In the diagram, the circles with centresThe Problem:P,Q, andRhav radii of 3, 2, and 1 respectively. Each circle touches the other two as shown. Find the area of triangle PQR.

I used the law of cosines to find angle P.

$\displaystyle c^2 = a^2 + b^2 - 2ac(cos \theta)$

$\displaystyle 3^2 = 4^3 + 5^2 -2(4)(5)(cos P)$

$\displaystyle 9 = 41 -40 (cos P)$

$\displaystyle -32 = -40 Cos P$

$\displaystyle -32/-40 = Cos P$

$\displaystyle 0.8 = Cos P $

$\displaystyle 0.8 (inverse cos) = P$

$\displaystyle P = 36.87^o$

I then thought of using the sine or cosine formula to find the height (I thought of drawing a height from the centre R to where the circles containing P and Q touch). But I don't know that it's a right-angle triangle. I am stuck at this point.