MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 10

One side of a rectangular piece of paper is

$6\text{ cm}$ , the adjacent sides being longer than

$6\text{ cm}$ . One corner of the paper is folded so that it sets on the opposite longer side. If the length of the crease is

$l$

$\text{cm}$ and it makes an angle

$\theta$ with the long side as shown, then

$l$ is

0%
$\cfrac { 3 }{ \sin ^{ 2 }{ \theta } \cos { \theta } }$
0%
$\cfrac { 6 }{ \sin ^{ 2 }{ \theta } \cos { \theta } }$
0%
$\cfrac { 3 }{ \sin { \theta } \cos { \theta } }$
0%
$\cfrac { 3 }{ \sin ^{ 3 }{ \theta } }$

Explanation

Let EC $=x$

In $\Delta BCE$ ,

$\displaystyle\frac{x}{l}=\sin \theta \Rightarrow x=l\sin \theta$

$EF=x=l\sin \theta$

In $\Delta ABF,$

$\angle ABF=90-2\theta$

$\angle AFB=2\theta$

$\sin 2\theta =\displaystyle\frac{6}{l\sin \theta}$

$\Rightarrow l=\displaystyle\frac{6}{\sin\theta .\sin 2\theta}$

$=\displaystyle\frac{6}{\sin \theta(2\sin \theta\cdot \cos\theta)}$

$l=\displaystyle\frac{3}{\sin^2\theta \cdot \cos\theta}$ .

The angle of elevation of the top of an unfinished pillar at a point 150 m from its base is 30

$^\circ$ . If the angle of elevation at the same point is to be 45

$^\circ$ , then the pillar has to be raised to a height of how many metres?

0%
59.4 m
0%
61.4 m
0%
62.4 m
0%
63.4 m

Explanation

$\Delta$ ABC

$\displaystyle tan \, \theta \, = \, \frac{AB}{BC}$

$\displaystyle tan \, 30^{\circ} \, = \, \frac{AB}{150}$

$AB = 150$

$\displaystyle \times \, \frac{1}{\sqrt{3}} \,$

$= \, 50 \sqrt{3} \, m$
If

$\theta \, = \, 45^{\circ}$

$\displaystyle tan \, 45^{\circ} \, = \, \frac{AB}{BC}$

$1 \, = \, \dfrac{AB}{150}$

$AB = 150$ m
Difference in height is

$150- 5\sqrt{3}$

$= 63.4$ m (approx.)

1161538_908313_ans_1bbbc7877adb4ad69567ce22206c5838.png

A man on the top of a bamboo pole observes that the angle of depression of the base and the top of another pole are

${60}^{o}$ and

${30}^{o}$ respectively. If the second pole stands

$5$ m above the ground level, then the height of the bamboo pole on which the man is sitting is

0%
$5$ m
0%
$2.5$ m
0%
$10$ m
0%
$12.5$ m

Explanation

$\triangle ABC,\quad \cfrac { x }{ b } =\tan { { 30 }^{ o } } \Rightarrow b=x\sqrt { 3 }$
In

$\triangle ADE\quad$

$\cfrac { x+5 }{ b } =\tan { { 60 }^{ o } } \Rightarrow \cfrac { x+5 }{ \sqrt { 3 } } =b$
Equating we get,

$3x=x+5\Rightarrow x=\dfrac52$

$\therefore$ height

$=2.5cm$

The angles of elevation of the top of a temple, from the foot and the top of a building

$30\ m$ high, are

$60^{\circ}$ and

$30^{\circ}$ respectively. Then height of the temple is

0%
$50\ m$
0%
$43\ m$
0%
$40\ m$
0%
$45\ m$

Explanation

$\triangle DBC$ ,

$\tan 60^{\circ} = \dfrac {x + 30}{BC}$

$\Rightarrow BC = \dfrac {x + 30}{\sqrt {3}} ..... (1)$

$\triangle DAE$ ,

$\Rightarrow \tan 30^{\circ}=\dfrac{DE}{AE}$

Now,

$AE = BC$

$\dfrac {x}{\dfrac {x + 30}{\sqrt {3}}} = \dfrac {1}{\sqrt {3}}$

$\Rightarrow \sqrt {3}x = \dfrac {x + 30}{\sqrt {3}}$

$\Rightarrow 3x = x + 30$

$\Rightarrow 2x = 30$

$\Rightarrow x = 15\ m$

$\therefore\ h = x + 30$

$= 15 + 30 = 45\ m$

A 25 m long ladder is placed against a vertical wall such that the foot of the ladder is 7 m from the feet of the wall. If the top of the ladder slides down by 4 cm, by how much distance will the foot of the ladder slide?

0%
$9\ m$
0%
$15\ m$
0%
$5\ m$
0%
$8\ m$
0%
$4\ m$

The angles of elevation of the top of a tower from two points A & B lying on the horizontal through the foot of the tower are respectively

$30^{\circ}$ and

$45^{\circ}$ of A & B are on the same side of the tower and AB = 48 m then the height of the tower is

0%
24( $\sqrt{3} + 1)$ m
0%
25 m
0%
26( $\sqrt{3} - 1)$ m
0%
23 m

Explanation

Let the heightof tower be h.

$tan 45^{\circ} = \cfrac{h}{BC}$

$l = \cfrac{h}{BC}$

h = BC (1)

$tan 30^{\circ} = \cfrac{h}{48 + BC}$

$\frac{1}{\sqrt{3}} = \cfrac{h}{48 + BC} \Rightarrow \cfrac{1}{\sqrt{3}} = \cfrac{h}{48 + h}$ [From eq. (1)]

48+h=

$\sqrt{3}h \Rightarrow 48 = \sqrt{3}h -h$

48 = h(

$\sqrt{3} - 1) \Rightarrow h = \cfrac{48}{(\sqrt{3} - 1)} = 24(\sqrt{3} + 1)$ m.

The angle of elevation of a tower from a point on the ground is

${30}^{o}$ . At a point on the horizontal line passing through the foot of the tower and

$100$ meters nearer to it, the angle of elevation is found to be

${60}^{o}$ , then the height of the tower is

0%
$50\sqrt { 3 }$ meters
0%
$\dfrac{50}{\sqrt { 3 }}$ meters
0%
$\dfrac{100}{\sqrt { 3 } }$ meters
0%
$100\sqrt { 3 }$ meters

Explanation

Let

$CD=x\ m$

$BC=100m$

From

$\triangle ACD,$

$tan60^o=\dfrac{AD}{CD}$

$\Rightarrow \sqrt{3}=\dfrac{AD}{x}$

$\Rightarrow AD=x\sqrt { 3 } ..........(i)$

From

$\triangle ABD$

$tan30^o=\dfrac{AD}{BD}$

$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{AD}{x+100}$

$\Rightarrow \sqrt{3}AD=x+100$

$\Rightarrow 3x=x+100$

$[By\ (i)]$

$\Rightarrow x=50$

Hence the height of the tower AD

$=50\times\sqrt{3}\ m=50\sqrt { 3 }\ m$

1005329_925128_ans_dc76016249bd4c89a4ef025f3094d352.png

The height of a tower is

$50\sqrt {3}\ m$ . The angle of elevation of a tower from a distance

$50\ m$ from its feet is

0%
$30^{\circ}$
0%
$45^{\circ}$
0%
$60^{\circ}$
0%
$90^{\circ}$

Explanation

It is given that the height of the tower is

$50\sqrt 3\text{ m}$ and the distance of the tower from the point is

$50\text{ m}$ .

Representing this through a diagram.

Now, In

$\triangle ABC$ ,

$\tan \theta = \dfrac {50\sqrt {3}}{50} = \sqrt {3}$

$\theta = 60^{\circ}$ .

