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CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 12 - MCQExams.com
CBSE
Class 10 Maths
Some Applications Of Trigonometry
Quiz 12
A
,
B
,
C
are three collinear points on the ground such that
B
lies between
A
and
C
and
A
B
=
10
m
.
If the angles of elevation of the top of a vertical tower standing at
C
are respectively
30
∘
and
60
∘
as seen from
A
and
B
,
then the height of the tower is:
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0%
5
√
3
m
0%
5
m
0%
10
√
3
3
m
0%
20
√
3
3
m
The angle of elevation of the top of a tower from a point A on the ground is
30
0
. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to
60
0
. The height of the tower is :
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0%
√
3
m
0%
5
√
3
m
0%
10
√
3
m
0%
20
√
3
m
Explanation
Let the height of the tower be 'h' m.
In right
△
D
C
B
tan
60
=
D
C
C
B
⇒
√
3
=
h
x
∴
x
=
h
√
3
In right
△
D
C
A
tan
30
=
D
C
C
A
⇒
1
√
3
=
h
x
+
20
⇒
h
√
3
+
20
=
√
3
h
∴
h
=
10
√
3
Hence the height of the tower is
10
√
3
m
.
The angles of elevation of the top of a tower from two points A and B lying on the horizontal plane through the foot of the tower are respectively
45
0
and
30
0
. If A and B are on the same side of the tower and
A
B
=
48
m
e
t
r
e
, then the height of the tower is:
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0%
24
(
√
3
+
1
)
metre
0%
24 metre
0%
14
√
2
metre
0%
96 metre
Explanation
Let
Q
A
=
x
m
Q
B
=
A
B
+
Q
A
=
(
48
+
x
)
m
In
\Delta PQA,\\ \angle PQA=90°,\angle PAQ=45°\\ \tan 45°=\cfrac{PQ}{QA}=\cfrac{PQ}{x}\\ \therefore PQ=xm\rightarrow (1)
In
\Delta PQB, \\ \angle PQB=90°,\angle PBQ=30°\\ \tan 30°=\cfrac{PQ}{QB}=\cfrac{PQ}{(48+x)}\\ \Rightarrow PQ=\cfrac{(48+x)}{\sqrt3}m\rightarrow(2)
From
(1)\&(2)
x=\cfrac{(48+x)}{\sqrt3}\Rightarrow x=\cfrac{48}{\sqrt3-1}m\\ \therefore\quad height\quad of\quad tower (PQ)=x=\cfrac{48}{\sqrt3-1}=24(\sqrt3+1)m
Option A is correct
At a point A, the angle of elevation of a tower of a lower is such that its tangent is
\dfrac{5}{{12}};
on walking 120 meters nearer the tower the tangent of the angle of elevation is
\dfrac{3}{4}.
The height of the tower is :
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0%
225 meters
0%
200 meters
0%
230 metres
0%
None of these
Explanation
Here AB is the tower, it is given that
\tan\phi =\dfrac{5}{12}
……….(i)
On walling
240
m near to tower, it reaches point C, then
\tan\theta =\dfrac{3}{4}
………(ii)
Let
BC=x
m,
AB=y
m
Now in
\Delta
ABC,
\tan\theta =\dfrac{y}{x}=\dfrac{3}{4}
\Rightarrow x=\dfrac{4}{3}y
Now
\tan\theta =\dfrac{y}{240+x}=\dfrac{5}{12}
=\dfrac{y}{240+\dfrac{4}{3}y}=\dfrac{5}{12}
\Rightarrow 12y=1200+\dfrac{20}{3}y
\Rightarrow 36y-20y=3600
\Rightarrow 16y=3600
\Rightarrow y=225
m.
A man is walking towards a vertical pillar in a straight path, at uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is
30^o
. 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
60^o
. Then the time taken(in minutes) by him, from B to reach the pillar, is?
