangle looks like this:
Code:

            8     A o  *  *  *  o B       * 60°    *       5 *    * 7         *  *           o           C

Formula: .$\displaystyle \text{Area} \;=\;\tfrac{1}{2}\,bc\sin A$

Therefore: .$\displaystyle A \;=\;\tfrac{1}{2}(5)(8)\sin60^o \;=\;\boxed{10\sqrt{3}\text{ cm}^2}$

• Apr 27th 2010, 06:01 AM
Dovud
Thank you!