iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 11th April Evening Shift
The angle of elevation of the top $\mathrm{P}$ of a tower from the feet of one person standing due South of the tower is $45^{\circ}$ and from the feet of another person standing due west of the tower is $30^{\circ}$. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to
A.
10
B.
$\frac{5}{2} \sqrt{5}$
C.
$5 \sqrt{5}$
D.
5
Correct Answer: A
Explanation:
Let's denote the person standing due south as S and the one standing due west as W. Also, let the tower be at point T.
From person S's perspective, we have a right triangle $\triangle SPT$. The height of the tower PT is given as 5m, which is the opposite side for angle S. The angle at S is $45^\circ$. From the tangent trigonometric ratio, we have :
$
\tan{45^\circ} = \frac{PT}{ST} = \frac{5}{ST}
$
Since $\tan{45^\circ}=1$, we get $ST = 5$m.
Similarly, from person W's perspective, we have a right triangle $\triangle WPT$. The angle at W is $30^\circ$. From the tangent trigonometric ratio, we have:
$\tan{30^\circ} = \frac{PT}{WT} = \frac{5}{WT}$
Since $\tan{30^\circ}=\frac{1}{\sqrt{3}}$, we get $WT = 5\sqrt{3}$m.
Since S and W are perpendicular to each other (one is due south and the other is due west), $\triangle SWT$ forms a right triangle.
We can find $SW$ (the distance between the two people) using the Pythagorean theorem:
Hence, the correct answer is 10 meters, which corresponds to Option A.
2023
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 6th April Morning Shift
From the top $\mathrm{A}$ of a vertical wall $\mathrm{AB}$ of height $30 \mathrm{~m}$, the angles of depression of the top $\mathrm{P}$ and bottom $\mathrm{Q}$ of a vertical tower $\mathrm{PQ}$ are $15^{\circ}$ and $60^{\circ}$ respectively, $\mathrm{B}$ and $\mathrm{Q}$ are on the same horizontal level. If $\mathrm{C}$ is a point on $\mathrm{AB}$ such that $\mathrm{CB}=\mathrm{PQ}$, then the area (in $\mathrm{m}^{2}$ ) of the quadrilateral $\mathrm{BCPQ}$ is equal to :
A.
$200(3-\sqrt{3})$
B.
$300(\sqrt{3}-1)$
C.
$300(\sqrt{3}+1)$
D.
$600(\sqrt{3}-1)$
Correct Answer: D
Explanation:
Given, $A B$ be a vertical wall of height $30 \mathrm{~m}$ and $P Q$ be a vertical tower.
Such that $\angle B Q A=\angle T A Q=60^{\circ}$ (Alternate angles)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 29th July Morning Shift
The angle of elevation of the top of a tower from a point A due north of it is $\alpha$ and from a point B at a distance of 9 units due west of A is $\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point B from the tower is 15 units, then $\cot \alpha$ is equal to :
A.
$\frac{6}{5}$
B.
$\frac{9}{5}$
C.
$\frac{4}{3}$
D.
$\frac{7}{3}$
Correct Answer: A
Explanation:
$NA = \sqrt {{{15}^2} - {9^2}} = 12$
${h \over {15}} = \tan \theta = {2 \over 3}$
$h = 10$ units
$\cot \alpha = {{12} \over {10}} = {6 \over 5}$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th July Evening Shift
A horizontal park is in the shape of a triangle $\mathrm{OAB}$ with $\mathrm{AB}=16$. A vertical lamp post $\mathrm{OP}$ is erected at the point $\mathrm{O}$ such that $\angle \mathrm{PAO}=\angle \mathrm{PBO}=15^{\circ}$ and $\angle \mathrm{PCO}=45^{\circ}$, where $\mathrm{C}$ is the midpoint of $\mathrm{AB}$. Then $(\mathrm{OP})^{2}$ is equal to :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 27th July Evening Shift
The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $45^{\circ}$. Let R be a point on AQ and from a point B, vertically above $\mathrm{R}$, the angle of elevation of $\mathrm{P}$ is $60^{\circ}$. If $\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d}$ and the area of the trapezium $\mathrm{PQRB}$ is $\alpha$, then the ordered pair $(\mathrm{d}, \alpha)$ is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 27th July Morning Shift
Let a vertical tower $A B$ of height $2 h$ stands on a horizontal ground. Let from a point $P%$ on the ground a man can see upto height $h$ of the tower with an angle of elevation $2 \alpha$. When from $P$, he moves a distance $d$ in the direction of $\overrightarrow{A P}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d=\sqrt{7} h$, then $\tan \alpha$ is equal to
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 25th July Morning Shift
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 29th June Evening Shift
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th June Morning Shift
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.
