we know that
The measure of the external angle is the semidifference of the arcs that it covers
so
∠MLN=(1/2)*(mayor arc MN-minor arc MN)
Let
x------> minor arc MN
major arc MN=360-x
substitute in the formula above
∠MLN=(1/2)*(mayor arc MN-minor arc MN)
75.86=(1/2)*(360-x-x)------> 151.72=(360-2x)------>360-151.72=2x
x=104.14°
therefore
the answer is
the measure of arc MN (minor arc) is 104.14°
