Find AB60°302045° h60° – InterviewSolution

1.

Find AB60°302045° h60°

Answer»

from triangle ACF , the value of AC = CF/tan45° = 20.

and assume a perpendicular distance from F, parallel to CD then the value of CD = tan 60°*(DE-CF) = √3*10 = 10√3.

And at last the value of DB = DE/tan60 = 30/√3 = 10√3.

so , total distance AB = AC+CD+DB = 20+10√3+10√3 = 20(1+√3).

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