question_answer

From the top of a 10 m high building, the angle of elevation of the top of a tower is \[60{}^\circ \] and the angle of depressionof its foot is \[45{}^\circ \]. Find the height of the tower. (Take \[\sqrt{3}=1.732\])

A)  27.32 m

B)  25.44 m

C)  19.96 m                      

D)  21.92 m

Correct Answer: A

Solution :

Let AB be the building and CD be the tower. Draw \[BE\bot CD.\] Then, CE = AB = 10 m, \[\angle EBD=60{}^\circ \]and \[\angle ACB=\angle CBE=45{}^\circ \] \[\frac{AC}{AB}=\cot 45{}^\circ =1\Rightarrow \frac{AC}{10}=1\] \[\Rightarrow \]   AC = 10 m From \[\Delta EBD,\]we have \[\frac{DE}{BE}=\tan 60{}^\circ =\sqrt{3}\Rightarrow \frac{DE}{AC}=\sqrt{3}\] \[\Rightarrow \]   \[\frac{DE}{10}=1.732\Rightarrow DE=17.32\,\,m\] Height of the tower \[CD=CE+DE=\text{(10+17}\text{.32)}\] = 27.32 m