amos leaves home and cycles for a distance of 35 km on a bearing of N30E. he changes direction and cycles in a direction of S60E before heading home on a bearing of S49W. to the nearest km, find the total distance amos travelled.

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 ]

Step-by-step explanation:

  21.3

step-by-step explanation:

multiplying the given equation by c/sin(25°), we get

  c = 9/sin(25°) ≈ 21.2958 ≈ 21.3

we assume "c" is the length of ab, the side opposite angle c in triangle abc.