The sides of a triangle are in the ratio 3:4:5

a) Where is its circumcentre located?

b) Into how many arcs do the sides of the triangle divide the circumcircle

c) What is theratio of the sum of the central angles of the shorter arcs and the cental angle of the longest arc?

$$\begin{gathered} Let the sides of a trinagle be 3x,4x,5x \\ Now \sin ce (3x{)}^2+(4x{)}^2=(5x{)}^2 \\ Hence it follows pythagoras theorem. \\ Hence the triangle is a right angled triangle \\ a) Now as we know that circumcentre of a right angled triangle is the centre of the hypotenuse of the triangle. \\ Hence circumcentre in this case is the centre of the side having length = 5x \end{gathered}$$


Now it is clearly seen from the figure that AC is the diameter of the circle.
AB,BC,AC divides the circumcircle into three arcs $\stackrel{⏜}{AB},\stackrel{⏜}{BC}and \stackrel{⏜}{CA}$ respectively.
Now  AC is the diameter of the circle . Central angle corresponding to the largest arc i.e$\stackrel{⏜}{CA}$ is $\angle COA$ and since COA is a straight line (as AC is the diameter) .Hence $\angle COA$= ​180$°$.
Now sum of the central angles of the smaller arcs = 360$°$-central angle of the larger arc=360$°$-180$°$=180$°$

$\therefore \frac{sum of central angles of smaller arcs}{central angle of larger arcs}=\frac{180°}{180°}=1$