# Tricky triogonometri...

• Aug 26th 2006, 07:24 AM
Neffets...:P
Tricky triogonometri...
How can I find angle v in this figure?

Info:

AO=OB=OP=OC=6370 km
• Aug 26th 2006, 07:25 AM
Glaysher
Tricky with the figure missing?
• Aug 26th 2006, 07:26 AM
Neffets...:P
continues....
PS=20240 km
CP = 2780 km

• Aug 26th 2006, 07:43 AM
CaptainBlack
Quote:

Originally Posted by Neffets...:P
PS=20240 km
CP = 2780 km

The key is that $\displaystyle \angle POC=(arcCP)/OC=2780/6370$ radian.

The rest is just fiddly trig.
• Aug 26th 2006, 07:54 AM
Neffets...:P
thanks
thanks mate! I think i'll manage it all now...

:D :D :D :D

You guys are GREAT!
• Aug 26th 2006, 11:48 PM
Neffets...:P
need more help...
:mad: i have tried a lot, but i have not been able to find the answer!

CP is a part of a circle...
the angle is on a tangent.

Can you guys show me how you find the answer?
I know I seem a bit stupid, but...

:D :D
• Aug 27th 2006, 12:01 AM
CaptainBlack
Quote:

Originally Posted by Neffets...:P
:mad: i have tried a lot, but i have not been able to find the answer!

CP is a part of a circle...
the angle is on a tangent.

Can you guys show me how you find the answer?
I know I seem a bit stupid, but...

:D :D

Drop a perpendicular from $\displaystyle C$ onto $\displaystyle OS$, call the new point at the foot of the perp. $\displaystyle Q$.

Now look at $\displaystyle \triangle OCQ$ you know $\displaystyle \angle QOC$, and $\displaystyle OC$, so you can find $\displaystyle OQ$ and $\displaystyle CQ$ using the $\displaystyle \sin$ and $\displaystyle \cos$ of $\displaystyle \angle QOC$ and $\displaystyle OC$.

Now switch attention to $\displaystyle \triangle CQS$. You know $\displaystyle SO$ and $\displaystyle OQ$, so you know $\displaystyle QS$. You already know $\displaystyle CQ$, so Pythagoras's theorem will now finish the problem.

RonL
• Aug 27th 2006, 04:22 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Drop a perpendicular from $\displaystyle C$ onto $\displaystyle OS$, call the new point at the foot of the perp. $\displaystyle Q$.

Now look at $\displaystyle \triangle OCQ$ you know $\displaystyle \angle QOC$, and $\displaystyle OC$, so you can find $\displaystyle OQ$ and $\displaystyle CQ$ using the $\displaystyle \sin$ and $\displaystyle \cos$ of $\displaystyle \angle QOC$ and $\displaystyle OC$.

Now switch attention to $\displaystyle \triangle CQS$. You know $\displaystyle SO$ and $\displaystyle OQ$, so you know $\displaystyle QS$. You already know $\displaystyle CQ$, so Pythagoras's theorem will now finish the problem.

RonL

I forgot:

Extent the tangent at $\displaystyle C$ to meet $\displaystyle OS$ at $\displaystyle T$, then $\displaystyle \triangle OCQ$ is similar to $\displaystyle \triangle CQT$. This will allow you to find $\displaystyle \angle CPS$, and you already have enough information to fins $\displaystyle \angle PSC$ which will allow you to find angle at $\displaystyle v$

RonL
• Aug 27th 2006, 07:55 AM
CaptainBlack
The attached scan shows all the relevant dimensions and angles filled
in on the diagram.

RonL
• Aug 28th 2006, 07:44 AM
Neffets...:P
Thanks Captain!
This forum is great! Now I can finally get help when I'm in a tricky situation...
Thanks a lot Captain Black!!!! :-D :-D

I couldn't find the answer with your first explanation, so I'm glad you did remember the rest...

Neffets...:P