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Amos leaves home and cycles for a distance of 35km on a bearing of N30°E. He changes direction of S60°E before heading home on a bearing of S49°W. To the nearest kilometer, find the total distance Amos traveled. Include workings.

Guest May 17, 2015

Best Answer 

 #1
avatar+78551 
+10

 

The angle between the departing vector and the returning vector is

 

(90 - 30 - (90-49))  =  19°

 

And when he truns S60E, the angle made by these two vectors is just (30 + 60) = 90°

 

And the remaining angle in the triangle is   just (180 - 90 - 19)  = 71°

 

Using the Law of Sines, we can find the second leg of his journey, thusly.....

35/sin71  = d /sin 19

d = 35sin19/sin71 = about 12.05km

 

And since this is a right triangle, the return vector is given by

√(35^2 + 12.05^2)  = about 37km

 

So the total trip length was 35 + 12.05 + 37 = about 84.05km   [84, to the nearest km ]

 

Here's an (approximate) pic......

 

 

CPhill  May 17, 2015
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1+0 Answers

 #1
avatar+78551 
+10
Best Answer

 

The angle between the departing vector and the returning vector is

 

(90 - 30 - (90-49))  =  19°

 

And when he truns S60E, the angle made by these two vectors is just (30 + 60) = 90°

 

And the remaining angle in the triangle is   just (180 - 90 - 19)  = 71°

 

Using the Law of Sines, we can find the second leg of his journey, thusly.....

35/sin71  = d /sin 19

d = 35sin19/sin71 = about 12.05km

 

And since this is a right triangle, the return vector is given by

√(35^2 + 12.05^2)  = about 37km

 

So the total trip length was 35 + 12.05 + 37 = about 84.05km   [84, to the nearest km ]

 

Here's an (approximate) pic......

 

 

CPhill  May 17, 2015

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