Choose the correct answer from the alternatives given.
A tree is broken by the wind. If the top of the tree struck the ground at an angle of

$30^\circ$ and at a distance of 30 m from the root, then the height of the tree is

0%
$15 \sqrt 3 m$
0%
$20 \sqrt 3 m$
0%
$25 \sqrt 3 m$
0%
$30 \sqrt 3 m$

Explanation

$\Delta$ ABC,

$tan 30^{\circ} = \frac{BC}{AB}$

$\Rightarrow BC = 30 \times \frac{1}{\sqrt{3}} = 10\sqrt{3}$ m
and

$cos 30^{\circ} = \frac{AB}{AC}$

$\Rightarrow AC = \frac{AB}{\sqrt{3} / 2} = \frac{2 \times 30}{\sqrt{3}} = 20 \sqrt{3}$
Height of the tree = BA = BC + CA
= 10

$\sqrt{3} + 10\sqrt{3} = 30\sqrt{3}$ m.
1161271_907945_ans_b039bd5bb6aa44aab740dcd641118dc4.png

The shadow of a stick of height

$1$ metre, when the angle of elevation of the sun is

${ 60 }^{ o }$ , will be

0%
$\cfrac { 1 }{ \sqrt { 3 } }$ m
0%
$\cfrac { 1 }{ 3 }$ m
0%
$\sqrt { 3 }$ m
0%
$3$ m

Explanation

$\dfrac { AB }{ BC } ={ \tan60 }^{ 0 }\quad ;\quad BC=\dfrac { AB }{ { \tan60 }^{ 0 } } =\dfrac { 1 }{ \sqrt { 3 } } m$

1005454_927155_ans_2f8318874d204441a9d5b80cebf330f6.jpg

If the angle of elevation of the sun at

$8:00$ O'clock is

$\alpha$ and at

$10:00$ O'clock is

$\beta$ then ______ holds.

0%
$\alpha > \beta$
0%
$\alpha < \beta$
0%
$\alpha \geq \beta$
0%
$\alpha = \beta$

If a flag-staff of

$6m$ height placed on the top of a tower throws a shadow of

$2\sqrt { 3 } m$ along the ground, then what is the angle that the sun makes with the ground?

0%
${ 60 }^{ o }$
0%
${ 45 }^{ o }\quad$
0%
${ 30 }^{ o }$
0%
${ 15 }^{ o }$

Explanation

Let the Angle be

$x$

and Length of shadow

$=2\sqrt3 m$

Length of Tower

$=6m$

So,

$\tan x=\dfrac{6}{2\sqrt3}\Rightarrow \tan x=\sqrt3\Rightarrow \angle x=60^o$

Hence, the answer is

$60^{o}$ .

The height of a chimney when it is found that on walking towards it

$100\ ft$ . in a horizontal line through its base the angular elevation of its top changes from

${30}^{o}$ to

${45}^{o}$ is ________

0%
$8\sqrt 3$
0%
$50(\sqrt 3+1)$
0%
$100(\sqrt 3+1)$
0%
None of these

The angles of depression of two points

$A$ and

$B$ on a horizontal plane such that

$AB = 200\ \text{m}$ from the top

$P$ of a tower

$PQ$ of height

$100\ \text{m}$ are

$45^\circ - \theta$ and

$45^\circ + \theta$ . If the line

$AB$ passes through

$Q$ the foot of the tower, then angle

$\theta$ is equal to:

0%
$45^\circ$
0%
$30^\circ$
0%
$22.5^\circ$
0%
$15^\circ$

Explanation

$AB = 200 = AQ -BQ$

$100\cot (45^\circ - \theta) - 100\cot (45^\circ + \theta)=200$

$200 = 100\left [ \dfrac{1 + \tan \theta}{1 - \tan \theta} - \dfrac{1 - \tan \theta}{1 + \tan \theta} \right ]$

$2 = \dfrac{4 \tan \theta}{1 - \tan^2 \theta} = \tan2 \theta$

$\theta = 22.5^\circ$
1080527_1006371_ans_652d5fed942c42b8b81110699bc14ba2.JPG

The angle of elevation of the top of a tower from the top and bottom of a building of height

$a$ are

$30^{\circ}$ and

$45^{\circ}$ respectively. If the tower and the building are at same level, the height of the tower is

0%
$\dfrac{\sqrt{3}a(1+\sqrt{3})}{2}$
0%
$a\left(\sqrt{3}+1\right)$
0%
$\sqrt{3}a$
0%
$a\left(\sqrt{3}-1\right)$

The angle of elevation of the top of an unfinished tower at a point

$120m$ from its base is

$45^o$ . How much higher must the tower be raised so that its angle of elevation becomes

$60^o$ at the same point?

0%
$90m$
0%
$92m$
0%
$97m$
0%
$87.84m$

Explanation

Let PA be the unfinished tower.

