This chapter is currently out of syllabus
1
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
Out of Syllabus
The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45o from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30o, then the distance (in m) of the foot of the tower from the point A is :
A
$$15\left( {1 + \sqrt 3 } \right)$$
B
$$15\left( {3 - \sqrt 3 } \right)$$
C
$$15\left( {3 + \sqrt 3 } \right)$$
D
$$15\left( {5 - \sqrt 3 } \right)$$
2
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
Out of Syllabus
ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot–1 (3$$\sqrt 2$$ ) and cosec–1 (2$$\sqrt 2$$ ) respectively, then the height of the tower (in metres) is :
A
$${{100} \over {3\sqrt 3 }}$$
B
25
C
20
D
10$$\sqrt 5$$
3
JEE Main 2019 (Online) 9th April Evening Slot
+4
-1
Out of Syllabus
Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15º with ground. Then the distance (in m) between the poles, is :-
A
$$5\left( {2 + \sqrt 3 } \right)$$
B
$${5 \over 2}\left( {2 + \sqrt 3 } \right)$$
C
$$10\left( {\sqrt3 - 1 } \right)$$
D
$$5\left( {\sqrt3 + 1 } \right)$$
4
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
Out of Syllabus
Two vertical poles of heights, 20 m and 80 m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :
A
12
B
16
C
15
D
18
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