JEE Main Height and Distance Practice Questions With Solutions

The angle of elevation of the top $\mathrm{P}$P of a tower from the feet of one person standing due South of the tower is ${45}^{\circ }$45^{\circ} and from the feet of another person standing due west of the tower is ${30}^{\circ }$30^{\circ}. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to

From the top $\mathrm{A}$A of a vertical wall $\mathrm{AB}$AB of height $30\mathrm{m}$30 \textrm{ } m, the angles of depression of the top $\mathrm{P}$P and bottom $\mathrm{Q}$Q of a vertical tower $\mathrm{PQ}$PQ are ${15}^{\circ }$15^{\circ} and ${60}^{\circ }$60^{\circ} respectively, $\mathrm{B}$B and $\mathrm{Q}$Q are on the same horizontal level. If $\mathrm{C}$C is a point on $\mathrm{AB}$AB such that $\mathrm{CB}=\mathrm{PQ}$CB = PQ, then the area (in ${\mathrm{m}}^2$m^{2} ) of the quadrilateral $\mathrm{BCPQ}$BCPQ is equal to :

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

A horizontal park is in the shape of a triangle $\mathrm{OAB}$OAB with $\mathrm{AB}=16$AB = 16. A vertical lamp post $\mathrm{OP}$OP is erected at the point $\mathrm{O}$O such that $\mathrm{\angle }\mathrm{PAO}=\mathrm{\angle }\mathrm{PBO}={15}^{\circ }$\angle PAO = \angle PBO = 15^{\circ} and $\mathrm{\angle }\mathrm{PCO}={45}^{\circ }$\angle PCO = 45^{\circ}, where $\mathrm{C}$C is the midpoint of $\mathrm{AB}$AB. Then $(\mathrm{OP}{)}^2$\left(\right. OP \left.\right)^{2} 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}^{\circ }$45^{\circ}. Let R be a point on AQ and from a point B, vertically above $\mathrm{R}$R, the angle of elevation of $\mathrm{P}$P is ${60}^{\circ }$60^{\circ}. If $\mathrm{\angle }\mathrm{BAQ}={30}^{\circ },\mathrm{AB}=\mathrm{d}$\angle BAQ = 30^{\circ} , AB = d and the area of the trapezium $\mathrm{PQRB}$PQRB is $\alpha$\alpha, then the ordered pair $(\mathrm{d},\alpha )$\left(\right. d , \alpha \left.\right) is :

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

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

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

Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $\frac{\pi }{8}$\frac{\pi}{8} and $\theta$\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 tan²$\theta$\theta is equal to