A) \[\frac{h\cot x}{\cot x+\cot y}\]
B) \[\frac{h\cot \,y}{\cot x+\cot y}\]
C) \[\frac{h\cot x}{\cot x-\cot y}\]
D) \[\frac{h\cot y}{\cot y-\cot y}\]
Correct Answer: C
Solution :
[c] Suppose the height of the building = BC Then, in \[\Delta ABC\] \[\cot y=\frac{AB}{BC}\]\[\Rightarrow \]\[AB=BC\cdot cot\,y\] (i) and in \[EDC,\]\[\cot x=\frac{ED}{CD}\] From Eq. (i), we get \[\cot x=\frac{BC\cot y}{CD}\] \[(\because \,\,ED=AB)\] \[\Rightarrow \] \[CD\cot x=h\cot y+CD\cot y\] \[(\because BC=h+CD)\] \[\Rightarrow \] \[CD=\frac{h\cot y}{\cot x-\cot y}\] \[\because \] \[BC=h+CD=\frac{h\cot y}{\cot x-\cot y}\] \[\Rightarrow \] \[BC=\frac{h\cot x}{\cot x-\cot y}\] |
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