MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 1 - MCQExams.com
CBSE
Class 10 Maths
Some Applications Of Trigonometry
Quiz 1
Let $$10$$ vertical poles standing at equal distances on a straight line, subtend the same angle of elevation $$\alpha$$ at a point $$O$$ on this line and all the poles are on the same side of $$O$$. If the height of the longest pole is $$h$$ and the distance of the foot of the smallest pole from $$O$$ is $$a$$; then the distance between two consecutive poles, is
Report Question
0%
$$\displaystyle \frac{h \sin \alpha + a \cos \alpha}{9 \cos \alpha}$$
0%
$$\displaystyle \frac{h \cos \alpha - a \sin \alpha}{9 \sin \alpha}$$
0%
$$\displaystyle \frac{h \sin \alpha + a \cos \alpha}{9 \sin \alpha}$$
0%
$$\displaystyle \frac{h \cos \alpha - a \sin \alpha}{9 \cos \alpha}$$
Explanation
Since all $$10$$ poles are subtending equal angles at $$O$$. Let the distance between two consecutive poles $$=d$$.
Distance form $$O$$ to the smallest pole $$=a$$
Total base distance in right angled triangle $$=a+9d$$
$$\tan\alpha =\cfrac { h }{ a+9d } $$
$$\quad 9d\tan\alpha +a\tan\alpha =h$$
$$\quad d=\cfrac { h-a\tan\alpha }{ 9\tan\alpha } $$
$$\quad =\cfrac { h-a\dfrac{\sin\alpha }{\cos \alpha} }{ 9 \dfrac{\sin\alpha }{\cos \alpha} } $$
$$=\cfrac { h\cos\alpha -a\sin\alpha }{ 9\sin\alpha } $$
A kite is flying at an inclination of $$60^\circ$$ with the horizontal. If the length of the thread is $$120\text{ m},$$ then the height at which kite is:
Report Question
0%
$$60\sqrt 3\text{ m}$$
0%
$$60\text{ m}$$
0%
$$\dfrac{60}{\sqrt 3}\text{ m}$$
0%
$$120\text{ m}$$
Explanation
Let at point $$A$$ kite is flying at a height $$AB$$ equal to $$h.$$
In $$\triangle ABC,$$
$$\begin{aligned}{}\sin 60^\circ &= \frac{{AB}}{{AC}}\\\\\frac{{\sqrt 3 }}{2} &= \frac{h}{{120}}\\\\h& = \frac{{120 \times \sqrt 3 }}{2}\\\\ &= 60\sqrt 3 \text{ m}\end{aligned}$$
Hence, the kite is flying at the height of $$60\sqrt3\text{ m}.$$
A vertical pole of height $$10$$ meters stands at one corner of a rectangular field. The angle of elevation of its top from the farthest corner is $$30^{\circ}$$, while that from another corner is $$60^{\circ}$$. The area (in $$m^{2})$$ of rectangular field is
Report Question
0%
$$\dfrac {200\sqrt {2}}{3}$$
0%
$$\dfrac {400}{\sqrt {3}}$$
0%
$$\dfrac {200\sqrt {2}}{\sqrt {3}}$$
0%
$$\dfrac {400\sqrt {2}}{\sqrt {3}}$$
Explanation
$$\tan 30^{\circ} = \dfrac {10}{AC} = \dfrac {1}{\sqrt {3}}$$
$$\therefore\sqrt{l^2+b^2}= AC = 10\sqrt {3} \text{ cm}$$ -----------(i)
$$\tan 60^{\circ} = \dfrac {10}{DC} = \sqrt {3}$$
$$\therefore DC = \dfrac {10}{\sqrt3}=l$$
In traingle DCE
$$l^2=\dfrac{100}{3}$$ -----------(ii)
$$b^2 = {300 - \dfrac {100}{3}}$$
$$b^2 = {\dfrac {800}{3}}$$
$$l^2.b^2=\dfrac{100}{3}\times \dfrac{800}{3}=\dfrac{80000}{9}$$
$$Area = lb=\dfrac{\sqrt{80000}}{{\sqrt9}} = \dfrac {200\sqrt {2}}{3} \text{m}^{2}$$.
A pole $$15$$ m long rests against a vertical wall at an angle of $$\displaystyle 30^{\circ}$$ with the ground How high up the wall does the pole reach?
