A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?

Given:

A man goes 15 metres due west and then 8 metres due north.

To do:

The distance from the starting point.

Solution:

Let the starting position of the man be $O$. The position when he moves $15\ m$ due West be $A$ and the final position when he moves $8\ m$ due North be $B$.

In $∆ABO$,

By Pythagoras theorem,

$BO^2 = AB^2 + AO^2$

$BO^2 = 8^2 + 15^2$

$BO^2 = 64 + 225$

$BO= \sqrt{289}$

$BO = 17\ m$

Therefore, the man is $17\ m$ far from the starting point.

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Updated on: 10-Oct-2022

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