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.
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
To prove that CH/AC = BH/AH, we can make use of similar triangles.
Looking at the diagram you provided, we can see that triangle CAH and triangle CBH share the angle at C. We can conclude that they are similar triangles.
Now, let's compare the corresponding sides of the similar triangles:
1) Side CH of triangle CAH corresponds to side BH of triangle CBH.
2) Side AC of triangle CAH corresponds to side AH of triangle CBH.
Since the triangles are similar, the corresponding sides are proportional. This implies that CH/AC = BH/AH.
Therefore, we have proven that CH/AC = BH/AH.
Now we can proceed with the rest of the problem:
We know that cos(gamma) = CH/AC.
We also know that cot(alpha) = BH/CH and tan(beta) = CH/AH.
Multiplying cot(alpha) and tan(beta), we get cot(alpha) * tan(beta) = (BH/CH) * (CH/AH) = BH/AH.
Thus, we can substitute CH/AC with BH/AH in the equation cos(gamma) = CH/AC to get cos(gamma) = cot(alpha) * tan(beta).
Hence, we have successfully shown that cos(gamma) = cot(alpha) * tan(beta), as required by the problem.
I hope this explanation helps you understand how to approach and solve the problem. Let me know if you have any further questions!