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0%
6
0%
10
0%
20
0%
5
Explanation
R.E.F image
let
AB = x,QB = y,PQ = z
In
\triangle PAQ
\tan 30^{\circ} = \dfrac{z}{x+y}
\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{z}{x+y}
\Rightarrow z = \dfrac{x+y}{\sqrt{3}}
In
\triangle PBQ
\tan 60^{\circ} = \dfrac{z}{y}
\Rightarrow z = \sqrt{3y}
\therefore \dfrac{x+y}{\sqrt{3}} = \sqrt{3y}
\Rightarrow x+y = 3y
\Rightarrow x = 3y-y
\Rightarrow x = 2y \Rightarrow y = \dfrac{x}{2}
So to go with x, it takes around 10 minutes.
with y, it takes around 5 minutes.
To cross a river a person covers a straight forward distance of 325 m along a bridge over the river. If bridge subtends 30
^{o}
angle with edge of the river, find the width of the river?
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0%
152.5
0%
155
0%
165
0%
162.5
In a storm, a tree got broken by the wind whose top meets the ground at an angle of 30
^{o}
, at a distance of 30 meters from the root. What was the height of the tree before breaking?
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0%
30\ m
0%
30\sqrt{3}\ m
0%
60\ m
0%
60\sqrt{3}\ m
Explanation
Based on the given information, we can draw the figure shown above.
Here, line segment
AB
represents the height of the tree.
The tree breaks at point
D
.
\therefore
Segment
AD
is the broken part of tree which then takes the position of
DC
.
\therefore AD = DC
Given
m\angle DCB = 30^\circ, BC=30\:m
In right angled
\Delta DBC
,
tan\:30^\circ=\dfrac{BD}{BC}
\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{BD}{30}
\Rightarrow BD=\dfrac{30}{\sqrt {3}}
\therefore BD=10\sqrt{3}\:m
Also
cos\:30^\circ=\dfrac{BC}{DC}
\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{30}{DC}
\Rightarrow DC=\dfrac{60}{\sqrt {3}}
\therefore DC=20\sqrt{3}\:m
Hence
AD = DC = 20\sqrt{3}\: m
From the figure,
AB = AD + DB
\therefore \ AB = 20\sqrt{3} + 10\sqrt{3}
\Rightarrow AB = 30\sqrt{3}\: m
Hence, the height of tree is
30\sqrt{3}\: m
.
The angles of elevation of the top of a tower from two points A and B lying on the horizontal through the foot of the tower are respectively
15^{\circ}
and
30^{\circ}
. If A and B are on the same side of the tower and AB = 48 m, Then the height of the tower is:
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0%
24\sqrt{3}
m
0%
24 m
0%
24\sqrt{2}
m
0%
96 m
Explanation
AB
is the tower
In
\triangle{ABC}, \tan{{15}^{\circ}}=\dfrac{AB}{AC}
\Rightarrow 2-\sqrt{3}=\dfrac{h}{d+48}
.....
\left(1\right)
\Rightarrow \tan{{30}^{\circ}}=\dfrac{AB}{BD}=\dfrac{h}{d}
\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{h}{d}
or
h=\dfrac{d}{\sqrt{3}}
Eqn
\left(1\right)\Rightarrow 2-\sqrt{3}=\dfrac{\dfrac{d}{\sqrt{3}}}{d+48}
\Rightarrow 2-\sqrt{3}=\dfrac{d}{\sqrt{3}\left(d+48\right)}
\Rightarrow 2\sqrt{3}-3=\dfrac{d}{\left(d+48\right)}
\Rightarrow \left(2\sqrt{3}-3\right)\left(d+48\right)=d
\Rightarrow d\left(2\sqrt{3}-3\right)+48\left(2\sqrt{3}-3\right)=d
\Rightarrow d\left(2\sqrt{3}-3-1\right)=-48\left(2\sqrt{3}-3\right)
\Rightarrow d\left(2\sqrt{3}-4\right)=48\left(3-2\sqrt{3}\right)
\Rightarrow 2d\left(1\sqrt{3}-2\right)=48\left(3-2\sqrt{3}\right)
\Rightarrow d=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(\sqrt{3}-2\right)}
We have
h=\dfrac{d}{\sqrt{3}}
\Rightarrow \sqrt{3}h=d
\Rightarrow \sqrt{3}h=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(\sqrt{3}-2\right)}
where
d=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(1\sqrt{3}-2\right)}
from above
\Rightarrow h=\dfrac{24\left(3-2\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}-2\right)}
\Rightarrow h=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(3-2\sqrt{3}\right)}
\therefore h=24
m
The shadow of a tree is
17\sqrt {3}\ \text{m}
. If the height of the tree is
17\ \text{m}
, then the sun's altitude is :
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0%
30^{\circ}
0%
45^{\circ}
0%
60^{\circ}
0%
90^{\circ}
Explanation
Length of the tree
= 17\sqrt{3}\ \text{m}
Height of the tree
= 17 \ \text{m}
then , the sun's altitude is
\tan \theta = \dfrac{\text{Height of tree}}{\text{Length of shadow}}
\tan\theta = \dfrac{17}{17\sqrt{3}}
\tan \theta = \dfrac{1}{\sqrt{3}}
\theta = 30^{\circ}
A tower substends an angle of
30^{o}
at a point on the same level as the foot of the tower. At a second point,
h
meter above first, point the depression of the foot of the tower is
60^{o}
, the horizontal distance of the tower from the point is _________________.
