For this case we use the following formula
Area of Sector = Area * radians of sector / 2 * pi radians
Where,
Area: it is the area of the complete circle.
We have then:
Area = pi * r ^ 2
Area = pi * (6) ^ 2
Area = 36pi
Substituting values:
5pi = 36pi * radians of sector / 2 * pi
Clearing:
radians of sector = ((5pi) * (2pi)) / (36pi)
radians of sector = (10pi ^ 2) / (36pi)
radians of sector = (10pi) / (36)
radians of sector = (10/36) pi
radians of sector = (5/18) pi
in degrees:
(5/18) pi * (180 / pi) = 50 degrees
Answer:
The measure of the central angle is:
50 degrees
Answer:


Where 
And replacing we got:

And solving we got:

Where 
And the possible solutions are:

Step-by-step explanation:
For this case we use the equation given by the image and we have:

We can rewrite the last expression like this if we multiply both sides of the equation by -1.

Now we can use the quadratic formula given by:

Where 
And replacing we got:

And solving we got:

Where 
And the possible solutions are:
