kasie-tutor
Banned
- Joined
- Jan 20, 2008
- Messages
- 36
Here's the problem I have a question about:
"Planning a railroad route. Shown in the figure is a proposed railroad route from town A to town C. The track will branch out from B to C at angle \(\displaystyle $\theta$\) to go around the city. Show that the total distance \(\displaystyle $d$\) from A to C is given by \(\displaystyle $d=20\tan\frac{1}{2}\theta+40$\)."
This question is in a section that introduces the trigonometry multiple and half angle formulas.
Based on the formula, I think distance \(\displaystyle $d$\) is not the segment AC, but is the sum of segments AB + BC. I found BC in terms of \(\displaystyle $\theta$\), which is \(\displaystyle $\frac{20}{\sin\theta}$\). This means that when \(\displaystyle $\theta$\) is 90 degrees, BC is 20.
Now I am having trouble relating AB to BC, so that I can add them and somehow show that they add to \(\displaystyle $d$\).
Another thing I tried is to re-write \(\displaystyle $d$\) like this...
\(\displaystyle $d=20\tan\frac{1}{2}\theta+40$\)
\(\displaystyle $d=20(\frac{1-\cos\theta}{\sin\theta})+40$\)
\(\displaystyle $\frac{d-40}{20}=\frac{1-\cos\theta}{\sin\theta}$\)
...but this didn't end up giving me any clues.
Can anyone see how to get AB in terms of \(\displaystyle $\theta$\) and point me in the right direction?
Thanks!
"Planning a railroad route. Shown in the figure is a proposed railroad route from town A to town C. The track will branch out from B to C at angle \(\displaystyle $\theta$\) to go around the city. Show that the total distance \(\displaystyle $d$\) from A to C is given by \(\displaystyle $d=20\tan\frac{1}{2}\theta+40$\)."
This question is in a section that introduces the trigonometry multiple and half angle formulas.
Based on the formula, I think distance \(\displaystyle $d$\) is not the segment AC, but is the sum of segments AB + BC. I found BC in terms of \(\displaystyle $\theta$\), which is \(\displaystyle $\frac{20}{\sin\theta}$\). This means that when \(\displaystyle $\theta$\) is 90 degrees, BC is 20.
Now I am having trouble relating AB to BC, so that I can add them and somehow show that they add to \(\displaystyle $d$\).
Another thing I tried is to re-write \(\displaystyle $d$\) like this...
\(\displaystyle $d=20\tan\frac{1}{2}\theta+40$\)
\(\displaystyle $d=20(\frac{1-\cos\theta}{\sin\theta})+40$\)
\(\displaystyle $\frac{d-40}{20}=\frac{1-\cos\theta}{\sin\theta}$\)
...but this didn't end up giving me any clues.
Can anyone see how to get AB in terms of \(\displaystyle $\theta$\) and point me in the right direction?
Thanks!