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0%
h\ \cos{60^{o}}
0%
(h/3)
\cot{30^{o}}
0%
(h/3)\ \cot{60^{o}}
0%
h
\cot{30^{o}}
On the level ground, the angle of elevation of the top of angle nearer to it the angle of elevation becomes
60 ^ { \circ } .
The height is _________________.
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0%
10 m
0%
15 m
0%
20 m
0%
none
If a tower
30\ m
high, casts a shadow
10\sqrt{3}\ m
long on the ground, then what is the angle of elevation of the sun?
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0%
60^{o}
0%
30^{o}
0%
45^{o}
0%
90^{o}
Explanation
Given, Length = 30 m (Tower)
Base =
10 \sqrt 3
m (Shadow)
We have to find the angle, lets take that A
So,
tan A =
\frac{Perpendicular}{Base}
tan A =
\frac{30}{10 \sqrt 3}
tan A =
\frac{3}{ \sqrt 3}
tan A =
\sqrt 3
So, tan A = tan
60^o
So, A =
60^o
An airplane flying horizontally at a height of
2500\sqrt{3} \text{ m}
above that ground, is observed to be at an angle of elevation
60^\circ
from the ground. After a flight of
25 \text{ sec}
the angle of elevation is
30^\circ
. Find the speed of the plane in
\text{ m/sec}
.
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0%
100\text{ m/sec}
0%
200\text{ m/sec}
0%
300\text{ m/sec}
0%
400\text{ m/sec}
Explanation
Let
P
and
Q
be the two positions of the airplane.
QB=2500\sqrt{3} \text{ m}
Let the speed of the aeroplane be
x \text{ m/sec}
.
In
\triangle ABQ
,
\begin{aligned}{}\tan {60^\circ} &= \frac{{QB}}{{AB}}\\\sqrt 3 & = \frac{{2500\sqrt 3 }}{x}\\x &= 2500\text{ m}\end{aligned}
In
\triangle APC
,
\begin{aligned}{}\tan {30^\circ} &= \frac{{2500\sqrt 3 }}{{x + y}}\\\frac{1}{{\sqrt 3 }} &= \frac{{2500\sqrt 3 }}{{x + y}}\\x + y &= 3 \times 2500\\2500 + y& = 7500\\y &= 5000\text{ m}\end{aligned}
To travel
5000 \text{ m}
the aeroplane takes $$25
\text{ sec}
$$.
