The distance he is from his house is mathematically given as
OB=8.718
The jogger's distance from his house
Question Parameters:
A jogger ran 6 miles due east of his house. Then he ran 4 miles at a heading of 30 east of north
The from his house after running 10 miles
AC/AB=cos60
Generally
[tex]OB=\sqrt{OC^2+BC^2}\\\\OB=\sqrt{6+2}^2+(2\sqrt{3})^2[/tex]
OB=8.718
Therefore,he is 8.718 miles distance from his house
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