Answer:
the transmitter signal is picked up for 42.63m of the drive.
Ans = 42.63m
Explanation:
Solution
Let triangle ABC formed by line from 56 miles south of transmitter, to the transmitter itself then to 53 miles west of the transmitter with sides
AB BC AC
Where AB = 56 miles
and BC = 53 miles
Therefore AC = Sqr((56m)^2 +(53m)^2) = 77.10m
The point of intersection of the radius of the radio transmitter signal and the triangle formed by the path of travel of the traveller and the lines AB and AC
To find the perpendicular line that can be drawn from C to AB we have from trigonometric relations
56 × sin(t) = 53 × sin (90 - t) because the traveller moves from directly south of the transmitter to directly west of the transmitter
Hence we have
56×sin(t) = 53×cos(t) because sin(90-t) = cos(t)
Rearranging 56×sin(t) = 53×cos(t) we have
1=(56×sin(t))/ (53×cos(t))
or (sin(t)/ cos(t))=1/(56/53)=53/56
That is tan(t)=53/56 and ACTAN(t) = 43.42°
Angle (t)
Drawing a perpendicular line from the point of the radio transmitter C to the travel path of the traveller AB and calling the point of Intersection E we have EC = 53×sin(43.42)=38.49m
It is seen that the distance from the point of intersection of the radius of the radio transmitter and intersection of the line CE and AB is EI1 where I1 is the first point of intersection of the radius of the radio transmitter and the line AB
EI1=Sqr((44m)^2-(38.49m)^2)
= 21.31m
Also since the triangles CEI1 and CEI2 are identical, it follows that EI1 = EI2 = 21.32m
The distance over which the traveller will be able to receive the signal from the radio transmitter while travelling from point A to point B is the distance I1 to I2 which is equal to 2*I
=2×21.31=42.63m
Ans = 42.63m