question_answer

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of \[R=8.3\text{ }J\text{ }mo{{l}^{-1}}\]at the/foot of the tower and the angle of elevation of the top of the tower from A or B is\[4.100\,kJ\,mo{{l}^{-1}}\]. The height of the tower is       AIEEE  Solved  Paper-2007

A)  \[3.7904\text{ }kJ\,mo{{l}^{-1}}\]

B)                         \[37.904\text{ }kJ\text{ }mo{{l}^{-1}}\]                

C)         \[41.00\text{ }kJ\text{ }mo{{l}^{-1}}\]  

D)         \[AgI{{O}_{3}}\]

Correct Answer: C

Solution :

Let h be the height of a tower \[\because \] \[\angle AOB={{60}^{o}}\] \[\therefore \] \[\Delta OAB\]is equilateral. \[\therefore \]\[OA=OB=AB=a\] Now, in\[\Delta OAC\], \[\tan {{30}^{o}}=\frac{h}{a}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{a}\Rightarrow h=\frac{a}{\sqrt{3}}\]