Lesson A radio transmitter accessibility area

A radio transmitter accessibility area

Problem 1

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).