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 (http://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, and we have different angles alpha & gamma, there are no equal sides either. I did try to prove if there are any similar triangles, but I couldn't.
I drew a diagram like this, according to this question (http://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, and we have different angles alpha & gamma, there are no equal sides either. I did try to prove if there are any similar triangles, but I couldn't.
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It's a long one. First you should draw the triangles on a piece of paper to understand the situation. Draw also a line from A perpendicular to BC and name that point "x".
Let's work only with that plane first. ABx is a rectangle triangle. We can say that :
sin2@ = Ax / a
Ax = a*sin2@
cos2@ = Bx / a
Bx = a * cos2@
We need xC so :
xC = 2a - (a * cos2@) --- coz BC = 2a
With the use of Pythagorus law we can say (with the other triangle) :
AC^2 = Ax^2 + xC^2
AC^2 = (a * sin2@)^2 + ((2a - (a * cos2@))^2
AC = sqrt[ (a^2 * sin^2(2@)) + 4a^2 - 4a^2(cos2@) + (a^2 * cos^2(2@)) ]
Taking out a^2
AC = a * sqrt[ sin^2(2@) + 4 - 4cos2@ + cos^2(2@) ]
If : sin^2(2@) + cos^2(2@) = 1
AC = a * sqrt[ 1 + 4 - 4cos2@ ]
(1) AC = a * sqrt[ 5 - 4cos2@ ]
Now let's work on the other plan ACD
tan@ = AD / AC
AD = tan@ * AC
We replace AC by the equation (1) found above :
AD = tan@ * a * sqrt[ 5 - 4cos2@ ]
AD = a * tan@ * sqrt[ 5 - 4cos2@ ]
Everything is the same except what's in the sqrt. Let's work only with wath is inside of the sqrt
5 - 4cos2@ ---> it should equal 1 + 8sin^2(2@)
cos2@ = cos^2(@) - sin^2(@)
5 - 4 (cos^2(@) - sin^2(@))
5 - 4cos^2(@) + 4sin^2(@)
since,
cos^2(@ + sin^2(@) = 1
cos^2(@) = 1 - sin^2(@)
we replace that
5 - 4cos^2(@) + 4sin^2(@)
5 - 4 (1 - sin^2(@)) + 4sin^2(@)
5 - 4 + 4sin^2(@) + 4sin^2(@)
1 + 8sin^2(@)
That is what you have in the square root
Let's work only with that plane first. ABx is a rectangle triangle. We can say that :
sin2@ = Ax / a
Ax = a*sin2@
cos2@ = Bx / a
Bx = a * cos2@
We need xC so :
xC = 2a - (a * cos2@) --- coz BC = 2a
With the use of Pythagorus law we can say (with the other triangle) :
AC^2 = Ax^2 + xC^2
AC^2 = (a * sin2@)^2 + ((2a - (a * cos2@))^2
AC = sqrt[ (a^2 * sin^2(2@)) + 4a^2 - 4a^2(cos2@) + (a^2 * cos^2(2@)) ]
Taking out a^2
AC = a * sqrt[ sin^2(2@) + 4 - 4cos2@ + cos^2(2@) ]
If : sin^2(2@) + cos^2(2@) = 1
AC = a * sqrt[ 1 + 4 - 4cos2@ ]
(1) AC = a * sqrt[ 5 - 4cos2@ ]
Now let's work on the other plan ACD
tan@ = AD / AC
AD = tan@ * AC
We replace AC by the equation (1) found above :
AD = tan@ * a * sqrt[ 5 - 4cos2@ ]
AD = a * tan@ * sqrt[ 5 - 4cos2@ ]
Everything is the same except what's in the sqrt. Let's work only with wath is inside of the sqrt
5 - 4cos2@ ---> it should equal 1 + 8sin^2(2@)
cos2@ = cos^2(@) - sin^2(@)
5 - 4 (cos^2(@) - sin^2(@))
5 - 4cos^2(@) + 4sin^2(@)
since,
cos^2(@ + sin^2(@) = 1
cos^2(@) = 1 - sin^2(@)
we replace that
5 - 4cos^2(@) + 4sin^2(@)
5 - 4 (1 - sin^2(@)) + 4sin^2(@)
5 - 4 + 4sin^2(@) + 4sin^2(@)
1 + 8sin^2(@)
That is what you have in the square root
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Hey, but I don't think we can give BC = 2a, AB = a, because there is no rational relationship between these two, isn't it?
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