Report Question
0%
$$5$$ m
0%
$$7$$ m
0%
$$7.5$$ m
0%
$$8$$ m
Explanation
As shown in the fig. let $$AB$$ be the length of the pole.
$$\displaystyle \sin 30^{\circ}=\frac{x}{15}$$
Or
$$x=15\displaystyle \sin 30^{\circ}\\=\frac{15}{2}\\=7.5$$m
the shadow cast by a tower is 30 m long when the elevation of sum is $$\displaystyle 30^{\circ} $$ If the elevation of sum is $$\displaystyle 60^{\circ} $$ then the length of the shadow is
Report Question
0%
30 m
0%
20 m
0%
10 m
0%
$$\displaystyle 10\sqrt{3} m $$
Explanation
$$\displaystyle \tan 30^{\circ}=\frac{h}{30}$$ or $$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{30}$$
$$\displaystyle \therefore h=\frac{30}{\sqrt{3}}$$
$$\displaystyle \tan 60^{\circ}=\frac{30/\sqrt{3}}{x}$$
or $$\displaystyle \sqrt{3}=\frac{30}{x\sqrt{3}}$$
$$\displaystyle \therefore $$ x=10
The angle of depression of a boat from the top of a cliff 300 m high is $$\displaystyle 60^{\circ} $$ The distance of the boat from the foot of the cliff is
Report Question
0%
$$\displaystyle 100 \sqrt{3} $$
0%
100
0%
$$\displaystyle 300 \sqrt{3} $$
0%
300
If the length of the shadow of a pole is equal to the height , of the pole, then the angle of the elevation of the sun is
Report Question
0%
$$ \displaystyle 30^{\circ} $$
0%
$$ \displaystyle 75^{\circ} $$
0%
$$ \displaystyle 60^{\circ} $$
0%
$$ \displaystyle 45^{\circ} $$
Explanation
Let AB=h be the pole and let $$\displaystyle \angle ACB=\theta $$
Let AC be the shadow
Hence, by hypothesis, CA=h
From right angled $$\displaystyle \bigtriangleup BCA $$
$$\displaystyle \tan \theta =\frac{h}{h}=1 $$
$$\displaystyle \therefore \theta =45^{\circ} $$
The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal when it is
Report Question
0%
Above the horizontal level
0%
Below the horizontal level
0%
At the horizontal level
0%
None of the above
Explanation
In the figure, $$AB$$ is a vertical object with $$A$$ as foot and $$B$$ as top.
$$P$$ is the point of observation.
So, $$PA$$ is the line of horizontal sight and we have to raise our head if we look at $$B$$.
The angle, described in this case $$=\angle PAB$$ when $$PB$$ is the line of sight.
But by definition, $$\angle APB$$ is the angle of elevation.
Hence, option A is correct.
From the point B a perpendicular BD is drawn on AC. If $$\cos 30^o=0.8$$, find the length of AD.
Report Question
0%
$$45$$
0%
$$30$$
0%
$$50$$
0%
$$80$$
Explanation
In $$ \displaystyle \bigtriangleup ' S ABC and ABD , \angle A $$ is common and $$ \displaystyle \angle ABC = \angle ADB $$
Hence $$ \displaystyle \bigtriangleup ' ABC and \bigtriangleup ABD $$ are equiangular
$$ \displaystyle \therefore \frac{AD}{AB}= \frac{AB}{AC} $$ OR
$$ \displaystyle AD=\frac{AB^{2}}{AC}=\frac{AC^{2}-BC^{2}}{AC} $$
$$ \displaystyle =AC\left [ 1-\left ( \frac{BC}{AC} \right )^{2} \right ] $$
$$ \displaystyle =\frac{BC}{\cos 30^{\circ}}\left [ 1-\left ( \cos 30^{\circ} \right )^{2} \right ] $$
$$ \displaystyle =\frac{100}{0.8}\left [ 1-\left ( 0.8 \right )^{2} \right ] $$
$$ \displaystyle =\frac{100\times 10}{8}\left [ 1-\frac{64}{100} \right ] $$
$$ \displaystyle =\frac{100\times 10}{8}\times \frac{36}{100}=45 $$
The length of the shadow of a pole is $$\displaystyle \sqrt{3} $$ times its height The elevation of the sum must be
Report Question
0%
$$\displaystyle 60^{\circ} $$
0%
$$\displaystyle 45^{\circ} $$
0%
$$\displaystyle 30^{\circ} $$
0%
$$\displaystyle 15^{\circ} $$
Explanation
$$\displaystyle \tan \theta =\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}$$
$$\displaystyle \therefore \theta =30^{\circ}$$
The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal when it is below the horizontal level
that is -
Report Question
0%
the case when we lower our head to look at the object.
0%
the case when we upper our head to look at the object.
0%
the case when we look straight to the object.
