question_answer

    ABC is a triangular park with\[AB=AC=100\text{ }m\]A clock tower is situated at the mid point of BC. The angle of elevation if the top of the tower at A and B are\[co{{t}^{-1}}3.2\]and\[\cos e{{c}^{-1}}2.6\] respectively. The height of the tower is

A)  16 m                                    

B)  25 m

C)  50 m                                    

D)  None of these

Correct Answer: B

Solution :

                Let OP be the clock tower standing at the midpoint\[O\]of side BC of\[\Delta ABC\]. Let\[\alpha =\angle PAO\]\[=co{{t}^{-1}}3.2\]and\[\beta =\angle PBO=\cos e{{c}^{-1}}2.6.\] Then,\[cot\text{ }\alpha =3.2\]and \[cosec\text{ }\beta =2.6\]. \[\therefore \]\[\cot \beta =\sqrt{\cos e{{c}^{2}}\beta -1}=\sqrt{{{(2.6)}^{2}}-1}=2.4\] In a triangle PAO and PBO, we have                         \[AO=h\cot \alpha =3.2\,h\] and        \[BO=h\,\cot B=2.4\,h\] In \[\Delta ABO,\text{ }A{{B}^{2}}=O{{A}^{2}}+O{{B}^{2}}\] \[\Rightarrow \]               \[{{100}^{2}}={{(3.2h)}^{2}}+{{(2.4h)}^{2}}\] \[\Rightarrow \]               \[{{100}^{2}}=16{{h}^{2}}\] \[\Rightarrow \]               \[{{h}^{2}}=625\]\[\Rightarrow \]\[h=25\,m\]