1
JEE Main 2023 (Online) 6th April Morning Shift
+4
-1

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

A
$$200(3-\sqrt{3})$$
B
$$300(\sqrt{3}-1)$$
C
$$300(\sqrt{3}+1)$$
D
$$600(\sqrt{3}-1)$$
2
JEE Main 2022 (Online) 29th July Morning Shift
+4
-1

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 :

A
$$\frac{6}{5}$$
B
$$\frac{9}{5}$$
C
$$\frac{4}{3}$$
D
$$\frac{7}{3}$$
3
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1

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

A
$$\frac{32}{\sqrt{3}}(\sqrt{3}-1)$$
B
$$\frac{32}{\sqrt{3}}(2-\sqrt{3})$$
C
$$\frac{16}{\sqrt{3}}(\sqrt{3}-1)$$
D
$$\frac{16}{\sqrt{3}}(2-\sqrt{3})$$
4
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1

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 $$45^{\circ}$$. Let R be a point on AQ and from a point B, vertically above $$\mathrm{R}$$, the angle of elevation of $$\mathrm{P}$$ is $$60^{\circ}$$. If $$\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d}$$ and the area of the trapezium $$\mathrm{PQRB}$$ is $$\alpha$$, then the ordered pair $$(\mathrm{d}, \alpha)$$ is :

A
$$(10(\sqrt{3}-1), 25)$$
B
$$\left(10(\sqrt{3}-1), \frac{25}{2}\right)$$
C
$$(10(\sqrt{3}+1), 25)$$
D
$$\left(10(\sqrt{3}+1), \frac{25}{2}\right)$$
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