${{3 - \sqrt 2 } \over 4}$
Correct Answer: C
Explanation:
${l \over {80}} = \tan \theta $ ..... (i)
${{2l} \over {80}} = \tan {\pi \over 8}$ ..... (ii)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Morning Shift
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 :
use $\tan \alpha = {l \over 6} \Rightarrow l = \sqrt {{{72} \over 5}} $
height of pole = $10l = 12\sqrt {10} $
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th August Evening Shift
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Evening Shift
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th July Morning Shift
A spherical gas balloon of radius 16 meter subtends an angle 60$^\circ$ at the eye of the observer A while the angle of elevation of its center from the eye of A is 75$^\circ$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :
A.
$8(2 + 2\sqrt 3 + \sqrt 2 )$
B.
$8(\sqrt 6 + \sqrt 2 + 2)$
C.
$8(\sqrt 2 + 2 + \sqrt 3 )$
D.
$8(\sqrt 6 - \sqrt 2 + 2)$
Correct Answer: B
Explanation:
O $\to$ centre of sphere
P, Q $\to$ point of contact of tangents from A
Let T be top most point of balloon & R be foot of perpendicular from O to ground.
From triangle OAP, OA = 16cosec30$^\circ$ = 32
From triangle ABO, OR = OA sin75$^\circ$ = $32{{\left( {\sqrt 3 + 1} \right)} \over {2\sqrt 2 }}$
So level of top most point = OR + OT
$ = 8\left( {\sqrt 6 + \sqrt 2 + 2} \right)$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 20th July Evening Shift
Let in a right angled triangle, the smallest angle be $\theta$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sin$\theta$ is equal to :
A.
${{\sqrt 5 + 1} \over 4}$
B.
${{\sqrt 5 - 1} \over 2}$
C.
${{\sqrt 2 - 1} \over 2}$
D.
${{\sqrt 5 - 1} \over 4}$
Correct Answer: B
Explanation:
Let a $\Delta$ABC having C = 90$^\circ$ and A = $\theta$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 18th March Evening Shift
A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be ${\pi \over 3}$. If the radius of the circumcircle of $\Delta$ABC is 2, then the height of the pole is equal to :
A.
${{1 \over {\sqrt 3 }}}$
B.
2${\sqrt 3 }$
C.
${\sqrt 3 }$
D.
${{{2\sqrt 3 } \over 3}}$
Correct Answer: B
Explanation:
Let PD = h, R = 2
As angle of elevation of top of pole from A, B, C are equal. So D must be circumcentre of $\Delta$ABC
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th February Morning Shift
A man is observing, from the top of a tower, a boat speeding towards the lower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30$^\circ$ (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is 45$^\circ$. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 24th February Evening Shift
The angle of elevation of a jet plane from a point A on the ground is 60$^\circ$. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30$^\circ$. If the jet plane is flying at a constant height, then its height is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 24th February Morning Shift
Two vertical poles are 150 m apart and the height of one is three times that of the other. If
from the middle point of the line joining their feet, an observer finds the angles of elevation of
their tops to be complementary, then the height of the shorter pole (in meters) is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Evening Slot
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 4th September Evening Slot
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th April Evening Slot
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th April Morning Slot
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 $
Correct Answer: C
Explanation:
$\Delta $APM
${h \over {AM}} = {1 \over {3\sqrt 2 }}$
$\Delta $BPM
${h \over {BM}} = {1 \over {\sqrt 7 }}$
$\Delta $ABM
AM2 + MB2 = (100)2
'
$ \Rightarrow $ 18h2 + 7h2 = 100 × 100
$ \Rightarrow $ h2 = 4 × 100
$ \Rightarrow $ h = 20
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th April Evening Slot
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 :-
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Evening Slot
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 :
$ \therefore $ Height of intersection point is 16 m
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th January Evening Slot
If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30o and the angle of depression of reflection of the cloud in the lake from P be 60o, then the height of the cloud (in meters) from
the surface of the lake is :
A.