0%
irrelevant defination
Explanation
option A is correct
the case when we lower our head to look at the object.
Report Question
0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.
0%
Assertion is correct, but Reason is incorrect.
0%
Assertion is incorrect, but Reason is correct.
Explanation
In $$\triangle ABC$$
$$\tan 30 = \frac{P}{B} = \frac{AB}{BC}$$
$$\frac{1}{\sqrt{3}} = \frac{AB}{20}$$
$$AB = \frac{20}{\sqrt{3}}$$
$$AB = 11.56$$ m
Both assertion and reason are correct.
In the given figure given, $$\beta$$ is _____
Report Question
0%
Angle of elevation
0%
Angle of depression
0%
Line of sight
0%
Horizontal level
Explanation
Here, $$A$$ is an object below the level of the eye and $$O$$ is eye.
Also, $$OB$$ is a horizontal line through $$O$$.
Thus $$\angle AOB = B$$ is called the angle of depression.
In the given figure, $$\alpha$$ is _____
Report Question
0%
Angle of elevation
0%
Angle of depression
0%
Line of sight
0%
Horizontal level
Explanation
The given object A is above the level of the eye O. Thus $$\alpha$$ is the angle of elevation.
So, option A is correct.
In the given figure, $$\beta$$ is _____.
Report Question
0%
Angle of elevation
0%
Angle of depression
0%
Line of sight
0%
Horizontal level
Explanation
The object A is below the level of eye O.
Thus, B is angle of depression.
So, option B is correct.
In case of angle of elevation, the observer has to look _____ to view the object.
Report Question
0%
Straight
0%
Anywhere
0%
Down
0%
Up
Explanation
In case of angle of elevation. The observer has to look up to view the object.
From a point P on a level ground, the angle of elevation of the top tower is $$30^{\circ}$$. If the tower is $$100\ m$$ high, the distance of point P from the foot of the tower is:
Report Question
0%
$$149\ m$$
0%
$$156\ m$$
0%
$$173\ m$$
0%
$$200\ m$$
Explanation
Let $$AB$$ be the tower.
Then, $$\angle APB = 30^{\circ}$$ and $$AB = 100\ m$$.
$$\dfrac {AB}{AP} = \tan 30^{\circ} = \dfrac {1}{\sqrt {3}}$$
$$\Rightarrow AP = (AB \times \sqrt {3})m$$
$$= 100\sqrt {3}m$$
$$= (100\times 1.73)m$$
$$= 173\ m$$.
The given figure represents
Report Question
0%
Angle of elevation
0%
Angle of depression
0%
Right angle
0%
Line of sight
Explanation
$$\textbf{Step 1: Figuring out what the figure represents}$$
$$\text{An angle of elevation is defined as an angle between the horizontal plane and line of sight from the}$$
$$\text{observer's eye to some object above}$$
$$\therefore\text{The figure represents the angle of elevation}$$
$$\textbf{Hence, the figure represents option A }$$
In angle of elevation the observer looks ____ to view the object whereas in angle of depression he looks ____ to view object.
Report Question
0%
Up, up
0%
Down, down
0%
Down, up
0%
Up, down
Explanation
In angle of elevation the observer looks up to view the object whereas in angle of depression he looks down to view the object.
If the height of a tower and the length of its shadow is equal, then the value of the angle of elevation of the sun is-
Report Question
0%
$$30^{\circ}$$
0%
$$45^{\circ}$$
0%
$$60^{\circ}$$
0%
None of these
Explanation
$$ \tan \theta = \dfrac{height}{shadow} $$
Here the height is equal to the shadow
$$\tan { \theta } =\dfrac { height }{ height } =1$$
$$ \tan \theta = 1 $$
$$ \theta = 45^{o} $$
If the length of the shadow of a pole is equal to the height of the pole, then the angle of elevation of the sun is
Report Question
0%
$${ 30 }^{ o }\quad $$
0%
$${ 75 }^{ o }$$
0%
$${ 60 }^{ o }\quad $$
0%
$${ 45 }^{ o }$$
Explanation
$$AC=BC$$
$$tan(B)=\dfrac{AC}{BC}$$
$$=\dfrac { AC }{ AC } =1$$
$$\angle B =tan^{-1}1$$
$$\angle B={ 45 }^{ 0 }$$
The heights of two poles are $$80$$ m and $$65$$ m. If the line joining their tops makes an angle of $$45^{\circ}$$ with the horizontal, then the distance between the poles is :
Report Question
0%
$$15$$ m
0%
$$22.5$$ m
0%
$$30$$ m
0%
$$7.5$$ m
Explanation
Let, $$AB$$ and $$CD$$ be the two poles of height $$80$$ m and $$65$$ m respectively.
