Question

Two boats left a base camp at bearing of S60°E and S50°W at speeds of 50km/h and 70km/h respectively. Find the distance between the both boats after two hours.

Answer 1

• 197.93 km

Explanation:

Let the base camp is point A and boats' locations after two hours are points B and C.

By connecting the three points together we get a triangle ABC with sides:

• AB = 50*2 = 100 km
• AC = 70*2 = 140 km

The angle between AB and AC is:

• 60 + 50 = 110 degrees (opposite directions from south)

We are looking for the distance BC, which can be found by using the law of cosines:

• BC² = AB² + AC² - 2AB*AC*cos ∠BAC

• BC² = 100² + 140² - 2*100*140*cos 110°
• BC² = 39176.56 (rounded)
• BC = √39176.56 = 197.93 km (rounded)

The distance between the boats is 197.93 km.