AB is a vertical tower. The point A is on the | vedclass

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

Question

(D) Let $AB = h$. Since $C$ is the midpoint of $AB$,$AC = \frac{h}{2}$ and $CB = \frac{h}{2}$.Given $AP = n \cdot AB = nh$.In $	riangle PAC$,$	an(\a

Vedclass · Accepted Answer

$n = (2n^2 + 1)	an \alpha$

Vedclass · Answer

(D) Let $AB = h$. Since $C$ is the midpoint of $AB$,$AC = \frac{h}{2}$ and $CB = \frac{h}{2}$.Given $AP = n \cdot AB = nh$.In $	riangle PAC$,$	an(\angle APC) = \frac{AC}{AP} = \frac{h/2}{nh} = \frac{1}{2n}$.In $	riangle PAB$,$	an(\angle APB) = \frac{AB}{AP} = \frac{h}{nh} = \frac{1}{n}$.We know $\alpha = \angle APB - \angle APC$.Therefore,$	an \alpha = 	an(\angle APB - \angle APC) = \frac{	an(\angle APB) - 	an(\angle APC)}{1 + 	an(\angle APB) \cdot 	an(\angle APC)}$.Substituting the values: $	an \alpha = \frac{\frac{1}{n} - \frac{1}{2n}}{1 + (\frac{1}{n})(\frac{1}{2n})} = \frac{\frac{1}{2n}}{1 + \frac{1}{2n^2}} = \frac{\frac{1}{2n}}{\frac{2n^2 + 1}{2n^2}} = \frac{1}{2n} \cdot \frac{2n^2}{2n^2 + 1} = \frac{n}{2n^2 + 1}$.Thus,$n = (2n^2 + 1)	an \alpha$.

$AB$ is a vertical tower. The point $A$ is on the ground and $C$ is the middle point of $AB$. The part $CB$ subtends an angle $\alpha$ at a point $P$ on the ground. If $AP = n \cdot AB$,then the correct relation is:

Similar Questions

In a $\triangle ABC$,the expression $\frac{a}{\tan A} + \frac{b}{\tan B} + \frac{c}{\tan C}$ is equal to:

The angles $\alpha, \beta, \gamma$ of a triangle satisfy the equations $2 \sin \alpha + 3 \cos \beta = 3 \sqrt{2}$ and $3 \sin \beta + 2 \cos \alpha = 1$. Then,angle $\gamma$ equals (in $^{\circ}$)

The equation $\sin 3\theta = 4 \sin \theta \sin 2\theta \sin 4\theta$ in the interval $0 \le \theta \le \pi$ has:

In $\triangle ABC$,if $a=2, b=\sqrt{6}$,and $c=\sqrt{3}+1$,then $\sin^2 C - \sin^2 A =$

The equation $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$ has a solution if

Vedclass Test Series

Exam Paper Generator

Online Exam Module