A) 415m
B) 315.4m
C) 200 m
D) 305 m
Correct Answer: B
Solution :
(b) According to the problem, from the figure, AB is the tower of height 200 m. The distance between the height is \[\text{1}00\text{1}=\text{91}\times \text{1}0+\text{91}\]. \[\text{91}0=\text{91}\times \text{1}0+0\] \[\therefore \] \[=\left( \frac{144}{48}+\frac{384}{48}+\frac{240}{48} \right)=3+8+5=16\] \[\frac{a}{b}\] ??(1) \[\frac{c}{d}=\frac{L.C.M.(a,c)}{H.C.F.(b,d)}\] \[\Rightarrow \] \[L.C.M.\] ?..(2) \[\frac{6}{14}and\frac{2}{7}\]The required distance \[\Rightarrow \] \[\frac{L.C.M.(6,2)}{H.C.F.(14,7)}=\frac{6}{7}\] \[\text{7}\times \text{13}+\text{13}=\text{1}0\text{4}=\text{23}\times \text{13}\]You need to login to perform this action.
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