# 3D trigonometry question

I'm really struggling with the following question:

A river running due east has straight parallel banks. A vertical post stands with its base, P, on the north side of the river. On the south bank are two surveyors, A who is to the east and B who is to the west of the post. A & B are at a distance 2/7a apart and the angle APB = 150 degrees. The angles of elevation from A and B of the top Q, of the post are 45 degrees and 30 degrees. Find in terms of a, the width of the river and the height of the post.

what I have figured out so far is that:
QP = AP (tan45 = 1)
PB = APsqrt3 (half equilateral)
BQ = 2AP (half equilateral)
AB is the longest side of the triangle ABP (angle APB = 150 degrees)

after struggling for 30mins I checked the answers at the back of the book and they are given as
a sqrt3/49 and 2a/7 sqrt7...certain things look familiar here but...

my problems trying to find just the width are as follows:
- if I was to assume that a is the length of either side opposite angle A in triangles ABP or APQ then this would be impossible because AB is 2/7a but is the longest side (opposite P which is 150 degrees). is a just an arbitrary value that AB is 2/7 of?

- If I create a line PR that is perpendicular to AB, PR would be the width of the river but how using pythagoras theorem, the sine, or cosine theorems can I make an equation with respect to a without knowing the relationship algebraically between AB and any other lines, the lengths of AR or BR, or any of the angles? (don't know the angles RPA or RPB)

- If I assume that 2/7a is a typo and that it should read 7/2a and decide that a is now the length of AP or BP this too would be impossible because
a + a sqrt3 is less than 7/2a and thus you can't make a triangle.

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VietDao29
Homework Helper
All the results the book provides are correct.
I'll give you a small hint. If you cannot find the width of the river first, then go finding the height of the post first. It can be found by using some trigonometric funtions (sin, cos, tan, ...) and the cosine law. Then you can find the width of the river easily (you will know 3 legs of the triangle, and 1 angle).
Viet Dao,

meh...I don't know whether to feel stupid for thinking there was something wrong with this problem, or proud of myself for solving it after I returned home from the pub last night. Just decided..to hell with it, lets give AP a value of 1 and see what happens...After giving AP a value of 1 I used the cosine rule to find the length of AB (2a/7 = sqrt7), divided this by sqrt seven to find PQ, then the sine rule to find angle B before getting the correct answer for the width of the river. Thanks for the reply anyway though Viet Dao Last edited: