question_answer

ABC is triangular park with AB = AC = 100 m. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower at\[A\]and\[B\]are \[{{\cot }^{-1}}3.2\] and \[\text{cose}{{\text{c}}^{-1}}2.6\]respectively. The height of the tower is[EAMCET 1992]

A) 50 m

B) 25 m

C) 40 m

D)  None of these

Correct Answer: B

Solution :

DP is a clock tower standing at the middle point D of BC. \[\angle PAD=\alpha ={{\cot }^{-1}}3.2\Rightarrow \cot \alpha =3.2\] and \[\angle PBD=\beta =\text{cose}{{\text{c}}^{-1}}2.6\Rightarrow \text{cosec}\beta =2.4\] \[\therefore \] \[\cot \beta =\sqrt{(\text{cose}{{\text{c}}^{\text{2}}}\beta -1)}=\sqrt{(5.76)}=2.4\] In the triangles \[PAD\] and \[PBD\], \[AD=h\]\[\cot \alpha =3.2h\]and \[BD=h\cot \beta =2.4h\] In the right angled \[\Delta ABD\], \[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\] \[\Rightarrow \] \[{{100}^{2}}=[{{(3.2)}^{2}}+{{(2.4)}^{2}}]{{h}^{2}}=16{{h}^{2}}\] \[\Rightarrow \]\[h=25m\].