Answer is 100√3 let ABC be a Δ and D be an external point and AC= 100 therefore ∠DAB=30=∠ABC. tan 30 = 100÷distance of the car from the building that is= 1/√3 = 100 / distance of the car from the building that is, distance = 100√3 m
Let AB be the building and B be the foot and A be the top of the building.
and C be the car the distance between the car and the foot of the building 
α=30°, AB=100 m
tan α=AB/BC
⇒tan 30°=1/√3=AB/BC
⇒BC=100√3=173.2 m
Ans) The distance between the foot of the building and the car is 173.2 m.
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