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DMill · May 3, 2009, 05:32 AM

I really do hope you can assist me with a question. I am helping some friends prepare for exams and I can't seem to get this question. Perhaps my diagram is off... Maybe the publisher missed some info??

A vertical pole AB is standing on level ground. L and M are two points 20m apart on the same side of a straight road running from East to West past the pole. The bearing of B, the foot of the pole, from L and M is 015 and 055 respectively. The angle of elevation of A from L is 17 degrees.

a) Draw a sketch of triangle BLM
b) Calculate the distance LB - supposed to be 17.9m
c) Calculate the height of the pole AB - I got that one out, using 17.9 to be the given correct answer of 5.47

Perito · May 3, 2009, 11:11 AM

angle LBM = 55-15 = 40
angle LMB = 35
angle BLM = 105

we use the Law of sines:

$\frac {MB}{sin(105)} = \frac {20}{sin(40)}$

$MB = \frac {20 sin(105)}{sin(40)} = 30.05$

$\frac {LB}{sin(35)} = \frac {20}{sin(105)}$

$LB = \frac {20 sin(35)}{sin(105)}$

$LB = 11.88$

I'm not sure where they got 17.9. Maybe my bearing is backward.

The second image is, to my mind, more correct since the "bearing" from L to B is 15 degrees (east of north) and the bearing from M to B is 55 degrees. I think the math is still the same, however and it doesn't give the 17.9

Unknown008 · May 4, 2009, 07:43 AM

Nope, the second one does give 17.9, I got it. Maybe you did a calculation mistake, Perito.

Perito · May 4, 2009, 07:51 AM

Undoubtedly.