Help needed on trigonometry question

Mr. Lapointe and Kimberlake are standing on the level ground Wrigley Field for the Pearl Jam Concert. They are both positioned directly at center stage watching the band rock out but Mr. Lapointe is 7.8m closer to the stage. Mr. Lapointe has to raise his head 15o to see the front of the stage, whereas Kimberlake only has to raise her head 10o. What is the elevation of the stage?

Hello again nada!

Let’s start perhaps by having you draw a “diagram” of the scenario in this problem. Maybe draw something on a piece of paper, take a picture with your phone, and upload it here, so I can get a better idea of your approach!

Wrong diagram, sorry this is the right one

Nice! That’s exactly the diagram that I have.
So basically now you need to figure out two things:

  1. where does trigonometry actually come in.
  2. what variables are you solving for.

Let h be the vertical distance from the ground to the stage (so you’re trying to find what h is), and let d be the horizontal distance from the stage to Mr. Lapointe.

So then we can form two equations with two unknowns:

\begin{cases} \tan (10^\circ) = \dfrac{h}{d+7.8} \\ \tan (15^\circ) = \dfrac{h}{d} \end{cases}

To solve this system of equations, use substitution or elimination, and use your calculator, and what you should get is that d = 15 and h = 4.

Does that make sense? With the diagram you made, you were on a very good track!

How would you solve the equations though, since there are multiple variables in just 1 equation?

There are two variables (h and d) in two equations.

\begin{cases} \tan (10^\circ) = \dfrac{h}{d+7.8} \\ \tan (15^\circ) = \dfrac{h}{d} \end{cases}

Solve the second one for d and then sub this into the other equation, and then solve for h:

\begin{align*} \tan (10^\circ) &= \dfrac{h}{\dfrac{h}{\tan (15^\circ)} + 7.8} \\ \dfrac{\tan (10^\circ)}{\tan (15^\circ)} &= \dfrac{h}{h+7.8\cdot\tan (15^\circ)} \\ (h+7.8\cdot\tan (15^\circ))\dfrac{\tan (10^\circ)}{\tan (15^\circ)} &= h \\ h\dfrac{\tan (10^\circ)}{\tan (15^\circ)} + (7.8\cdot\tan (15^\circ))\cdot \dfrac{\tan (10^\circ)}{\tan (15^\circ)} &= h \\ h \left(\dfrac{\tan (10^\circ)}{\tan (15^\circ)} - 1\right) &= -7.8\cdot \tan (10^\circ) \\ h &= \dfrac{-7.8\cdot \tan (10^\circ)}{\dfrac{\tan (10^\circ)}{\tan (15^\circ)} - 1} \\ h &\approx 4 \end{align*}\tag*{}

But then wouldn’t the solving the first equation just give me d x tan50/h?

Sorry, I’ll try to be more explicit.

So we have these two equations \begin{cases} \tan (10^\circ) = \dfrac{h}{d+7.8} ~~(1) \\ \tan (15^\circ) = \dfrac{h}{d} ~~~~~~~~~~~~(2)\end{cases}

Take equation (2) and solve for d. So \tan (15^\circ) = \dfrac{h}{d} \implies d = \dfrac{h}{\tan (15^\circ)}.

Then simply plug this into the equation (1) and solve for h: \tan (10^\circ) = \dfrac{h}{(d)+7.8} \implies \tan (10^\circ) = \dfrac{h}{\left( \dfrac{h}{\tan (15^\circ)}\right) + 7.8}

Oh ok thanks.

No problem! I really hope I’m making sense. Feel free to keep asking questions if something is unclear. I want to be helpful! There are no stupid questions

This topic was automatically closed 24 hours after the last reply. New replies are no longer allowed.