Embibe Experts Solutions for Chapter: Heights and Distances, Exercise 1: JEE Main - 27 July 2022 Shift 1

A vertical pole fixed to the horizontal ground is divided in the ratio $3: 7$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 \mathrm{m}$ away from the base of the pole, then the height of the pole (in meters) is :

Let $AB$ and $PQ$ be two vertical poles, $160\mathrm{m}$ apart from each other. Let $C$ be the middle point of $B$ and $Q$, which are feet of these two poles. Let $\frac{\pi }{8}$ and $\theta$ be the angles of elevation from $C$ to $P$ and $A$, respectively. If the height of pole $PQ$ is twice the height of pole $AB$, then ${\tan }^2\theta$ is equal to

From the base of a pole of height $20$ meter, the angle of elevation of the top of a tower is $60°$. The pole subtends an angle $30°$ at the top of the tower. Then the height of the tower is:

A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower in two parts such that $QR=15\mathrm{m}$. If from a point $A$ on the ground the angle of elevation of $R$ is $60°$ and the part $PR$ of the tower subtends an angle of $15°$ at $A$, then the height of the tower is 

Let a vertical tower $AB$ of height $2h$ stands on a horizontal ground. Let from a point $P$ on the ground a man can see upto height $h$ of the tower with an angle of elevation $2\alpha$. When from $P$, he moves a distance $d$ in the direction of $\overrightarrow{AP}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d=\sqrt{7}h$, then $\tan \alpha$ is equal to

The angle of elevation of the top $P$ of a vertical tower $PQ$ of height $10$ from a point $A$ on the horizontal ground is $45°$. Let $R$ be a point on $AQ$ and from a point $B$, vertically above $R$, the angle of elevation of $P$ is $60°$. If $\angle BAQ=30°,AB=d$ and the area of the trapezium $PQRB$ is $\alpha$, then the ordered pair $\left(d,\alpha \right)$ is

A horizontal park is in the shape of a triangle $OAB$ with $AB=16$. A vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO=\angle PBO=15°$ and $\angle PCO=45°$, where $C$ is the midpoint of $AB$. Then ${\left(OP\right)}^2$ is equal to

The angle of elevation of the top of a tower from a point $A$ due north of it is $\alpha$ and from a point $B$ at a distance of $9$ units due west of $A$ is ${\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point $B$ from the tower is $15$ units, then $\cot \alpha$ is equal to