So, the speed of the plane will be,
\dfrac{5000}{25}=200\text{ m/sec}
A chinny of
20\ mt
height, standing on the top of a building subtends an angle whose tangent is
\dfrac{1}{6}
at a distance of
70\ mt
from the foot of the building. The height of the building is
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0%
50\ mt
0%
100\ mt
0%
13.749\ mt
0%
\dfrac{80}{\sqrt{3}}mt
The angles of elevation of the top of a mountain from two points A and B on a horizontal line are
{15}^{0}
and
{75}^{0}
.If AB=650 mt,then the height of the mountain is
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0%
\cfrac {325 \sqrt {3}} {3} mt
0%
\cfrac {325 \sqrt {3}} {2} mt
0%
\cfrac {324 \sqrt {3}} {3} mt
0%
\cfrac {324 \sqrt {3}} {2} mt
The angles of elevation of the top of a building from the top and bottom of a tree are
x
and
y
respectively. If the height of the tree is
h
meter then, in meter, the height of the building is:
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0%
\dfrac{h cot x}{cot x+ cot y}
0%
\dfrac{h cot y}{cot x+ cot y}
0%
\dfrac{h cot x}{cot x - cot y}
0%
\dfrac{h cot y}{cot x- cot y}
Explanation
In
\triangle AED
\tan x =\dfrac {DE}{AE}\quad \therefore \ \tan x =\dfrac {H-h}{AE}
------- (1)
AE=\dfrac {H-h}{\tan x}
------(A)
In
\triangle DBC
\tan y=\dfrac {DC}{BC}\quad \therefore \ \tan y =\dfrac {H}{AE}
------- (2)
AE=\dfrac {H}{\tan y}
------ (B)
Comparing
A
and
B
\dfrac {H-h}{\tan x}=\dfrac {H}{\tan y}
(H-h)\tan y=H.\tan x
H\ \tan y-H\ \tan x=h\ \tan y
H=\dfrac { h\tan { y } }{ \tan { y } -\tan { x } }
=\dfrac { \dfrac { h }{ \cot { y } } }{ \dfrac { 1 }{ \cot { y } } -\dfrac { 1 }{ \cot { x } } }
=\dfrac { \dfrac { \dfrac { h }{ \cot { y } } }{ \cot { x } -\cot { y } } }{ \cot { x } .\cot { y } }
=\dfrac { h\cot { x } }{ \cot { x } -\cot { y } }
The shadow of the tower becomes 60 meters longer when the altitude of the sun changes from
45^{\circ} \ \text to\ 30^{\circ}
. Then the height of the tower is:
Report Question
0%
20(\sqrt{3}+1)m
0%
24(\sqrt{3}+1)m
0%
30(\sqrt{3}+1)m
0%
30(\sqrt{3}-1)m
Explanation
In
\Delta TOW
,
\tan 45^o = \dfrac{h}{x}
\therefore x = h
In
\Delta TXW
,
\tan 30^o = \dfrac{h}{60 + x}
\dfrac{1}{\sqrt{3}} = \dfrac{h}{60 + h}
60 + h = \sqrt{3} h
60 = h (\sqrt{3} - 1)
h = \dfrac{60}{\sqrt{3} - 1} = \dfrac{60 (\sqrt{3} + 1)}{3 - 1} = 30 (\sqrt{3} + 1) m
The upper part of a tree broken over by the wind makes an angle of
30^{o}
with the ground, and the distance from the
root to the point where the top of the tree meets the ground is
15\ m
. The present height of the tree is
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0%
15\ m
0%
10\sqrt{3}\ m
0%
20\ m
0%
None\ of\ these
Explanation
Height of the tree is =x
\begin{array}{l} \tan \theta =\dfrac { P }{ B } \\ \tan { 30^{ 0 } } =\dfrac { x }{ { 15 } } \\ \therefore x=15\times \dfrac { 1 }{ { \sqrt { 3 } } } \\ =8.66 \end{array}
A ladder is inclined to a well at an angle
\alpha
to the horizontal. Its foot is drawn up to a distance and as such, it slides a distance
b
along the well and makes angle
\beta
with the horizontal, then
a
is equal to:
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0%
\frac{b}{2}\tan (\alpha + \beta )
0%
b\tan (\alpha + \beta )
0%
b\tan (\frac{{\alpha + \beta }}{2})
0%
\frac{b}{2}\tan (\frac{{\alpha + \beta }}{2})
Explanation
R.E.F image
tan\alpha =\frac{a}{\sqrt{l^{2}-a^{2}}}
tan\beta =\frac{b+a}{\sqrt{l^{2}-(b+a)^{2}}}
on solving,
a = b tan
(\alpha +\beta )
A flagstaff
5
m high is placed on a building
25
m high. If the flag and building be subtend equal angels on the observer at a height
30
m, then the distance between observer and the top of flag is
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0%
\dfrac {5\sqrt {3}}{2}
0%
5\sqrt {\dfrac {2}{3}}
0%
5\sqrt {\dfrac {3}{2}}
0%
\dfrac {5\sqrt {2}}{3}
Explanation
Given,
∠DAE = ∠ DAC = \theta
(Let's assume)
Thus,
∠EAC = 2\theta
Let's assume the distance between the building and the observer is
'x' m
.
