SOLUTION: A small radio transmitter broadcasts in a 21 mile radius. If you drive along a straight line from a city 25 miles north of the transmitter to a second city 29 miles east of the tra

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Question 1206903: A small radio transmitter broadcasts in a 21 mile radius. If you drive along a straight line from a city 25 miles north of the transmitter to a second city 29 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

Circle equation
(x-h)^2 + (y-k)^2 = r^2
where,
(h,k) = center
r = radius

Let's place the radio transmitter at the origin. Meaning h = 0 and k = 0.
The radius is r = 21 in this case.
The circle equation will then update to x^2+y^2 = 441
Points inside the circle, or on the boundary, will get the radio signal.

A = (0,25) is where you start from since it is 25 miles north of the transmitter.
B = (29,0) is where you are driving to, which is 29 miles east of the transmitter.
I'll skip a few steps, but you should find the equation of line AB is y = (-25/29)x + 25

We have this system of equations

Use substitution to plug the second equation into the first one to end up with

Skipping a few more steps, the solutions to that equation are approximately: x = 5.48588 and x = 19.24127
These are the x coordinates of the intersection points of the line and circle.
Since we're looking at approximate solutions, it seems reasonable to assume your teacher will allow you to use a graphing calculator to make quick work of this equation.
Note: The exact solutions involve very large messy numbers.

If , then points on line AB are inside the circle or on the circle's boundary.
Otherwise, you'll be outside of the circle and won't pick up the signal.

Use those x values to determine the corresponding paired y values.
x = 5.48588 leads to y = 20.2708
Let point C be located at (5.48588, 20.2708)
x = 19.24127 leads to y = 8.4127
Let point D be located at (19.24127, 8.4127)

Diagram

The diagram was made with GeoGebra which is a useful tool to verify many types of math problems.

Use the distance formula to determine these approximate segment lengths
AB = 38.28838
CD = 18.1611
Then,
CD/AB = 18.1611/38.28838 = 0.47432

For roughly 47.432% of the trip, you'll be able to pick up this particular radio signal.

Answer by ikleyn(52780)   (Show Source):
You can put this solution on YOUR website!
.
A small radio transmitter broadcasts in a 21 mile radius. If you drive along a straight line
from a city 25 miles north of the transmitter to a second city 29 miles east of the transmitter,
during how much of the drive will you pick up a signal from the transmitter?
~~~~~~~~~~~~~~~~~

        This problem admits simple and elegant geometric solution without using equations.

Let A be the point 25 miles north from the transmitter. Let B be the point 29 miles east from the transmitter. Let O be the circle of the radius 21 miles with the center at the transmitter location. Then we have a right angled triangle OAB and straight line AB, intersecting this triangle in points B and C. They want we find the length of the chord BC. My designations are close to that on the plot of the other tutor. Draw the perpendicular OE from the center/vertex O to line BC. This perpendicular is the height (the altitude) in triangle OAB on the hypotenuse. The area of this triangle can be computed as half the product of its legs or as half the product of the hypotenuse and the height. So, we can write this equation |AB|*h = |OA|*|OB|, (1) where h is the height OE. The hypotenuse |AB| is = 38.2884 miles (rounded). From (1) h = . It gives the distance from the center O to line AB. Then the chord length is |BC| = 2*|BE| = 2*|CE| = = = 18.1611 miles (rounded). Now we have all the numbers and the ANSWER: the entire trip is about 38.2884 miles; the part where the signal is accessible is about 18.1611 miles; the ratio of these values is = 0.4743 (rounded).

Solved.

At the beginning, I said that the equations are not used in the solution, but it is not precisely correct.
One equation (1) is still used, but in minimal degree.
The rest is pure geometry.