question_answer

The angle of elevation of the top of a tower from the bottom of a building is twice that from its top. What is the height of the building, if the height of the tower is 75 m and the angle of elevation of the top of the tower from the bottom of the building is \[60{}^\circ \]?

A)  25 m

B)  37.5 m

C)  50 m   

D)  60 m

Correct Answer: C

Solution :

[c] \[2\theta =60{}^\circ \]\[\Rightarrow \]\[\theta =30{}^\circ \]

In \[\Delta ABC,\]\[\tan 60{}^\circ =\frac{AB}{BC}\]

\[\sqrt{3}=\frac{75}{BC}\]

\[BC=\frac{75}{\sqrt{3}}\]                  . (i)

In \[\Delta ADE,\]\[\tan 30{}^\circ =\frac{AD}{DE}\]

\[\frac{1}{\sqrt{3}}=\frac{75-h}{DE}\]               (ii)

\[\because \]       \[BC=DE\]

\[\therefore \]      \[\frac{75}{\sqrt{3}}=\sqrt{3}\,\,(75-h)\]

\[\Rightarrow \] \[75=225-3h\]\[\Rightarrow \]\[h=50\,m\]