+0

# 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

0
643
1

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.

May 17, 2015

#1
+94526
+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......

May 17, 2015

#1
+94526
+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