question_answer

The angles of elevation of the top of a building from the ground floor and first floor of another building are\[\text{6}0{}^\circ \]and\[\text{45}{}^\circ \] respectively. If the first floor is 40 m above the ground floor, what is the height of the building?

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 \[AD=\frac{bc}{\sqrt{{{b}^{2}}+{{c}^{2}}}}\]