Find the distance of the plane from sign a to the nearest kilometer

Question

A pilot is flying over a straight highway. He determines the angles of depression to two signs, which are 3 km apart, two be 32∘ and 56∘, as shown in the figure below. Find the distance of the plane from sign 𝐴 to the nearest kilometer.

Answer 1

ΔPAB by Sine Law: PA/sin32° = AB/sin(56° - 32°); PA = ABsin32˜/sin24° = 3sin32°/sin24° km = 3.900km = 4 km;

Answer 2

There wasn't a diagram, so I will try to create one.

P --------------------------C

| . .

| . .

| . .

| . .

| . .

O---------A----------------B

P represents the position of the plane. O is the point directly below the plane. A is sign A. B is sign B. C is the point directly above B. The dotted lines are the lines of sight from P to A and B.

Angle CPB is 32°. Angle CPA is 56°. PC is parallel to OB. From a theorem in geometry, angle PBO is 32° and angle PAO is 56°. (Theorem: If parallel lines are cut by a transversal, the alternate interior angles are equal.)

tan 32° = PO/OB. tan 56° = PO/OA. OB = 3 km + OA.

Solving the first two equations for PO gives PO = OB tan 32 = OA tan 56. Substituting 3 + OA for OB gives

(3 + OA) tan 32 = OA tan 56. Therefore OA = 3 tan 32 / (tan 56 - tan 32) = 2 km.

This is my interpretation of the diagram. If it differs from your diagram, this solution may help you obtain an answer. Check all math.