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:

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

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) $a^2-b^2=c(a cos B - b cos A)$
b) $c= a cos B + b cos A$

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

Any help would be greatly appreciated!

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

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

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

The triangle looks like this:

Code:

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

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

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

Thank you!