45
B.
42
C.
50
D.
60
Correct Answer: C
Explanation:
tan 30o = ${x \over y} \Rightarrow y = \sqrt 3 x\,\,\,\,....(i)$
tan 60o = ${{25 + x + 25} \over y}$
$ \Rightarrow \,\,\sqrt 3 y = 50 + x$
$ \Rightarrow $ $3x = 50 + x$
$ \Rightarrow $ x = 25 m
$ \therefore $ Height of cloud from surface
= 25 + 25 = 50m
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Morning Slot
Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30o at B. The height (in m) of the lamp-post is -
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30o to 45o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles
of elevation of the top of the tower at P, Q and R are respectively 45$^\circ $, 30$^\circ $ and 30$^\circ $, then the height of the tower (in m) is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Evening Slot
A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. Fromthe top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feetof the towers T1 and T2 is :
A.
$10\sqrt 2 $
B.
$10\sqrt 3 $
C.
$20\sqrt 3 $
D.
$20\sqrt 2 $
Correct Answer: C
Explanation:
Let the distance between T1 and T2 be x
From the figure
EA = 60 m (T1) and DB = 80 m (T2)
$\angle DEC = \theta $ and $\angle BEC = 2\theta $
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Morning Slot
An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt 3 $ kmabove it, is obsered at an elevation of ${60^o}$ from a point on the ground. If, after five seconds, its elevation from the same point, is ${30^o}$, then the speed (in km / hr) of the aeroplane, is :
A.
1500
B.
1440
C.
750
D.
720
Correct Answer: B
Explanation:
For $\Delta $OA, A, OA1 = ${{\sqrt 3 } \over {\tan {{60}^o}}}$ = 1 km
As, a distance of 3 $-$ 1 = 2 km is convered in 5 seconds.
Therefore the speed of the plane is
${{2 \times 3600} \over 5}$ = 1440 km/hr
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point
on the ground such that AP = 2AB. If $\angle $BPC = $\beta $, then tan$\beta $ is equal to:
A.
${1 \over 4}$
B.
${2 \over 9}$
C.
${4 \over 9}$
D.
${6 \over 7}$
Correct Answer: B
Explanation:
Let the height of tower $AB = x$ and $LCPA = \propto $
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 10th April Morning Slot
The angle of elevation of the top of a vertical tower from a point A, due east of it is 45o. The angle of elevation of the top of the same tower from a point B, due south of A is 30o. If the distance between A and B is $54\sqrt 2 \,m,$ then the height of the tower (in metres), is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Offline)
A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30o. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60o. Then the time taken (in minutes) by him, from B to reach the pillar, is :
A.
6
B.
10
C.
20
D.
5
Correct Answer: D
Explanation:
According to given information, we have the following figure
Now, from $\triangle A C D$ and $\triangle B C D$, we have
$\tan 30^{\circ}=\frac{h}{x+y}$
and $\tan 60^{\circ}=\frac{h}{y}$
$\Rightarrow h=\frac{x+y}{\sqrt{3}}\quad...(i)$
and $h=\sqrt{3} y\quad...(ii)$
From Eqs. (i) and (ii),
$\frac{x+y}{\sqrt{3}}=\sqrt{3} y \Rightarrow x+y=3 y$
$\Rightarrow x-2 y=0 \Rightarrow y=\frac{x}{2}$
$\because$ Speed is uniform.
$\therefore$ Distance $y$ will be cover in $5 \mathrm{~min}$.
$\because$ Distance $x$ covered in $10 \mathrm{~min}$.
$\therefore$ Distance $\frac{x}{2}$ will be cover in $5 \mathrm{~min}$.
2015
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2015 (Offline)
If the angles of elevation of the top of a tower from three collinear points $A, B$ and $C,$ on a line leading to the foot of the tower, are ${30^ \circ }$, ${45^ \circ }$ and ${60^ \circ }$ respectively, then the ratio, $AB:BC,$ is :
A.
$1:\sqrt 3 $
B.
$2:3$
C.
$\sqrt 3 :1$
D.
