A) 40 m
B) 42 m
C) 45 m
D) 47 m
Correct Answer: B
Solution :
[b] \[AB=h\] (height of the tower) |
\[BD=36m;\] \[BC=49m\] |
\[\angle D=47{}^\circ ;\] \[\angle C=43{}^\circ \] |
Now, in \[\Delta ABD,\] |
\[\tan 47{}^\circ =\frac{h}{36m}\] ? (i) |
and in \[\Delta ABC,\] \[\tan 43{}^\circ =\frac{h}{49m}\] |
\[\tan (90{}^\circ -47{}^\circ )=\frac{h}{49}\] |
\[\therefore \cot 47{}^\circ =\frac{h}{49}\] (ii) |
Multiplying equations (i) and (ii) |
\[\tan 47{}^\circ .\cot 47{}^\circ =\frac{h}{36}\times \frac{h}{49}=1=\frac{{{h}^{2}}}{36\times 49}\] |
\[h=6\times 7=42m\] |
\[\therefore \] Option [b] is correct |
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