Solving a Trigonometry Problem: Uncovering "h

  • Thread starter Thread starter ooovooos
  • Start date Start date
  • Tags Tags
    Trigonometry
AI Thread Summary
The discussion centers on solving for the variable "h" in a trigonometry problem involving two triangles and the tangent function. Participants express confusion about the setup and the complexity of the equations derived from the tangent ratios of 30° and 60°. They attempt to relate the lengths of the sides using known values and trigonometric identities but find the algebra challenging. Despite the difficulties, there is encouragement to persist with the problem-solving process, emphasizing that the approach is fundamentally correct. The conversation highlights the common struggle with re-engaging in trigonometry after a long break.
ooovooos
Messages
2
Reaction score
0

Homework Statement


[PLAIN]http://img530.imageshack.us/img530/5062/dgsdgsdgs.jpg
what is h??

Homework Equations


tan x = sin x/ cos x
cos x = adj/hyp
sin x = opp/hyp


The Attempt at a Solution


uh I am not even really sure where to start. i haven't done trig in a really really long time. I am guessing you have to use trig. and i set the area under the dotted triangle as x so the entire bottom length would be 100 + X but I am not even sure how that helps.
 
Last edited by a moderator:
Physics news on Phys.org
Add one unknown - length of the leg adjacent to 60° angle. You have two unknowns - h and the leg. Can you express tangents of both angles using these unknowns (and known length 100)?
 
^ no offense but the "clue" you gave was so small that I'm just back where I started...confused...didnt really push me in any particular direction. -_-
tan 30 = h / (100 + x) = sqrt 3 over 3
tan 60 = h / x = sqrt 3?
i don't see how that helps.
i tried solving for h using the one that equals tan 60..that turns out to be x times sqrt 3...i plugged that into the first equation...and that just seems way too complicated. I did this type of question two years ago in HS...i am so sure this question isn't supposed to be this complicated.
 
ooovooos said:
i tried solving for h using the one that equals tan 60..that turns out to be x times sqrt 3...i plugged that into the first equation...and that just seems way too complicated. I did this type of question two years ago in HS...i am so sure this question isn't supposed to be this complicated.

Is this the equation you got?

\frac{x\sqrt{3}}{100 + x} = \frac{\sqrt{3}}{3}

It requires a bit of algebra, but it's not bad.
 
ooovooos said:
i tried solving for h using the one that equals tan 60..that turns out to be x times sqrt 3...i plugged that into the first equation...

You got to two equations in two unknowns, and you are solving them using correct approach. Just don't give up.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top