Posted by Jessica on Tuesday, July 5, 2011 at 4:23am.
This is the problem given: The side of a hill faces due south and is inclined to the horizontal at an angle alpha. A straight railway upon it is inclined at an angle beta to the horizontal. If the bearing of the railway is gamma east of north, show that cos(gamma) = cot(alpha)tan(beta)
I drew a diagram like this, according to this question (imageshack.us/photo/myimages/30/trigonometryquestion.png/).
We have cos(gamma) = CH/AC
cot(alpha) = BH/CH
tan(beta) = CH/AH,
so cot(alpha)xtan(beta) = BH/AH
I'm stuck at proving CH/AC = BH/AH
Is there anyway around this? I'll appreciate any help from you all. Thank you very much.
Note: AC is not perpendicular to BC.

math  MathMate, Tuesday, July 5, 2011 at 8:37am
From the diagram, I am not sure if you have the same interpretation of the question as I have.
There is a slope facing south, at α with the horizontal.
The railway is at an angle of γ east of north (on the horizontal projection).
The angle β is the angle the railway makes with the horizontal.
If you look at the situation in plan view, north towards the top of the paper, we see a line at γ towards east. Let the railway line length be L, and denote A by the south end of the line, and B the north end of the line.
Let the elevation of A (south) be zero.
Then B is h above A, where
h=Lsinβ.
Now we will calculate h in a different way by projecting the point B to a northsouth line on the slope, call it B'.
B' should be also h above point A, since the side of the hill faces northsouth.
Now calculate h by first projecting L onto the horizontal plane, then project the resulting line to the NS line, and finally multiply by tanα to get the height h.
h=Lcosβcosγtanα
Thus
h=Lsinβ=Lcosβcosγtanα
transpose cosγ to the left, and cancel common factor L, we get:
cosγ=(sinβ/cosβ)cotα
=tanβcotα. QED
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