A small radio transmitter broadcasts in

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 63 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

Answer 1

To find how much of the transmission area the car traverse the point of entrance and point of exit must be determined.

Draw and label a diagram. The transmitter is at (0, 0).

The north town is at (0, 79) and the east town AT (63, 0)

The car travels on the path of y = -70x/63 + 79

The edge of transmission is at x² + y² = 53²

y² = 4900x²/3969 + 11060x/63 + 6241

y² = 100x²/81 + 1580x/9 + 6241

x² + 100x²/81 + 1580x/9 + 6241 = 2809

81x² + 100x² + 142200x + 505521 = 227529

181x² + 142200x + 277911 = 0

This leads (if my computation is correct) to a rather messy irrational number.

Answer 2

Hello, Dezarae,

The new city is 70 miles north and 63 miles east. It says the driver goes a straight line from one to the other. That makes this a right triangle:

So all we need is the Pythagorean theorem to calculate the total distance of the straight line between the two cities, D.

D^2 = (63)^2 + (70)^2

or D = 94.2 miles. The range is 63 miles, so 63/94.2 would be the percentage of the distance the car is within range. That is 66.9%.

A note: the problem does not actually ask for a percentage of distance driven. Technically, once the driver goes 63 miles away from the transmitter, he is out of range. Perhaps it is a trick problem?

Bob