Transmitter Question

Question

A small radio transmitter broadcasts in a 53 mile radius. If you drive along a straight line from a city 70 miles north of the transmitter to a second city 74 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

Answer 1

It's right triangle 70 by 74 by sqr(70^2 + 74^2)

as you drive north, you pick up the signal for 53 miles then lose it as you head further north

70-53 + 74 =91 miles of the drive you have no signal

the total drive = 70+74 = 144 miles

91/144 is the percent of the drive with no signal

91/144 = about 63.2% of the drive is with no signal

but you probably really meant to go north, then to go southeast to a point directly east of the transmitter

the part with no signal is where a circle intersects a line from (0,70) to (74,0)

the circle is x^2 + y^2 = 53^2

the line has slope -70/74 =-35/37

y=-35/37x + b

70=35/37(0) + b

b = 70

y=(-35/37)x +70 is the line

x^2 + [(-35/37)x + 70]^2 = 53^2

It's a quadratic equation, solve for x, you get 2 solutions,

x^2 + (35/37)^2x + 2(70)(-35/37)x + 4900 = 2809

(1+35/37)^2(x^2) - 4900/37x + 4900

72/37)^2(x^2) - (4900/37)x + 1191 = 0

(5184/37^2)(x^2) -(4900/37)x + 1191 = 0

multiply by 37^2

5184x^2 - 181300x + 1630479 = 0

use the quadratic formula

x = 181,300/10368 + or - (1/10368)sqr(181300^2 -4(5184)(1,630,479))

you get 2 solutions, plug each into the equation for the line to solve for y

you get 2 points. calculate the distance between those 2 points, that's where you receive the signal

use the distance formula: d^2 = sum of squares of the y differences and the x differences