Question

A man goes 15 m due West and then 8 m due North. How far he is from the starting point? * 1 point

Answer 1

Answer:

17 m. due north west

Step-by-step explanation:

By applying Pythagoras Theorem ,

$ac = \sqrt{ {15}^{2} + {8}^{2} } = \sqrt{225 + 64} = \sqrt{289} = 17$

So the man is 17 m. away from his innitial point due north west.

Answer 2

$\bf{\underline{Given}}:-$

• AB = 15 m
• BC = 8 m
• ∠ABC = 90°

To find :-

• How far he is from the starting point = ?

Solution :-

By using Pythagoras theorem,

In right ∆ ABC

AC = √(AB)² + (BC)²

⤇ AC = √(15)² + (8)²

⤇ AC = √225 + 64

⤇ AC = √289

⤇ AC = 17 m

Hence,the man is 17 m away from the starting point.

More Information :-

Pythagoras theorem :-

⤇ In a right triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Converse of Pythagoras theorem :-

⤇ In a triangle,if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.