question_answer

From the top of a building 30 m high the top and bottom of a tower are observed to have angles of depression \[30{}^\circ \]and \[45{}^\circ ,\]respectively. Then, the height of the tower is

A)  \[30\left( 1+\frac{1}{\sqrt{3}} \right)m\] 

B)  \[30\left( 1-\frac{1}{\sqrt{3}} \right)\,\,m\]

C)  \[30\left( \sqrt{3}+1 \right)\,\,m\]

D)  \[30(\sqrt{3}-1)\,\,m\]

Correct Answer: B

Solution :

Here, AB be the building and CD is the tower. So,       AB = 30 m, Let        DC = h Then, as            DE ||AC so AE = DC = A \[\therefore \]                              \[BE=(30-h)\,\,m\] Here, \[\cot 45{}^\circ =\frac{AC}{BA}\]                       (in\[\Delta ABC\]) \[\frac{AC}{30}=1\]\[\Rightarrow \]\[AC=30\,\,m\] Now, in\[\Delta BDE,\]\[\tan 30{}^\circ =\frac{BE}{DE}\] \[\therefore \]      \[CD=AE=AB-BE=30-\frac{30}{\sqrt{3}}\] \[h=30\left( 1-\frac{1}{\sqrt{3}} \right)\,\,m\]