Let the transmitter be located at (0,0)
The equation for its effective coverage area will be a circle centered at (0,0) with a radius of 27
So....the equation is
x^2 + y^2 = 27^2
x^2 + y^2 = 729
Let the point 35 miles north of the transmitter be (0, 35)
Let the point 31 miles east of the trasmitter be (31,0)
The line connecting these (the line of travel) will have a slope of [ 35 - 0] / [ 0 - 31] = -35/31
The equation of this line will be
y = (-35/31)x + 35
We could solve this algebraically, but a graph seems easier to deal with
See here : https://www.desmos.com/calculator/ksd2p5jdwl
The signal will be received betwen the points (8.221 , 25.718) and (26.523, 5.055)
The distance between these points is
√ [ ( 26.523 - 8.221)^2 + ( 25.718 - 5.055)^2 ] ≈ 27.6 miles (1)
The distance between (0, 35) and (31,0) = √ [ ( 31)^2 + ( 35)^2 ] ≈ 46.8 miles (2)
So....you will receive the signal about (1)/ (2) = 27.6 / 46.8 ≈ 59% of the drive