$\sqrt 3 :\sqrt 2 $
Correct Answer: C
Explanation:
As $PB$ bisects $\angle APC,$ therefore $AB$ $:$ $BC$ $=PA:PC$
Also in $\Delta APQ,\sin {30^ \circ } = {h \over {PA}} \Rightarrow PA = 2h$
and in $\Delta CPQ,$ $\sin {60^ \circ } = {h \over {PC}} \Rightarrow PC = {{2h} \over {\sqrt 3 }}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2014 (Offline)
A bird is sitting on the top of a vertical pole $20$ m high and its elevation from a point $O$ on the ground is ${45^ \circ }$. It files off horizontally straight away from the point $O$. After one second, the elevation of the bird from $O$ is reduced to ${30^ \circ }$. Then the speed (in m/s) of the bird is :
A.
$20\sqrt 2 $
B.
$20\left( {\sqrt 3 - 1} \right)$
C.
$40\left( {\sqrt 2 - 1} \right)$
D.
$40\left( {\sqrt 3 - \sqrt 2 } \right)$
Correct Answer: B
Explanation:
Let the speed be $y$ $m/sec$.
Let $AC$ be the vertical pole of height $20$ $m.$
Let $O$ be the point on the ground such that $\angle AOC = {45^ \circ }$
Let $OC = x$
Time $t=1$ $s$
From $\Delta AOC,\,\,\tan {45^ \circ } = {{20} \over x}\,\,\,\,\,\,\,.....\left( i \right)$
and from $\Delta BOD,\,\,\tan {30^ \circ } = {{20} \over {x + y}}...\left( {ii} \right)$
$AB$ is a vertical pole with $B$ at the ground level and $A$ at the top. $A$ man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is ${60^ \circ }$. He moves away from the pole along the line $BC$ to a point $D$ such that $CD=7$ m. From $D$ the angle of elevation of the point $A$ is ${45^ \circ }$. Then the height of the pole is :
A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $AB(=a)$ subtends an angle of ${60^ \circ }$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is ${30^ \circ }$. The height of the tower is :
A.
$a/\sqrt 3 $
B.
$a\sqrt 3 $
C.
$2a/\sqrt 3 $
D.
$2a\sqrt 3 $
Correct Answer: A
Explanation:
In the $\Delta AOB,\,\,\angle AOB = {60^ \circ },$ and
$\angle OBA = \angle OAB$
(since $OA=OB=AB$ radius of same circle).
$\therefore$ $\Delta AOB$ is a equilateral triangle.
Let the height of tower is $h$
Given distance between two points $A$ & $B$ lie on boundary of
circular park, subtends an angle of ${60^ \circ }$ at the foot of the tower
is $AB$ i.e. $AB$$=a.$ A tower $OC$ stands at the center of a circular
park. Angle of elevation of the top of the tower from $A$ and $B$ is ${30^ \circ }$ .
In $\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$
A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is ${60^ \circ }$ and when he retires $40$ meters away from the tree the angle of elevation becomes ${30^ \circ }$. The breadth of the river is :
A.
$60\,\,m$
B.
$30\,\,m$
C.
$40\,\,m$
D.
$20\,\,m$
Correct Answer: D
Explanation:
From the figure
$\tan {60^ \circ } = {y \over x}$
$ \Rightarrow y = \sqrt {3x} .......\left( 1 \right)$
$\sqrt 3 x = {{x + 40} \over {\sqrt 3 }} \Rightarrow x = 20m$
2020
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Morning Slot
Let AD and BC be two vertical poles at A and B respectively on a horizontal ground. If
AD = 8 m, BC = 11 m and AB = 10 m; then the distance (in meters) of a point M on AB from the point
A such that MD2 + MC2 is minimum is ______.
Correct Answer: 5
Explanation:
(MD)2
+ (MC)2
= h2
+ 64 + (h – 10)2
+ 121
= 2h2
– 20h + 64 + 100 + 121
= 2(h2
– 10h) + 285
= 2(h – 5)2
+ 235
It is minimum if h = 5.
2020
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Morning Slot
The angle of elevation of the top of a hill from a point on the horizontal plane passing through the
foot of the hill is found to be 45o. After walking a distance of 80 meters towards the top, up a slope
inclined at an angle of 30o to the horizontal plane, the angle of elevation of the top of the hill
becomes 75o. Then the height of the hill (in meters) is ____________.
Correct Answer: 80
Explanation:
sin 30o = ${x \over {80}}$ $ \Rightarrow $ x = 40
cos 30o = ${y \over {80}}$ $ \Rightarrow $ y = $40\sqrt 3 $