Tricky triogonometri...

**Neffets...:P** · August 26th 2006, 07:24 AM

How can I find angle v in this figure?

Info:

AO=OB=OP=OC=6370 km

**Glaysher** · August 26th 2006, 07:25 AM

Tricky with the figure missing?

**Neffets...:P** · August 26th 2006, 07:26 AM

PS=20240 km
CP = 2780 km

Thanks for all answers

**CaptainBlack** · August 26th 2006, 07:43 AM

Originally Posted by Neffets...:P

PS=20240 km
CP = 2780 km

Thanks for all answers

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

The rest is just fiddly trig.

**Neffets...:P** · August 26th 2006, 07:54 AM

thanks mate! I think i'll manage it all now...

You guys are GREAT!

**Neffets...:P** · August 26th 2006, 11:48 PM

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...

**CaptainBlack** · August 27th 2006, 12:01 AM

Originally Posted by Neffets...:P

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...

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

**CaptainBlack** · August 27th 2006, 04:22 AM

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

**CaptainBlack** · August 27th 2006, 07:55 AM

The attached scan shows all the relevant dimensions and angles filled
in on the diagram.

RonL

**Neffets...:P** · August 28th 2006, 07:44 AM

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

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thanks

need more help...

Thanks Captain!

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