Let B be the point of observation i.e. 120 m away from the base of the tower.

Now, AB = 120m

Let,

$\angle{ABP} = 45°$

Let h m be the height by which the unfinished tower be raised such that its angle of elevation of the top from the same point becomes 60°.

Let CA = h

$\& \; \angle{ABC} = 60°$

In triangle ABP,

$\tan{45°} = \cfrac{PA}{AB}$

$\Rightarrow \; 1 = \cfrac{PA}{120}$

$\Rightarrow \; PA = 120M$

Now, in triangle ABC,

$\tan{60°} = \cfrac{CA}{AB}$

$\sqrt{3} = \cfrac{120 + h}{120}$

$h + 120 = 120\sqrt{3}$

$h = 120 (\sqrt{3} - 1)$

$h = 120 (1.732 - 1) \; \rightarrow (as \; \sqrt{3} = 1.732)$

$h = 120 \times 0.732$

$h = 87.84m$

927612_1001554_ans_433d79b718804439ac7373cdbf47bdf5.jpeg

A ladder rests against a wall at an angle

$\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distanced down the wall making an angle

$\beta$ with the horizontal, then

0%
$a = b\, tan \dfrac{\alpha + \beta}{2}$
0%
$a = b\, cot \dfrac{\alpha + \beta}{2}$
0%
$a \, tan \dfrac{\alpha - \beta}{2}$
0%
None

Explanation

$∆ OQB, cos β = \dfrac{OB}{BQ}$

⇒

$OB = ℓ cos β . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Similarly in

$∆ OPA, cos α = \dfrac{OA}{PA}$

$⇒ OA = ℓ cos α . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

$Now a = OB – OA = ℓ (cos β – cos α) . . . . . . . . . . . . . . . . . . (3)$

Also from

$∆ OAP, OP = ℓ sin α$

And in

$\triangle OQB; OQ = ℓ sin β$

$∴ b = OP – OQ = ℓ (sin α – sin β) . . . . . . . . . . . . . . . . . . . . . . (4)$

Dividing eq.

$(3)$ by

$(4)$ we get

$\dfrac{a}{b} =\dfrac{cos \beta – cos\alpha}{sin {\alpha} – sin {\beta}}$

$=\dfrac{2sin{\dfrac{\alpha+\beta}{2}}.sin{\dfrac{\alpha-\beta}{2}}}{2cos{\dfrac{\alpha+\beta}{2}}.sin{\dfrac{\alpha-\beta}{2}}}$

⇒

$\dfrac{a}{b} = tan \dfrac{(α + β)}{2}$

Thus ,

$a = b$

$tan \dfrac{(α + β)}{2}$ is proved

The angle of elevation of the top of a tower standing on a horizontal plane from point A is

$\alpha$ . After walking a distance d towards the foot of the tower, the angle of elevation is found to be

$\beta$ . The height of the tower is

0%
$\dfrac{d \,sin\, \alpha\, sin\, \beta}{sin\, (\beta - \alpha)}$
0%
$\dfrac{d \,sin\, (\beta - \alpha)}{sin\, \alpha\,sin\, \beta}$
0%
$\dfrac{d \,sin\,\alpha\, sin\, \beta}{sin\, (\alpha\,- \beta)}$
0%
$\dfrac{d\, sin\, (\alpha\, - \beta)}{sin\, \alpha\,sin \, \beta}$

Shadow of a vertical pillar is equal to the height of the pillar then the angle of elevation of sun will be :

0%
$90^{o}$
0%
$60^{o}$
0%
$45^{o}$
0%
$30^{o}$

Explanation

In right angle Δ

$ABC,$

$AB$ represents the vertical pole and

$BC$ represents the shadow on the ground.

$θ$ represents angle of elevation the sun.

$\tan θ = \dfrac h x$

But length of shadow

$=$ height of the tower, i.e.

