Answer
The distance between the transmitter and point A is 1.91 miles
Work Step by Step
Let point C be the position of the transmitter. The points ABC form a triangle.
We can find the angle at point A:
$A = 90^{\circ}-48^{\circ} = 42^{\circ}$
We can find the angle at point B:
$B = 302^{\circ}-270^{\circ} = 32^{\circ}$
We can find the angle at point C:
$C = 180^{\circ}-42^{\circ} -32^{\circ} = 106^{\circ}$
We can use the law of sines to find the distance between the transmitter and point A:
$\frac{AC}{sin~B} = \frac{AB}{sin~C}$
$AC = \frac{AB~sin~B}{sin~C}$
$AC = \frac{(3.46~mi)~sin~32^{\circ}}{sin~106^{\circ}}$
$AC = 1.91~mi$
The distance between the transmitter and point A is 1.91 miles
