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.
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......
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......