A) 60 m
B) 75 m
C) 30 m
D) 45 m
Correct Answer: A
Solution :
[a] Let the height of the tower is h m From figure, In\[\Delta BMQ,\]\[\tan 30{}^\circ =\frac{90-h}{x}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}=\frac{90-h}{x}}\] (i) In \[\Delta APQ,\]\[\tan 60{}^\circ =\frac{90}{x}\] \[\Rightarrow \] \[x=\frac{90}{\sqrt{3}}=30\sqrt{3}\] On putting the value of x in Eq. (i), We get \[\frac{1}{\sqrt{3}}=\frac{90-h}{30\sqrt{3}}\] \[\Rightarrow \] \[90-h=30\] \[\Rightarrow \] \[h=90-30=60\,m\] |
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