Question

A girl walks 5km due north,then in the direction of 60° of northeast towards her final destination.If the magnitude of her resultant displacement is 8km,how far she must have travelled in north east direction?

Answer 1

Answer:

4.2 km

Explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as isosceles triangle, scalene triangle, equilateral triangle and right angled triangle.

Sine rule states that for a triangle with angles A,B and C and corresponding opposite sides a, b, and c, the following formula holds:

$\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}$

Let the starting point be A, hence the side opposite to the  angle = a = distance travelled in north east direction.

Let the point of 60° of northeast be B = 90° + (90° - 60°) = 120°, hence the side opposite to the  angle = b = resultant displacement = 8 km.

Let C be the endpoint, hence the side opposite to the  angle = c = distance travelled north = 5 km

Using sine rule:

$\frac{b}{sin(B)}=\frac{c}{sin(C)} \\\\\frac{8}{sin(120)}=\frac{5}{sin(C)} \\\\sin(C)=\frac{5*sin(120)}{8} =0.5412\\\\C=sin^{-1}(0.5412)=32.8^o$

∠A + ∠B + ∠C = 180° (sum of angle in a triangle)

∠A + 120 + 32.8 = 180

∠A = 27.2°

$\frac{b}{sin(B)}=\frac{a}{sin(A)} \\\\\frac{8}{sin(120)}=\frac{a}{sin(27.2)} \\\\a=\frac{8*sin(27.2)}{sin(120)} \\\\a=4.2\ km$

Therefore she travelled 4.2 km in the north east direction