• Support PF! Buy your school textbooks, materials and every day products Here!

Trigo/Geom. problem

  • Thread starter Mentallic
  • Start date
  • #1
Mentallic
Homework Helper
3,798
94

Homework Statement


D is the top of a tower h metres high. A is the foot of the tower. The angles of depression of the points B and C from D are [itex]45+\theta[/itex] and [itex]45-\theta[/itex] respectively. Show that [itex]BC=2h.tan2\theta[/itex]

http://img43.imageshack.us/img43/9195/trigo.th.jpg [Broken]

The Attempt at a Solution



[tex]\angle ADB=90-(45+\theta)=45-\theta[/tex]

Therefore, [tex]tan(45-\theta)=\frac{AB}{h}[/tex]

Then, [tex]AB=h.tan(45-\theta)[/tex]

[tex]\angle CDA=90-(45-\theta)=45+\theta[/tex]

Therefore, [tex]tan(45+\theta)=\frac{AB+CD}{h}[/tex]

Then, [tex]AB=h.tan(45+\theta)-CD[/tex]

So I equated [tex]h.tan(45-\theta)=h.tan(45+\theta)-CD[/tex]

And finally, through expanding using the tangent sum formula and 'compressing' back into the double tangent formula, I resulted in [itex]BC=2h.tan2\theta[/itex]


My problem is that this way to do it took far too long. There must be an easier way but I cannot seem to find it. I need your help :smile:
 
Last edited by a moderator:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,795
925
Are B and D assumed to be in a straight line with A?
 
  • #3
Mentallic
Homework Helper
3,798
94
Yes. The diagram is simply a right-angled triangle with another line going from D to a point B on the opposite line of the triangle.

EDIT: Re-reading your question Hallsofivy, I think I may need to post a diagram to clear things up.
 
  • #5
Mentallic
Homework Helper
3,798
94
Bump. To re-iterate: I doubt such a question given to our class to answer would require over a page of algebraic manipulations, and this is why I'm thinking there is a much easier way to use that [itex]2\theta[/itex] in the diagram directly, and not obtain it through the tangent double angle formula.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,795
925
If the angle of depression to B is 45+[itex]\theta[/itex], then the angle DB makes with DA is 90- (45+ [itex]\theta-[/itex])= 45-[itex]\theta[/itex] and the angle DC makes with DA is 90- (45- [itex]\theta[/itex]= 45+[itex]\theta[/itex]. CA/h= tan(45+\theta) and BA/h= tan(45-[itex]\theta[/itex]) so CB= CA- BA= h(tan(45+[itex]\theta[/itex])- tan(45-[itex]\theta[/itex])). Now use the identities for tan(a+ b) and tan(a-b).
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi Mentallic! :smile:

(have a theta: θ :wink:)
… Then, [tex]AB=h.tan(45-\theta)[/tex]

My problem is that this way to do it took far too long. There must be an easier way but I cannot seem to find it. I need your help :smile:
yeees :redface:

having got your AB = h.tan(45º - θ),

you never actually said AC = h.tan(45º + θ) …

which gives you the answer much quicker! :smile:
 
  • #8
Mentallic
Homework Helper
3,798
94
Oh thats much quicker! Thanks to both of you :smile:

Still curious though, can this problem be solved without the trig identities? Such as using the [itex]2\theta[/itex] angle and finding its tangent etc.?
 

Related Threads for: Trigo/Geom. problem

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
20
Views
1K
Replies
1
Views
2K
Replies
2
Views
2K
Top