Two walkers set off at the same time from a crossroad and walk along flat straight roads inclined to each other at 70°.If they both walk at a speed of 8km/h, find their distance apart 10 minutes later

Answer 1

First, we need to find the distance each walker has traveled in 10 minutes, which is 1/6 of an hour.

Distance = speed x time
Distance = 8 km/h x 1/6 hour
Distance = 8/6 km = 4/3 km

Now, we can calculate the distance between the two walkers using the Law of Cosines:

c^2 = a^2 + b^2 - 2ab*cos(C)

Let's assume the starting point of the two walkers is the vertex of an equilateral triangle, where each side length is 4/3 km.

c^2 = (4/3)^2 + (4/3)^2 - 2(4/3)(4/3)cos70°
c^2 = 16/9 + 16/9 - (64/9)(cos70°)
c^2 = 32/9 - (64/9)(0.3420)
c^2 = 32/9 - 21.8880/9
c^2 = (32 - 21.8880)/9
c^2 = 10.1120/9
c^2 ≈ 1.12
c ≈ √1.12
c ≈ 1.06 km

Therefore, the distance between the two walkers 10 minutes later is approximately 1.06 km.