Question

Radio direction finders are set up at points A and B, which are 2.00 mi. apart on an east-west line. From A it is found that the bearing of the signal from a radio transmitter is N 36° 20’ E, while from B the bearing of the same signal is N 43° 40’ W. Find the distance of the transmitter from B.

Answer 1

We can solve this problem using trigonometry and the properties of triangles.

Let C be the location of the radio transmitter. Then, ACB is a triangle with sides AC = x (the distance from A to the transmitter), BC = y (the distance from B to the transmitter), and AB = 2.00 mi.

We can use the fact that the sum of the interior angles of a triangle is 180 degrees to find the angle at C:

angle ACB = 180 degrees - angle BCA - angle CAB

From the information given in the problem, we know that:

angle CAB = N 36° 20' E

angle BCA = N 43° 40' W

To add or subtract angles, we need to convert them to a common direction. We can do this by adding or subtracting 180 degrees, or by using the fact that 1 degree = 60 minutes (') and 1 minute = 60 seconds ("). Therefore:

angle CAB = 36 degrees + 20/60 degrees = 36.3333... degrees

angle BCA = 180 degrees - (43 degrees + 40/60 degrees) = 136.6666... degrees

Substituting these values into the equation for angle ACB, we get:

angle ACB = 180 degrees - 136.6666... degrees - 36.3333... degrees = 7.0000... degrees

Now, using the law of sines, we can write:

x / sin(angle CAB) = 2.00 mi / sin(angle ACB)

y / sin(angle BCA) = 2.00 mi / sin(angle ACB)

Solving for x and y, we get:

x = 2.00 mi * sin(angle CAB) / sin(angle ACB) = 2.00 mi * sin(36.3333... degrees) / sin(7.0000... degrees) = 9.0734... mi

y = 2.00 mi * sin(angle BCA) / sin(angle ACB) = 2.00 mi * sin(136.6666... degrees) / sin(7.0000... degrees) = 1.1878... mi

Therefore, the distance of the transmitter from B is y = 1.1878... mi (rounded to 4 decimal places).