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/my-images/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 north-south line on the slope, call it B'.

B' should be also h above point A, since the side of the hill faces north-south.

Now calculate h by first projecting L onto the horizontal plane, then project the resulting line to the N-S 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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