A) 16
B) 18
C) 50
D) 15
Correct Answer: A
Solution :
We put one pole at origin. \[BC=80m,OA=20m\] Line OC and AB intersect at M. To find: Length of MN. Eqn of OC:\[y=\left( \frac{80-0}{a-0} \right)x\]\[\Rightarrow \]\[y=\frac{80}{a}x\]?.(1) Eqn of AB: \[y=\left( \frac{20-0}{0-a} \right)\left( x-a \right)\] \[\Rightarrow \]\[y=\frac{-20}{a}\left( x-a \right)\] ?(2) AtM: (1)=(2) \[\Rightarrow \]\[\frac{80}{a}x=\frac{-20}{a}\left( x-a \right)\]\[\Rightarrow \]\[\frac{80}{a}x=\frac{-20}{a}x+20\] \[\Rightarrow \]\[x=\frac{a}{5}\]\[\therefore \]\[y=\frac{80}{a}\times \frac{a}{5}=16\]You need to login to perform this action.
You will be redirected in
3 sec