This chapter is currently out of syllabus
1
JEE Main 2022 (Online) 25th July Morning Shift
+4
-1
Out of Syllabus

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 $$Q R=15 \mathrm{~m}$$. If from a point $$A$$ on the ground the angle of elevation of $$R$$ is $$60^{\circ}$$ and the part $$P R$$ of the tower subtends an angle of $$15^{\circ}$$ at $$A$$, then the height of the tower is :

A
$$5(2 \sqrt{3}+3) \,\mathrm{m}$$
B
$$5(\sqrt{3}+3) \,\mathrm{m}$$
C
$$10(\sqrt{3}+1) \,\mathrm{m}$$
D
$$10(2 \sqrt{3}+1) \,\mathrm{m}$$
2
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1
Out of Syllabus

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

A
$$15\sqrt 3$$
B
$$20\sqrt 3$$
C
20 + $$10\sqrt 3$$
D
30
3
JEE Main 2022 (Online) 28th June Morning Shift
+4
-1
Out of Syllabus

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 $${\pi \over 8}$$ 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 tan2$$\theta$$ is equal to

A
$${{3 - 2\sqrt 2 } \over 2}$$
B
$${{3 + \sqrt 2 } \over 2}$$
C
$${{3 - 2\sqrt 2 } \over 4}$$
D
$${{3 - \sqrt 2 } \over 4}$$
4
JEE Main 2021 (Online) 31st August Morning Shift
+4
-1
Out of Syllabus
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 m away from the base of the pole, then the height of the pole (in meters) is :
A
12$$\sqrt {15}$$
B
12$$\sqrt {10}$$
C
8$$\sqrt {10}$$
D
6$$\sqrt {10}$$
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