According to the picture,
$$AE=AB-CD=$$$$(80-65)\ m=15\ m$$ [Since $$CD=EB$$]
Now, from $$\Delta$$ $$ACE$$ we have,
$$\tan 45^o=\dfrac{AE}{EC}$$
$$\Rightarrow 1=\dfrac{15}{EC}$$
$$\Rightarrow EC=15$$
Now, $$BD=EC=15\ m$$.
Hence, the distance between the two poles is $$15\ m$$.
At an instant, the length of the shadow of a pole is $$\sqrt{3}$$ times the height of the pole. The angle of elevation of the sun is
Report Question
0%
$$30^{\circ}$$
0%
$$45^{\circ}$$
0%
$$60^{\circ}$$
0%
$$75^{\circ}$$
Explanation
Let the height of pole be $$h$$ metres and Angle of elevation be $$\theta.$$
According to question,
Shadow of pole $$=\sqrt{3}h$$ metres
We know that,
$$\tan \, \theta \, = \, \dfrac{perpendicular}{base}$$
$$\Rightarrow\tan \, \theta \, = \, \dfrac{h}{\sqrt{3}h}$$
$$\Rightarrow \displaystyle \tan \, \theta \, = \, \frac{1}{\sqrt{3}} $$
$$\Rightarrow \tan\theta = \, \tan \, 30^{\circ}$$
$$\Rightarrow\theta \, = \, 30^{\circ}$$
Hence, the elevation of the sun is $$30^o$$
Choose the correct answer from the alternative given.
A $$10$$ m long ladder is placed against a wall. It is inclined at an angle of $$30$$ to the ground. The distance (in m) of the foot of the ladder from the wall is?
Given $$(\sqrt{3}$$ = 1.732)
Report Question
0%
$$7.32$$
0%
$$8.26$$
0%
$$8.66$$
0%
$$8.16$$
Explanation
AC $$= 10m$$
BC =:?
$$\angle\, $$ACB = $$\theta$$ =$$30^{\circ}$$
From $$\Delta$$ABC,
$$\cos$$ $$\theta$$ = $$\dfrac{BC}{AC}$$
$$\cos$$ $$30^{\circ}$$ =$$\dfrac{BC}{10} \, \Rightarrow \, \dfrac{\sqrt{3}}{2} \, = \, \dfrac{BC}{10}$$
$$BC = 5$$$$\sqrt{3}$$ $$= 5$$ $$\times$$ $$1.732 = 8.660$$ metre
If the ratio of the height of tower and the length of its shadow is $$1: \sqrt{3}$$, then the angle of elevation of the sum has measure_________.
Report Question
0%
$$60$$
0%
$$45$$
0%
$$30$$
0%
$$90$$
Explanation
Let $$\overline{AB}$$ is a tower and $$\overline{BC}$$ is its shadow.
Given that , $$\dfrac{AB}{BC}=\dfrac{1}{\sqrt{3}}$$
Let the angle of elevation of the Sun is $$\theta$$
$$\therefore m\angle ACB = \theta$$
Now, $$\tan \theta =\dfrac{AB}{BC} $$
$$\therefore \tan \theta =\dfrac{1}{\sqrt{3}}=\tan 30^o$$
$$\therefore \theta =30^o$$
$$\therefore$$ the angle of elevation of the Sun $$=30^o$$
A $$20 m$$ pole casts a $$5 m$$ long shadow. If at the same time of the day, a building casts a shadow of $$20 m$$, how high is the building?
Report Question
0%
$$400$$ m
0%
$$4$$ m
0%
$$80$$ m
0%
$$100$$ m
Explanation
Let the angle subtended between pole and shadow be $$\theta$$
$$\Rightarrow$$ $$\tan\theta = \dfrac{20}{5} = 4$$
Similarly,
At same time of the day, angle subtended will remain same as position of sun is fixed.
Given that the s
hadow is $$20m$$ long.
Let height of the building be $$x$$
$$\Rightarrow$$ $$\tan\theta = \dfrac{x}{20} = 4$$
$$\Rightarrow x = 80$$ m
A vertical post $$15$$ ft high is broken at a certain height and its upper part, not completely separated, meets the ground at an angle of $$30^{\circ}$$. Find the height at which the post is broken.
Report Question
0%
$$10\ ft$$
0%
$$5\ ft$$
0%
$$15\sqrt {3} (2 - \sqrt {3})ft$$
0%
$$5\sqrt {3} ft$$
Explanation
Given that, $$AB = Post = 15\ feet, \angle CDB = 30^{\circ}$$ and the post breaks at point $$C$$.
