This chapter is currently out of syllabus
1
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
The angle of elevation of the summit of a mountain from a point on the ground is 45°. After climbing up one km towards the summit at an inclination of 30° from the ground, the angle of elevation of the summit is found to be 60°. Then the height (in km) of the summit from the ground is :
A
$${1 \over {\sqrt 3 - 1}}$$
B
$${{\sqrt 3 + 1} \over {\sqrt 3 - 1}}$$
C
$${1 \over {\sqrt 3 + 1}}$$
D
$${{\sqrt 3 - 1} \over {\sqrt 3 + 1}}$$
2
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
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
$$400\sqrt 3 $$
3
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
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
$$15\left( {5 - \sqrt 3 } \right)$$
4
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
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$$\sqrt 5 $$
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