A man goes ^@ 16 \space m^@ due west and then ^@16 \space m^@ due north. How far is he from the starting point?


Answer:

^@ 22.63 \space m^@

Step by Step Explanation:
  1. Let us consider a man starting from ^@ A ^@ and goes from ^@ A ^@ to ^@B ^@ and then from ^@B^@ to ^@C^@, as shown in the figure.
    Then, @^ \begin{aligned} AB = 16 \space m , BC = 16 \space m \space and \space \angle ABC = 90^\circ \end{aligned} @^ B A C N W 16 m 16 m
  2. From right ^@\Delta ABC^@, we have @^ \begin{aligned} AC^2 =&\space AB^2 + BC^2 \\ =&\space [ (16)^2 + (16)^2 ] \space m^2 \\ =&\space ( 256 + 256 ) \space m^2 \\ =&\space 512 \space m^2 \\ \therefore \space AC = &\space \sqrt{ 512 } \space m^2 = 22.63 \space m \end{aligned} @^ Hence, the man is ^@ 22.63 \space m^@ away from the starting position.

You can reuse this answer
Creative Commons License

What they say