# Tricky triogonometri...

• August 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
• August 26th 2006, 07:25 AM
Glaysher
Tricky with the figure missing?
• August 26th 2006, 07:26 AM
Neffets...:P
continues....
PS=20240 km
CP = 2780 km

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

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

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

The rest is just fiddly trig.
• August 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!
• August 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
• August 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 $C$ onto $OS$, call the new point at the foot of the perp. $Q$.

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

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

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

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

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

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

RonL

I forgot:

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

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

RonL
• August 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