$h = x.$

$\tan θ = \dfrac h h = 1$

$\tan θ = \tan 45°$

$θ = 45°$

Thus, angle of elevation of the sun

$= 45°.$

1486011_1230570_ans_210b0256d5694d03af242c94be365cc4.jpg

A vertical pole PS has two marks at Qand A such that the portions PQ, PR and PS subtend angles

$\alpha,\beta, \gamma$ at a point on the ground distant x from the pole. If PQ = a, PR =b, PS = c and

$\alpha + \beta + \gamma$ = 180; then

$x^{2}$ =

0%
$\dfrac{a}{a + b + c}$
0%
$\dfrac{b}{a + b + c}$
0%
$\dfrac{c}{a + b + c}$
0%
$\dfrac{abc}{a + b + c}$

Explanation

PQ = a = x tan

$\alpha$ , PR = b = x tan

$\beta$
PS = c = tab

$\gamma$
a + b + c = x ( tan

$\alpha$ + tan

$\beta$ + tan

$\gamma$ )
abc =

$x^3$ .tan

$\alpha$ .tan

$\beta$ . tan

$\gamma$
Dividing abc/a + b +c =

$x^2$ .1

$\Rightarrow$ (d)
when

$\alpha$ +

$\beta$ +

$\gamma$ =

$180^0$ then
tan

$\alpha$ + tan

$\beta$ + tan

$\gamma$ = tan

$\alpha$ .tan

$\beta$ . tan

$\gamma$
i.e.

$S_1 = S_3$
1083719_1006531_ans_6b46a971d86143df80b6c149a425aabc.JPG

$ABC$ is a triangular park with

$AB = AC =100\ m$ .

$A$ clock tower is situated at the mid-point of

$BC$ . The angles of elevation of the top of the tower at

$A$ and

$B$ are

$\cot^{-1} 3.2$ and

$cosec^{-1} 2.6$ respectively. The height of the tower is

0%
$16\ m$
0%
$25\ m$
0%
$50\ m$
0%
none of these

Explanation

$cot \alpha = 3.2 , cot \beta = 2.6$
The triangle being isosceles, AB = AC the median AO is perpendicular to base BC

$OA = h cot \alpha = 3.2 h$

$OB = h cot \beta = h \sqrt{(cosec^2 \beta - 1)} = (2.4) h$

$100^2 = OA^2 + OB^2 = h^2 \left[ \dfrac{32^2 + 24^2}{100} \right]$

$100 \times 10 = 4 \sqrt{8^2 + 6^2} h= 4\times10\times h$

$\therefore h = 25 m$
1084782_1006601_ans_39d79d0eed8740fa8e4cbd89612854c0.png

A tower subtends 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 the first, the depression of the foot of the tower is

$60^o$ . The horizontal distance of the tower from the point is

0%
$h\ \cot 60^o$
0%
$\dfrac{1}{3}$ $h\ \cot 30^o$
0%
$\dfrac{h}{3}$ $\cot 60^o$
0%
$h\ \cot 30^o$

Explanation

x = h cot

$60^0$ = h tan

$30^0$
= h .(1/

$\sqrt{3})$ =

$(1/3).h.$

$\sqrt{3}$ =

$\dfrac{1}{3}$

$h cot 30$
1105989_1006562_ans_b5a3e9e0ea3a4a5aa3323583f1d639bd.JPG

A vertical tower stands on a declivity which is inclined at

$15^o$ to the horizon. From the foot of the tower a man ascends the declivity for

$80$ feet and then finds that the tower subtends at angle of

$30^o$ . The height of the tower is-

0%
$20 ( \sqrt{6} - \sqrt{2})$
0%
$40 ( \sqrt{6} - \sqrt{2})$
0%
$40 ( \sqrt{6} + \sqrt{2})$
0%
$none\ of\ these$

Explanation

From triangle ABC by sine formula

$\dfrac{BC}{sin 75^o} = \dfrac{AB}{sin 30^o}$

$h = 80 \dfrac{sin 30^o}{sin (45^o + 30^o)} = \dfrac{40 \times 2 \sqrt{2}}{[ \sqrt{(3)} + 1]}$

$= \dfrac{40 . 2 \sqrt{2} ( \sqrt{3} - 1)}{3-1} = 40 \sqrt{2} ( \sqrt{3} - 1) = 40 (\sqrt{6} - \sqrt{2})$ .
1084791_1006603_ans_a5bbddeec0ab4146b6992ec405051f99.png

