Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.

Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.

Assertion is correct, but Reason is incorrect.

Assertion is incorrect, but Reason is correct.

MCQ Questions for 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

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:

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

$h.$

$\triangle ABC,$

$\begin{aligned}{}\sin 60^\circ &= \frac{{AB}}{{AC}}\\\\\frac{{\sqrt 3 }}{2} &= \frac{h}{{120}}\\\\h& = \frac{{120 \times \sqrt 3 }}{2}\\\\ &= 60\sqrt 3 \text{ m}\end{aligned}$

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

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?

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.

$\displaystyle \sin 30^{\circ}=\frac{x}{15}$

$x=15\displaystyle \sin 30^{\circ}\\=\frac{15}{2}\\=7.5$ m

1296929_380080_ans_f4accd53dbd246818eef04295860ab7b.png

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

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

1297469_380194_ans_95bc268185f740bd8ce15d21b6ff9a19.bmp

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

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

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

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.

0%
$45$
0%
$30$
0%
$50$
0%
$80$

Explanation

$\displaystyle \bigtriangleup ' S ABC and ABD , \angle A$ 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

$\displaystyle AD=\frac{AB^{2}}{AC}=\frac{AC^{2}-BC^{2}}{AC}$

$\displaystyle =AC\left [ 1-\left ( \frac{BC}{AC} \right )^{2} \right ]$

$\displaystyle =\frac{BC}{\cos 30^{\circ}}\left [ 1-\left ( \cos 30^{\circ} \right )^{2} \right ]$

$\displaystyle =\frac{100}{0.8}\left [ 1-\left ( 0.8 \right )^{2} \right ]$

$\displaystyle =\frac{100\times 10}{8}\left [ 1-\frac{64}{100} \right ]$

$\displaystyle =\frac{100\times 10}{8}\times \frac{36}{100}=45$

1033135_374004_ans_381dd2298c8c4657aa14f22ad9e6c801.png

The length of the shadow of a pole is

$\displaystyle \sqrt{3}$ times its height The elevation of the sum must be

0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 45^{\circ}$
0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 15^{\circ}$

Explanation

$\displaystyle \tan \theta =\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}$

$\displaystyle \therefore \theta =30^{\circ}$
1297475_380185_ans_46ce8938791e4d3fac450ca5eaa0ea10.bmp

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 -

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

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

$\triangle ABC$

$\tan 30 = \frac{P}{B} = \frac{AB}{BC}$

$\frac{1}{\sqrt{3}} = \frac{AB}{20}$

$AB = \frac{20}{\sqrt{3}}$

$AB = 11.56$ m
Both assertion and reason are correct.

In the given figure given,

$\beta$ is _____

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 _____

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 _____.

0%
Angle of elevation
0%
Angle of depression
0%
Line of sight
0%
Horizontal level

In case of angle of elevation, the observer has to look _____ to view the object.

0%
Straight
0%
Anywhere
0%
Down
0%
Up

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:

0%
$149\ m$
0%
$156\ m$
0%
$173\ m$
0%
$200\ m$

Explanation

Let

$AB$ be the tower.
Then,

$\angle APB = 30^{\circ}$ and

$AB = 100\ m$ .

$\dfrac {AB}{AP} = \tan 30^{\circ} = \dfrac {1}{\sqrt {3}}$

$\Rightarrow AP = (AB \times \sqrt {3})m$

$= 100\sqrt {3}m$

$= (100\times 1.73)m$

$= 173\ m$ .
1278198_539622_ans_d5488f80416f4835bd8185c0afe5b61a.jpg

The given figure represents

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.

0%
Up, up
0%
Down, down
0%
Down, up
0%
Up, down

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-

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

0%
${ 30 }^{ o }\quad$
0%
${ 75 }^{ o }$
0%
${ 60 }^{ o }\quad$
0%
${ 45 }^{ o }$

Explanation

$AC=BC$

$tan(B)=\dfrac{AC}{BC}$

$=\dfrac { AC }{ AC } =1$

$\angle B =tan^{-1}1$

$\angle B={ 45 }^{ 0 }$

1175066_926746_ans_ef474a28a7f44758a585421d9a8435fb.jpg

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 :

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

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

$30^o$

1135746_900156_ans_005bb954081a4d9b89de726e3361b8d9.png

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)

0%
$7.32$
0%
$8.26$
0%
$8.66$
0%
$8.16$

Explanation

$= 10m$
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

1137272_901109_ans_ea76166f8d9147e28800364126b9da5a.png

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_________.

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

$=30^o$

$20 m$ 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?

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 shadow 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.

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

$\sqrt{3}$ .

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

$BD=x=OM=AE$

$\triangle CAE,50-h=x\tan 15^{\circ}$

$\triangle CBD,50+h=x\tan 45^{\circ}$

$\implies \dfrac{50-h}{50+h}=\dfrac{\tan 15^{\circ}}{\tan 45^{\circ}}$

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

$\text{False}$

1495781_1007250_ans_7086408e4e58405dbd19fc982bc51015.png

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

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

$\theta$ =

$5\sqrt{\dfrac{3}{2}}$
1105997_1006566_ans_0bf668fd5e464b2f8d019694ecced15a.png

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

$\dfrac{a sin \theta sin \phi}{ \sqrt{( sin (\phi + \theta) sin ( \phi - \theta))} }$

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

$\implies BQ^2=a^2+CQ^2$

$\implies h^2(\cot^2 \theta-\cot^2 \phi)=a^2$

$\implies h=\dfrac{a}{\sqrt{\cot^2 \theta-\cot^2 \phi}}=\dfrac{a\sin \theta\sin \phi}{\sqrt{\cos^2 \theta\sin^2 \phi-\cos^2 \phi\sin^2 \theta}}=\dfrac{a\sin \theta\sin \phi}{\sqrt{\sin (\theta+\phi)\sin (\theta-\phi)}}$

1495768_1007636_ans_c095ce9267844609b79680b96bcf92b7.png

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 1 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 1 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.