Trigonometric Equations Question 120
Question: 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 $ cose{c^{-1}}2.6 $ respectively. The height of the tower is
[EAMCET 1992]
Options:
A) 50 m
B) 25 m
C) 40 m
D) None of these
Show Answer
Answer:
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 =cose{c^{-1}}2.6\Rightarrow cosec\beta =2.4 $
$ \therefore $ $ \cot \beta =\sqrt{(cose{c^{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 $ , $ AB^{2}=AD^{2}+BD^{2} $
$ \Rightarrow $ $ 100^{2}=[{{(3.2)}^{2}}+{{(2.4)}^{2}}]h^{2}=16h^{2} $
$ \Rightarrow $ $ h=25m $ .
BETA
