Proof

[intervade]

Ok, this is a problem I've been working on for a while.. I guess I'm just not sure where to start even.

Shown in the figure is a proposed railroad route through three towns located at points A, B and C. The track will branch out from B towards C at angle ?.
Show that the total distance d from A, B to C, along the proposed route is given by d = 20tan[(1/2)?] + 40

So.. where do I go to "prove" this? I've made an attempt at an identity d = 20[ (1-cos?)/sin? ] + 40 .. d = (20-20cos?)/sin? + 40

I don't know even know if I'm going the right direction...

 

[galactus]

Yes, you're on the right track. Keep going.

The distance from B to C can be gotten from \(\displaystyle 20csc({\theta})\)

The distance from A to B, would be \(\displaystyle 40-20cot({\theta})\)

The total distance, d, is then \(\displaystyle 20csc({\theta})+40-20cot({\theta})\)

\(\displaystyle =40+20(\underbrace{csc({\theta})-cot({\theta})}_{\text{this is tan(@/2)}})\)

But, as I identicated by the underbrace, \(\displaystyle csc{\theta}-cot{\theta}=tan(\frac{\theta}{2})\)

So, we have \(\displaystyle d=40+20tan\frac{\theta}{2}\)

The above identity can be easily derived by noting the identity \(\displaystyle tan\frac{\theta}{2}=\frac{1-cos{\theta}}{sin{\theta}}=csc{\theta}-cot{\theta}\)

We could also use \(\displaystyle \frac{20}{sin{\theta}}+40-\frac{20}{tan{\theta}}\) and go from there instead of using csc and cot.

 

[intervade]

ok, can you explain to me why 20/sin? is BC and (40 - 20/tan?) is AB? Are you just taking it right for the formula or is there some sort of equation I don't know yet? Also, is there a way I could prove this with law of sine/cosines or would that just be a waste of time?

 

[galactus]

Look at the triangle. Those are just the trig to use with the triangle. \(\displaystyle 40-20cot{\theta}\) is the side length adjacent to theta. See it?.

Let's say the adjacent side to the angle theta is x. Since the entire distance from A to the right angle of the triangle is 40, wouldn't the distance AB be 40-x?.

In this case, the \(\displaystyle x=20cot{\theta}\). Therefore, the distance AB is \(\displaystyle 40-20cot{\theta}\).

\(\displaystyle 20csc{\theta}\) is the length BC. Add them up for the total distance. I started here and worked up to the identity for \(\displaystyle tan\frac{\theta}{2}\)

Yes, as I mentioned at the end of the post, you can use sin and tan. Afterall, \(\displaystyle \frac{1}{sin{\theta}}=csc{\theta}, \;\ \frac{1}{tan{\theta}}=cot{\theta}=\frac{cos{\theta}}{sin\theta}}\)

It's all explained pretty well the the first post. It's nothing fancy, just some basic right triangle trig.

I don't think you have to prove the identity \(\displaystyle tan\frac{\theta}{2}=\frac{1-cos{\theta}}{sin{\theta}}\). Just use it.

 

[dddassdd]

Vertical pole and angle of inclination

a vertical post is fixed in horizontal field ABCD which makes four different inclinations alpha, beta, gamma, delta then prove
Cot^2 alpha-Cot^2 beta = Cot^2 delta-Cot^2 gamma can u solve this

 

[Deleted member 4993]

Re: Vertical pole and angle of inclination

a vertical post is fixed in horizontal field ABCD which makes four different inclinations alpha, beta, gamma, delta then prove
Cot^2 alpha-Cot^2 beta = Cot^2 delta-Cot^2 gamma can u solve this

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