A man goes 15 m due west and then 8 m due north.Find distance from the starting point
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Answered by
1
Step-by-step explanation:
first the man goes to the west 15 m from point A to point B and then it turns north from point B and walks 8 m to point C.
This forms a right angled triangle ABC
So,by pythagoras theorem
AC^{2} = AB^{2}+ BC^{2}AC
2
=AB
2
+BC
2
SO,
AC^{2} = 15^{2} + 8^{2}AC
2
=15
2
+8
2
AC^{2} = 289AC
2
=289
AC^{2}=225+64AC
2
=225+64
AC= \sqrt{289}AC=
289
AC = 17 metres.
So the man is now 17metres away from his starting point.
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0
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