# Is there enough info given in question

• Feb 2nd 2011, 07:23 AM
orlacoon
Is there enough info given in question
A ship which leaves a port, p, and travels south west, with constant speed, observes the flash of a lighthouse, l, ina direction S60W. After travelling for 10km, to a point,t, the ship observes the flash of the same lighthouse, due west.

Q1)Calculate how far the ship is from the lighthouse at this time i.e calculate length tl.

The answer given for this is 5.18km- Having sketched the diagram, I can't see that enough details have been supplied to use sine rule or otherwise to calculate this.

Q2) Calculate how close to the lighthouse the ship will pass.

The answer given here is 3.66km. Once again I cant figure this out. I took the point at which the ship passes the lighthouse to be the point at which t moves under the point l and sketched the resultant triangle. I used the answer from the first part to solve the triangle but had no joy.

Any ideas??
• Feb 2nd 2011, 08:46 AM
Unknown008
If you make your sketch, and label the points 1: the point the ship is when it first see the flash, 2 where it sees the flash for the second time and L the position of the lighthouse, you get:

Angle L12 = 15 degrees
Angle 1L2 = 90 - 60 = 30 degrees
Hence, angle 12L = 180 - (60 + 15) = 105 degrees

Side 12 = 10 km

Use the sine rule.

Let D be the distance you're looking for, that is when the ship is at 1.

$\displaystyle \dfrac{D}{\sin(105)} = \dfrac{10}{\sin(30)}$

Get this one first. (Smile)
• Feb 2nd 2011, 09:18 AM
tom@ballooncalculus
Quote:

Originally Posted by Unknown008
Angle 1L2 = 90 - 30 = 60 degrees

How so? I'm getting this, with your labels...

http://www.ballooncalculus.org/draw/misc/trig.png

... and therefore $\displaystyle \dfrac{D}{\sin(15)} = \dfrac{10}{\sin(30)}$

which is coming out right.
• Feb 2nd 2011, 09:23 AM
Unknown008
Oops, brain fart, I thought 60 degrees W of south means the upper angle 30 degrees, then 60 degrees will be the lower angle without... well, you see (Itwasntme)
• Feb 2nd 2011, 11:16 AM
orlacoon
Where are you getting the 15 degree angle from? Thanks, by the way!
Quote:

Originally Posted by tom@ballooncalculus
How so? I'm getting this, with your labels...

http://www.ballooncalculus.org/draw/misc/trig.png

... and therefore $\displaystyle \dfrac{D}{\sin(15)} = \dfrac{10}{\sin(30)}$

which is coming out right.

How do you get the 15 degree angle? Can work it out from there. Thanks for the help.
• Feb 2nd 2011, 01:01 PM
tom@ballooncalculus
Sw = S45W (Wink)
• Feb 3rd 2011, 12:09 AM
orlacoon
Oh, should have realised that! Was driving me mad! Thanks again.(Happy)