This chapter is currently out of syllabus
1
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
$${{3 - \sqrt 2 } \over 4}$$
2
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$$\sqrt {10}$$
3
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
Out of Syllabus
Two poles, AB of length a metres and CD of length a + b (b $$\ne$$ a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan$$\angle$$ACB = $${1 \over 2}$$, then :
A
x2 + 2(a + 2b)x $$-$$ b(a + b) = 0
B
x2 + 2(a + 2b)x + a(a + b) = 0
C
x2 $$-$$ 2ax + b(a + b) = 0
D
x2 $$-$$ 2ax + a(a + b) = 0
4
JEE Main 2021 (Online) 26th August Evening Shift
+4
-1
Out of Syllabus
A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = $$\sqrt 5$$ inches and $$\angle$$PCB = tan-1(2). The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is :

A
$${\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
B
tan$$-$$1(1)
C
$${\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
D
$${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$
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