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Two ships leave a Port O. One ship travels on a bearing of 3400 to a point P which is 50 km from O. The other ship travels on a bearing of 0600 to a point Q, 85 km from O. (a) Draw a diagram to represent the position of the port and the two ships. On your diagram, carefully label North, the given angles and the distances travelled. (b) Calculate the distance PQ in km (c) Determine the bearing of P from Q.

Guest Jan 28, 2015

Best Answer 

 #1
avatar+80804 
+10

 

 

 

Two ships leave a Port O. One ship travels on a bearing of 3400 to a point P which is 50 km from O. The other ship travels on a bearing of 0600 to a point Q, 85 km from O. (a) Draw a diagram to represent the position of the port and the two ships. On your diagram, carefully label North, the given angles and the distances travelled. (b) Calculate the distance PQ in km (c) Determine the bearing of P from Q.

 

I'm assuming that 3400 = 340.0    and that 0600 = 60.0

Then the angle between them is just 80°

We can use the Law of Cosines to find PQ =

PQ^2  = 85^2 + 50^2 - 2(85)(50)cos(80)

PQ^2 = 8248.9904898305

PQ = about 90.824 km

And the bearing from P to Q is given by [90 + sin-1(4.485/90.824)]° = [90 + 2.83]° = 92.83°

Here's a diagram.....("North" is at the "top' )

 

CPhill  Jan 28, 2015
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4+0 Answers

 #1
avatar+80804 
+10
Best Answer

 

 

 

Two ships leave a Port O. One ship travels on a bearing of 3400 to a point P which is 50 km from O. The other ship travels on a bearing of 0600 to a point Q, 85 km from O. (a) Draw a diagram to represent the position of the port and the two ships. On your diagram, carefully label North, the given angles and the distances travelled. (b) Calculate the distance PQ in km (c) Determine the bearing of P from Q.

 

I'm assuming that 3400 = 340.0    and that 0600 = 60.0

Then the angle between them is just 80°

We can use the Law of Cosines to find PQ =

PQ^2  = 85^2 + 50^2 - 2(85)(50)cos(80)

PQ^2 = 8248.9904898305

PQ = about 90.824 km

And the bearing from P to Q is given by [90 + sin-1(4.485/90.824)]° = [90 + 2.83]° = 92.83°

Here's a diagram.....("North" is at the "top' )

 

CPhill  Jan 28, 2015
 #2
avatar+91412 
+5

You are becoming a GeoGebra expert Chris.  

Melody  Jan 29, 2015
 #3
avatar+80804 
+5

I wish I knew how to construct "3-D" objects with it.....I haven't found the trick for that!!!

 

CPhill  Jan 29, 2015
 #4
avatar+91412 
0

I haven't tried that yet either.   

Melody  Jan 29, 2015

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