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An helicopter flies in a strange pattern, all at the same altitude of 5 km: 1) it flies 29 km in a direction 30.0° east of north, 2) it flies 20 km due south, and 3) it flies 10 km in a direction 30.0° north of west. How far did the helicopter end up from its starting point?

 

A.59.0 km

B.11.7 km

C.18.8 km

D.23.0 km

E.5.0 km

 Jun 4, 2016

Best Answer 

 #2
avatar+118658 
+5

If you want explanation you can ask. ://

 Jun 4, 2016
 #1
avatar+118658 
+5

29-10sqrt3 =11.7km

 Jun 4, 2016
 #2
avatar+118658 
+5
Best Answer

If you want explanation you can ask. ://

Melody  Jun 4, 2016
 #3
avatar+129840 
+5

Here's my approach to this one

 

Let him start at(0,0)......first, he flies to the point given by:  (29cos 60, 29sin60)  = (14.5, 25.1)

 

Then ....he flies 20 km directly south  and ends up at (14.5, 5.1)

 

Then......he flies to the point given by (14.5 + 10 cos150, 5.1 + 10 sin150) = (5.83975, 10.1)

 

The distance from this point to the origin [ where he started]  =

 

sqrt [ 5.83975^2 + 10.1^2]  ≈  11.67 km ≈ 11.7 km

 

[ Just as Melody said.....!!!!  ]

 

 

cool cool cool

 Jun 4, 2016
 #4
avatar+118658 
0

Hi Chris,

My logic wasn't that complicated.

I did a rough sketch on a peice of paper and realised there was a 90, 60, 30 degree triangle there.

The hypotenuse is 20 and one side is 10 so the other is 10sqrt3.

So the distance from the beginning is 29-10sqrt3 = 11.7 km

 

Here is a more precise diagram.

 

 Jun 4, 2016

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