The angle of elevation of the top of a tower at a point A on the ground is

$30$ . On walking

$20$ metres towards the tower, the angle of elevation is

$60$ , The height of the tower

0%
$50(\sqrt 3+1)$
0%
$10\sqrt 3$
0%
$8\sqrt 3$
0%
None of these

The shadow of a tower standing on a level ground is found to be

$60$ meters longer when the suns altitude is

$30$ than when it is

$45$ . The height, of the tower is______

0%
$30(\sqrt{3} + 1)$
0%
$30(\sqrt{3} - 1)$
0%
$20(\sqrt{3} + 1)$
0%
None of these

From the top of a $25\ \text{m}$ high pillar at the top of tower, angle of elevation is same as the angle of depression of foot of tower, then height of tower is:

0%
$25\ \text{m}$
0%
$100\ \text{m}$
0%
$75\ \text{m}$
0%
$50\ \text{m}$

Explanation

$OB$ is the length of the pillar and

$AD$ is the length of the tower.

Let

$\theta$ be angle of elevation to top of tower and angle of depression to foot of tower.

Height of pillar

$=h=25\ \text{m}$

Height of tower

$=h+d$

$\tan \theta=\cfrac{h}{BC}=\cfrac{d}{BC}$

$\therefore h=d$

$\therefore$ Height of tower

$=h+d=2h=50\ \text{m}$

1023362_1023000_ans_06637b720ea84d32a634d2afbdf5cf74.png

A uniform rod is rested on a wall and its lower end is tied to the wall with the help of a horizontal string of negligible mass as shown. Length of rod is

$L$ and height of the wall is

$h$ . What will be tension in the string? (all surface are frictionless)

0%
$\dfrac { mg\ell\sin { \theta } }{ 2h\cos { \theta } }$
0%
$\dfrac { mg\ell\cos ^{ 2 }{ \theta } }{ 2h\sin { \theta } }$
0%
$\dfrac { mg\ell\sin { \theta } \cos ^{ 2 }{ \theta } }{ 2h }$
0%
$\dfrac { mg\ell }{ 2h\sin { \theta } \cos ^{ 2 }{ \theta } }$

A boy playing on the roof top of a 10m high building throws a ball with a speed of 10m/s at an angle of

$30_{0}$ with the horizontal. How far from the throwing point will the ball be at the height of 10m from the ground.

$\sin { { 30 }^{ 0 } } =\quad \frac { 1 }{ 2 } \\ \cos { { 30 }^{ 0 } } =\quad \frac { \sqrt { 3 } }{ 2 }$

0%
8.66 m
0%
5.20 m
0%
4.33 m
0%
2.60 m

Explanation

The correct option is A

The ball will be at point P when it is at a height of

$10$ m from the ground. so, we have to find the distance OP which can be calculated directly by considering it is as a projectile on a leveled plane(OX)-

OP=R=

$\frac{u^2sin2\theta}{g}$

$\frac{10^2\times sin(2\times 30^0)}{10}$

$\frac{10\sqrt 3}{2}$ =

$5\sqrt3$ =

$8.66$

969127_1032623_ans_5e537fc8ad394d5d8a24ca4fa4ea06ae.PNG

Find the angle of depression of a boat from the bridge at a horizontal distance of

$25$ m from the bridge, if the height of the bridge is

$25$ m.

0%
$60^0$
0%
$30^0$
0%
$45^0$
0%
none

Explanation

$AB=BC=25m$

$\angle B=90^o$

$\angle BAC=\theta$

Now, in

$\triangle ABC$

$\angle BAC=\angle DCA$ (alternate angle)

$\Rightarrow \tan \theta=\cfrac {AB}{BC}$

$\Rightarrow \tan\theta=1$

$\theta=45^o$ (

$\because \tan 45^0=1$ )

1006203_1051275_ans_909bbf47b55c49e6af89b652651685c4.png

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a

$20\ meters$ high building are

${45}^{o}$ and

${60}^{o}$ respectively. Then the height of the tower is

0%
$20(\sqrt{3}-1)$
0%
$20(\sqrt{3}+1)$
0%
$\dfrac{20}{\sqrt{3}-1}$
0%
$\dfrac{20}{\sqrt{3}+1}$

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

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Practice Class 10 Maths Quiz Questions and Answers

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

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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