Question

A man walks 8km North and then 5km in a direction 60° East of North. Find the distance from his starting point

Answer 1

Answer:

The starting point is S, from which the man walks 8 km in the north direction uptp B. Then he turn 60 degrees in the clockwise direction to north and travels for 5 km to reach the end point E.

The required distance between starting and ending points is denoted by SE.

Let us first observe that by the geometry of the figure ASBE (which is clearly a rectangle), AS = BC and BS = AC = 8 km.

Now, In triangle EBC, sin30 = EC/EB

OR EC = EB*sin30 = 5*(1/2) = 2.5 km.

Also, cos30 = BC/EB

OR BC = AS = (√3/2)*5 = 5√3/2 = 4.33 km.

AE = AC + EC = 8 + 2.5 = 10.5 km.

Required value ES can now be found by using Pythagoras theorem.

ES² = AS² + AE²

OR ES² = (10.5²) + (4.33²)

OR ES² = 129

OR ES = 11.35 km.

Thus, the distance from the start point = 11.35 in the North East direction.

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