This chapter is currently out of syllabus
1
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
Out of Syllabus
The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30°. If the angle of depression of the image of C in the lake from the point P is 60°,then PC (in m) is equal to :
A
$$200\sqrt 3$$
B
400
C
100
D
$$400\sqrt 3$$
2
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
$$15\left( {5 - \sqrt 3 } \right)$$
3
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$$\sqrt 5$$
4
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
$$5\left( {\sqrt3 + 1 } \right)$$
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