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

#1**+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

#1**+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