Let a vertical tower AB have its end A on the | vedclass

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

Question

(D) Let $\angle APC = \alpha$. In $	riangle APC$,$	an \alpha = \frac{AC}{AP}$.Since $C$ is the mid-point of $AB$,$AC = \frac{1}{2} AB$. Given $AP =

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(D) Let $\angle APC = \alpha$. In $	riangle APC$,$	an \alpha = \frac{AC}{AP}$.Since $C$ is the mid-point of $AB$,$AC = \frac{1}{2} AB$. Given $AP = 2AB$,we have $	an \alpha = \frac{\frac{1}{2} AB}{2 AB} = \frac{1}{4}$.Now,consider $	riangle ABP$. $\angle BAP = 90^{\circ}$. $\angle BPC = \beta$ and $\angle APC = \alpha$,so $\angle BAP = \alpha + \beta$.$	an(\alpha + \beta) = \frac{AB}{AP} = \frac{AB}{2 AB} = \frac{1}{2}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we have:$\frac{1}{2} = \frac{\frac{1}{4} + 	an \beta}{1 - \frac{1}{4} 	an \beta}$.Multiplying both sides by $4(1 - \frac{1}{4} 	an \beta)$:$2(1 - \frac{1}{4} 	an \beta) = 1 + 4 	an \beta$$2 - \frac{1}{2} 	an \beta = 1 + 4 	an \beta$$1 = 4.5 	an \beta$$1 = \frac{9}{2} 	an \beta$$	an \beta = \frac{2}{9}$.

Let a vertical tower $AB$ have its end $A$ on the level ground. Let $C$ be the mid-point of $AB$ and $P$ be a point on the ground such that $AP = 2AB$. If $\angle BPC = \beta$,then $\tan \beta$ is equal to:

Similar Questions

From a $60 \ m$ high tower,the angles of depression of the top and bottom of a house are $\alpha$ and $\beta$ respectively. If the height of the house is $\frac{60 \sin(\beta - \alpha)}{x}$,then $x =$

$A$ person observes the angle of elevation of a building as $30^\circ$. The person proceeds towards the building with a speed of $25(\sqrt{3} - 1) \, m/hour$. After $2 \, hours$,he observes the angle of elevation as $45^\circ$. The height of the building (in meters) is:

$A$ ladder rests against a wall making an angle $\alpha$ with the horizontal. The foot of the ladder is pulled away from the wall through a distance $x$,so that it slides a distance $y$ down the wall making an angle $\beta$ with the horizontal. The correct relation is

The angle of elevation of the top of a tower from a point $20 \ m$ away from its base is $45^\circ$. The height of the tower is ....... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module