ABC is a triangular park with AB = AC = 100 , | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $DP = h$ be the height of the clock tower standing at the mid-point $D$ of $BC$.Given $\angle PAD = \alpha = \cot^{-1} 3.2 \Rightarrow \cot \a

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Let $DP = h$ be the height of the clock tower standing at the mid-point $D$ of $BC$.Given $\angle PAD = \alpha = \cot^{-1} 3.2 \Rightarrow \cot \alpha = 3.2$.Given $\angle PBD = \beta = \csc^{-1} 2.6 \Rightarrow \csc \beta = 2.6$.Then $\cot \beta = \sqrt{\csc^2 \beta - 1} = \sqrt{(2.6)^2 - 1} = \sqrt{6.76 - 1} = \sqrt{5.76} = 2.4$.In right-angled triangles $	riangle PAD$ and $	riangle PBD$:$AD = h \cot \alpha = 3.2h$ and $BD = h \cot \beta = 2.4h$.In the right-angled $	riangle ABD$ (since $AD \perp BC$ in isosceles $	riangle ABC$ where $AB=AC$):$AB^2 = AD^2 + BD^2$$100^2 = (3.2h)^2 + (2.4h)^2$$10000 = (10.24 + 5.76)h^2$$10000 = 16h^2$$h^2 = \frac{10000}{16} = 625$$h = 25 \, m$.

$ABC$ is a triangular park with $AB = AC = 100 \, m$. $A$ clock tower is situated at the mid-point $D$ of $BC$. The angles of elevation of the top of the tower at $A$ and $B$ are $\cot^{-1} 3.2$ and $\csc^{-1} 2.6$ respectively. The height of the tower is .... $m$.

Similar Questions

In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to

In $\Delta ABC,$ if $\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in $H.P.,$ then $a, b, c$ will be in

In a triangle $ABC$,evaluate the expression $\frac{a(r_1 r + r_2 r_3)}{r_1 - r + r_2 + r_3}$.

The number of solutions to $\sin x = \frac{6}{x}$ with $0 \leq x \leq 12 \pi$ is

In a triangle $ABC$,$\left(\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\right)^2 \leq$

Vedclass Test Series

Exam Paper Generator

Online Exam Module