A) 54.64m
B) 94.64m
C) 40m
D) 109.3m
Correct Answer: B
Solution :
Let the height of the building be TS. According to the problem, from the figure, the height of the building is \[\therefore \]. RS is the distance between the first floor and ground floor. Therefore, \[\text{7}\times \text{13}+\text{13}\] \[\therefore \]\[\text{224}=\text{12}0\times \text{1}+\text{1}0\text{4}\] \[\text{12}0=\text{1}0\text{4}\times \text{1}+\text{16}\]\[\text{1}0\text{4}=\text{16}\times \text{6}+\text{8}\] ...... (1) and\[16=8\times 2+0\] \[\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{\sqrt{{{\cos }^{2}}\theta }}{\sin \theta }=\frac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta }\]\[\text{3}0{}^\circ \] \[\text{3}0{}^\circ \]Height of the tower \[AD=\frac{bc}{\sqrt{{{b}^{2}}+{{c}^{2}}}}\]You need to login to perform this action.
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