Thus, for the triangle
ΔACE
;
EC = ED + DC = 5+25 = 30m
Opposite/adjacent
= \dfrac{EC}{EA} = \dfrac{30}{x} = tan2\theta
For the
ΔEDA;
Opposite/adjacent
= \dfrac{ED}{EA} = \dfrac{5}{x} = tan\theta
Now,
tan2\theta = \dfrac{2tan\theta }{ (1-tan^2\theta)}
or,
\dfrac{30}{x} = \dfrac{2(\dfrac{5}{x}) }{ (1 - \dfrac{25}{x^2})}
[As
tan\theta = \dfrac{5}{x}
and
tan2\theta = \dfrac{30}{x}]
or,
3 =\dfrac{ 1 }{ (1-\dfrac{25}{x^2}}
or,
3x^2 - 75 = x^2
or,
2x^2 = 75
or,
x = 5\sqrt{(\dfrac{3}{2})} = 6.124m
Thus the distance between the building and the observer is
5\sqrt{(\dfrac{3}{2})} m
The length of the shadow of a vertical tower on level ground increase by 10 meters when the altitude of the sun changes from
45^o
to
30^o
. Then the height of the tower is:
Report Question
0%
5\sqrt{3}
0%
10(\sqrt{3} + 1) meter
0%
5(\sqrt{3} + 1)
meter
0%
10\sqrt{3} meter
Explanation
Let\text{ the distance when the sun elevation is 45 be x}
\text{whereas when sun elevation is 30 the distance is 10+x}
\text{let h be the height of the tower}
\text{when the sun is at elevation of 45}
\text{tan45=h/x}
\text{1=h/x}
\text{h=x}\,\,\,\,\,\,\,.........\left( i \right)
when\,the\,sun\,elevation\,is\,{{30}^{0}}
\tan {{30}^{0}}=h/\left( x+10 \right)
\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+10}
\Rightarrow h=\frac{x+10}{\sqrt{3}}\,\,\,\,\,\,......\left( ii \right)
from\,\left( i \right)and\,\left( ii \right)
x=\frac{x+10}{\sqrt{3}}
\Rightarrow \sqrt{3}x=x+10
\Rightarrow x\left( \sqrt{3}-1 \right)=10
\Rightarrow x=\frac{10}{\sqrt{3}-1}
\,\,\,\,\,\,\,\,\,\,\,=\frac{10}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}
\,\,\,\,\,\,\,\,\,\,=5\left( \sqrt{3}+1 \right)
hence\,height=5\left( \sqrt{3}+1 \right)m
A tower is
50\sqrt{3}
m high. The angle of elevation of its top from a point 50 m away from its foot has measure............degree
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0%
65
0%
60
0%
30
0%
15
A ladder leans against a wall making an angle of measure
60^\circ
with the ground. If the foot of the ladder is
3
metre away from the wall, then the length of the ladder is ..... metre.
Report Question
0%
\dfrac{3}{2}
0%
3\sqrt{3}
0%
6
0%
\dfrac{3\sqrt{3}}{2}
Explanation
Given distance between the foot ofthe ladder and wall is
3\ m
Let the length of ladder be
l
\implies \cos 60^{\circ}=\dfrac{3}{l}
\implies l=3\text{sec} 60^{\circ}
\implies l=6
So the length of the ladder is
6\ m
The tops of two towers of height x and y standing on a level round subtends angle of 30 and 60 respectively at the line joining their feet then x:y is.
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0%
1 : 2
0%
2 : 1
0%
1 : 3
0%
3 : 1
Explanation
\tan 30^o=\dfrac{x}{a}
\tan 60^o=\dfrac{y}{a}
x=a\tan 30^o
y=a\tan 60^o
\dfrac{x}{y}=\dfrac{\tan 30^o}{\tan 60^o}=\dfrac{1}{3}
x : y =1 : 3
.
The measure of the angle of elevation of sun increases from
30\ to\ 45
, then the length of the shadow of a
100\ meter
high building reduces by
x\ meter
. Then
x
is..... meter.
