Help solve a question on sine rule, please

P,Q and R are three positions at the top of the three buildings respectively. The buildings lie on a straight line. The angles of depression of P and Q from R are 10 degrees and 40 degrees respectively. Given that the distance PQ and PR are 30m and 50m respectively, find the angle of elevation of P from Q.

http://www.flickr.com/photos/13256126@N0…

Let O be a point at same height as R
∠ORP = 10° (angle of depressions of P from R)
∠ORQ = 40° (angle of depressions of Q from R)
∠PRQ = 40° − 10° = 30°
http://i46.tinypic.com/2r3w6qb.png

By Sine Rule:
sin(∠PQR) / PR = sin(∠PRQ) / PQ
sin(∠PQR) = sin(∠PRQ) * PR/PQ = sin(30) * 50/30 = 1/2 * 5/3 = 5/6

Two possible triangles:
∠PQR = arcsin(5/6) = 56.44°
∠PQR = 180° − arcsin(5/6) = 123.56°

Since angle of elevation from Q to R = 40°, then angle of elevation from Q to P
= 180° − 40° − ∠PQR
= 140° − ∠PQR

When ∠PQR = 56.44°
angle of elevation from Q to P = 140° − 56.44° = 83.56°

When ∠PQR = 123.56°
angle of elevation from Q to P = 140° − 123.56° = 16.44°

http://i50.tinypic.com/2vsqbz4.png

angle R = 40-10 = 30 degrees
30/sin30 = 50/sinQ
sinQ = (50sin30)/30 = .8333333
angle Q = arcsin .83333333 = 123.56 degrees in quadrant 2
so angle P = 26.44 degrees
So angle of depression between P and Q is 26.44 - 10 = 16.44 degrees