Let the height at which post is broken be $$x$$ feet.
$$\Rightarrow AC = CD = (15 - x) feet$$
From $$\triangle BCD$$,
$$\sin 30^{\circ} = \dfrac {BC}{CD}$$
$$\Rightarrow \dfrac {1}{2} = \dfrac {x}{15 - x} \Rightarrow 2x = 15 - x\Rightarrow 3x = 15 \Rightarrow x = 5$$ feet
Hence, the height at which the post is broken is $$5$$ feet.
A vertical tower 50 ft. high Stands on a sloping ground. The foot of the tower is at the same level as the middle part of a vertical flag pole. From the top of the tower the angle of depression of the top and bottom of the pole are $$15^{o}$$ and $$45^{o}$$ respectively. Then the length of the pole is 100 $$\sqrt{3}$$.
Report Question
0%
True
0%
False
Explanation
$$OC=50$$ is the tower standing on a sloping ground $$OB$$
$$BA=2 h$$ is the flag pole whose mid point $$M$$ is in level with $$O$$
The angles of depression of $$A$$ and $$B$$ as seen from $$C$$ ,the top of the tower , are $$15^{\circ},45^{\circ}$$ respectively
Let $$BD=x=OM=AE$$
In $$\triangle CAE,50-h=x\tan 15^{\circ}$$
In $$\triangle CBD,50+h=x\tan 45^{\circ}$$
$$\implies \dfrac{50-h}{50+h}=\dfrac{\tan 15^{\circ}}{\tan 45^{\circ}}$$
Applying compoundo and dividendo rule
$$\implies \dfrac{2h}{100}=\dfrac{\tan 45^{\circ}-\tan 15^{\circ}}{\tan 45^{\circ}+\tan 15^{\circ}}=\dfrac{\sin 45^{\circ}\cos 15^{\circ}-\sin 15^{\circ}\cos 45^{\circ}}{\sin 45^{\circ}\cos 15^{\circ}+\sin 15^{\circ}\cos 45^{\circ}}=\dfrac{\sin 30^{\circ}}{\sin 60^{\circ}}=\dfrac{1}{\sqrt{3}}$$
$$2 h=\dfrac{100}{\sqrt{3}}$$
So the given relation is $$\text{False}$$
A flag-staff 5 m high stands on a building 25 m high. To an observer at a height of 30 m the flag-staff and the building subtend equal angles. The distance of the observer from the top of the flag-staff is
Report Question
0%
$$\dfrac{5}{2}$$
0%
$$5\sqrt{\dfrac{3}{2}}$$
0%
$$5\sqrt{\dfrac{2}{3}}$$
0%
none of these
Explanation
AB = 25 m is building, BC = 5 m is flag-staff. O is an observer at a height of 30 m. OC = x is the distance of the observer from the top of flag-staff AB and BC subtend the same angle at O.
tan $$\theta$$ = 5/x, tan 2$$\theta$$ = 30/x
Eliminate x. $$\dfrac{tan\,2\theta}{tan\,\theta}$$ = 6
1- $$tan^2 \theta$$ = 1/3
tan $$\theta$$ = $$\sqrt{2/3}$$
x = 5 cot $$\theta$$ = $$5\sqrt{\dfrac{3}{2}}$$
The angle of elevation of a certain peak when observed, from each end of a horizontal base line of length 2a is found to be $$\theta$$. When observed from the mid-point of the base the angle of elevation is $$\phi$$. Then the height of the peak is
$$\dfrac{a sin \theta sin \phi}{ \sqrt{( sin (\phi + \theta) sin ( \phi - \theta))} }$$
Report Question
0%
True
0%
False
Explanation
Let$$h$$ be the height of the peak
From the figure
$$AQ=BQ=h\cot \theta,CQ=h\cot \phi$$
$$CQ$$ is perpendicular to the base $$AB$$ of a isosceless triangle
$$\implies BQ^2=a^2+CQ^2$$
$$\implies h^2(\cot^2 \theta-\cot^2 \phi)=a^2$$
$$\implies h=\dfrac{a}{\sqrt{\cot^2 \theta-\cot^2 \phi}}=\dfrac{a\sin \theta\sin \phi}{\sqrt{\cos^2 \theta\sin^2 \phi-\cos^2 \phi\sin^2 \theta}}=\dfrac{a\sin \theta\sin \phi}{\sqrt{\sin (\theta+\phi)\sin (\theta-\phi)}}$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 10 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