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0%
100
0%
100\sqrt {3}
0%
100(\sqrt {3}-1)
0%
\dfrac {100}{\sqrt {3}}
A balloon is connected to a meteorological ground station by a cable of length 215 m inclined at
60^o
to the horizontal. Determine the height of the balloon from the ground. Assume that there is no stack in the cable.
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0%
175 m
0%
186 m
0%
190 m
0%
204 m
Explanation
Let
A
be the position of the balloon.
In right triangle
ABC,
\sin 60^o=\dfrac{AB}{AC}
AB=215 \sin 60^o
AB=215 \times \dfrac{\sqrt 3}{2}
AB =186 \ m
The upper
\dfrac 34 ^{th}
portion of a vertical pole subtends an angle
\tan^{-1} \bigg(\dfrac 35\bigg)
at a point in the horizontal plane through its foot and at a distance
40\ m
from the foot. A possible height of the vertical pole is:
Report Question
0%
40 m
0%
60 m
0%
80 m
0%
20 m
Explanation
Here,
\alpha =A- \beta
\therefore \beta = A - \alpha
Hence
\tan \beta = \dfrac{{\tan A + \tan \alpha }}{{1 - \tan A\tan \alpha }}
\Rightarrow \dfrac{3}{5} = \dfrac{{\dfrac{h}{{40}} + \left(- {\dfrac{{ h}}{{160}}} \right)}}{{1 - \left( {\dfrac{h}{{40}}} \right)\left( -{\dfrac{{ h}}{{160}}} \right)}}
\Rightarrow {h^2} - 200h + 6400=0
\Rightarrow \left( {h - 40} \right)\left( {h - 160} \right) = 0
\Rightarrow h = 40\,\,or\,\,h = 160
A 80 m long ladder is leaning on a wall. If the ladder makes an angle of
45^o
with the ground, find the distance of the ladder from the wall.
Report Question
0%
40\sqrt2 \ m
0%
20 \sqrt 3 \ m
0%
50 \sqrt 3 \ m
0%
None of these
Explanation
Here,
\cos \theta = \dfrac{Base}{Hypotenuse}
\cos 45^o=\dfrac{Base}{80}
Base = 80 \cos 45^o=\dfrac{80}{\sqrt 2}=40\sqrt 2
Therefore, Distance of the ladder from the wall
=40\sqrt 2 \text{ m}
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 to change from
30^{\circ}
to
45^{\circ}
: then after this , the time taken (in min) by the car to reach the foot of the tower, is
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0%
9(1+\sqrt{3})
0%
18(1+\sqrt{3})
0%
27(1+\sqrt{3})
0%
18(1+\sqrt{9})
A flag staff of height
'h'
stands in the centre of a rectangular field and subtends angle of
15^{o}
and
45^{o}
at the mid point of its sides. Find the distance between them
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0%
h\sqrt{2+\sqrt{3}}
0%
2h\sqrt{2+\sqrt{3}}
0%
4h\sqrt{2+\sqrt{3}}
0%
3h\sqrt{2+\sqrt{3}}
Explanation
\begin{array}{l} Q\tan { 15^{ \circ } } =\dfrac { h }{ { OQ } } \\ h\cot { 15^{ \circ } } =OQ \\ \dfrac { h }{ { 2-\sqrt { 3 } } } \, \, \, \left[ { \tan { { 15 }^{ \circ } } =2-\sqrt { 3 } } \right] \\ OP=\dfrac { h }{ { \tan { { 45 }^{ \circ } } } } =h \\ PQ=\sqrt { O{ P^{ 2 } }+O{ Q^{ 2 } } } \\ =h\sqrt { { { \left( { \dfrac { 1 }{ { 2-\sqrt { 3 } } } } \right) }^{ 2 } }+1 } \\ l\sqrt { { { \left( { \sqrt { 3 } +2 } \right) }^{ 2 } }+1 } \\ l\sqrt { 3+4+4\sqrt { 3 } +1 } \\ h\sqrt { 8+4\sqrt { 3 } } \\ 2h\sqrt { \sqrt { 3 } +2 } \\ 2h\sqrt { 2+\sqrt { 3 } } \\ Hence,\, option\, \, A\, is\, the\, \, correct\, answer. \end{array}
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Practice Class 10 Maths Quiz Questions and Answers
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