A) 4 m
B) 6 m
C) 7.5 m
D) 36 m
Correct Answer: B
Solution :
Let h be the height of the tower and \[\angle CBD=\theta \] |
\[\angle DAC=90{}^\circ -\theta \] |
\[\therefore \] In \[\Delta BCD,\] |
\[\tan \theta =\frac{CD}{BC}\] |
\[\Rightarrow \] \[\tan \theta =\frac{h}{3}\] ?(i) |
and in \[\Delta ACD,\] |
\[\tan \,\,(90{}^\circ -\theta )=\frac{CD}{AC}\] |
\[\Rightarrow \]\[\cot \theta =\frac{h}{12}\] ?(ii) |
From Eqs. (i) and (ii), we get |
\[\tan \theta \cdot \cot \theta =\frac{h}{3}\cdot \frac{h}{12}\] |
\[\Rightarrow \] \[{{h}^{2}}=36\]\[\Rightarrow \]\[h=6\,\,\text{m}\] |
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