MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 15

Alex observed that the angle of elevation to the top of

$800-foot$ Mount Colin was

$23^{\circ}$ . To the nearest foot, how much closer to the base of Mount Colin must Alex move so that his angle of elevation is doubled?

0%
$200$
0%
$400$
0%
$489$
0%
$1112$

Explanation

Let the initial position of Alex be C and final position be D.

Height of tower

$=800ft$

Angle of elevation

$=\angle ACD$

$=23^o$

In triangle

$\triangle ABC$

$\tan 23^o=\dfrac{AB}{BC}$

$\Rightarrow 0.424=\dfrac{800}{BC}$

$\Rightarrow BC=\dfrac{800}{0.424}=1886.8$

$\triangle ABD$

$\tan D=\dfrac{AB}{BD}$

$$\Rightarrow \tan 46 =\dfrac{800}{BD}$$

$\Rightarrow BD=\dfrac{800}{\tan 46}=\dfrac{800}{1.035}$

$=772.95$

Distance to be travelled

$DC=BC-BD$

$=1886.8-772.95$

$$=1113.85 \approx 1112$$

2114940_535025_ans_d6fb9011505e49ae807524667871942e.png

The angle of elevation of the top of a vertical tower from two points at distances

$a$ and

$b$

$(a>b)$ from the base and in the same line with it, are complimentary. If

$\theta$ is the angle subtended at the top of the tower by the line joining these points, then

$\sin{\theta}$ is equal to

0%
$\cfrac { a+b }{ a-b }$
0%
$\cfrac { a-b }{ a+b }$
0%
$\cfrac { (a-b)b }{ a+b }$
0%
$\cfrac { a-b }{ (a+b)b }$

Explanation

$In\quad \triangle CAD,\\ \tan { \phi } =\frac { AC }{ b } \quad \_ \_ \_ (1)$

$In\quad \triangle CAB,\\ \tan { (90-\phi ) } =\dfrac { AC }{ a } \quad \\ \Rightarrow \cot { \phi } =\frac { AC }{ a } \quad \_ \_ \_ \_ (2)$

From (1) and (2),

$\Rightarrow \dfrac { AC }{ b } =\dfrac { a }{ AC }$

$\Rightarrow{ AC }^{ 2 }=ab$

${ \Rightarrow AC }=\sqrt { ab }$

$\sin { \theta } =-\sin { (90-2\phi )=-\cos { 2\phi } }$

$=-1+2\sin ^{ 2 }{ \phi }$

$=-1+\frac { 2a }{ a+b }$

$\sin { \phi } =\sqrt { \dfrac { ab }{ ab+{ b }^{ 2 } } }$ and

$\sin ^{ 2 }{ \phi } =\frac { a }{ a+b }$ we get

$\sin \theta = \dfrac{a-b}{a+b}$

The angle of elevation and the angle of depression are

$30^o$ and

$30^o$ respectively when seen from the top of the first building to the top and base of the second building. If the distance between the bases of two building is 12 m, then find the height of big building.

0%
$16 \sqrt {3} m$
0%
$12 \sqrt {3} m$
0%
$14 \sqrt {3} m$
0%
$20 \sqrt {3} m$

Explanation

Given that the angle of elevation is

$30^{0}$ and the angle of depression is

$30^{0}$ when seen from the top of first building to the top and base of second building.

Let the height of first building be

$x$ and the height of second building be

$y$

From angle of depression ,we have

$tan(30^{0})=\frac{12}{x}$

$\Rightarrow x=12 cot(30^{0})=12 \sqrt3$

From angle of elevation , we have

$tan(30^{0})=\frac{y-x}{12}$

$\Rightarrow y-x=12 tan(30^{0})=4 \sqrt3$

$\Rightarrow y=x+ 4 \sqrt3=16 \sqrt3$

The top of a hill observed from the top and bottom of a building of height

$'h'$ is at angles of elevation

$p$ and

$q$ , respectively. The height of hill is ?

0%
$\dfrac {h\cot q}{\cot q - \cot q}$
0%
$\dfrac {h\cot p}{\cot p - \cot q}$
0%
$\dfrac {h\tan p}{\tan p - \tan q}$
0%
None of these

Explanation

Let

$AD = h$ is the height of the building and

$EB$ is height of the hill.
and

$\tan q = \dfrac {h + x}{y}$
and

$\tan p = \dfrac {x}{y}$

$\Rightarrow y = x\cot p$
Now,

$\tan q = \dfrac {h + x}{x\cot p}$

$\Rightarrow x \cot p = (h + x) \cot q$

$\Rightarrow x = \dfrac {h\cot q}{\cot p - \cot q}$

$\therefore h + x = h + \dfrac {h\cot q}{\cot p - \cot q}$

$= \dfrac {h\cot p}{\cot p - \cot q}$

Elevation angle of the top of the mirror from the foot of the tower of height

$h$ is

$\alpha$ and the tower subtend an angle

$\beta$ at the top of the mirror. Then, height of mirror is

0%
$\dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) - \cot \alpha}$
0%
$\dfrac {h\tan (\alpha - \beta)}{\tan (\alpha - \beta) - \tan \alpha}$
0%
$\dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) + \cot \alpha}$
0%
None of the above

Explanation

Let

$AB$ be the tower of height

$h$ and

$PQ$ be the mirror of height

$x.$

$\therefore \angle PBQ = \alpha$ and

$\angle APB = \beta$

$\triangle APB,$

$\angle PBA = 90^o-\alpha$

$\angle PAB = \angle PAC+90^o$

So, by using angle sum property,

$\angle APB+\angle PBA+\angle PAB=180^o$

$\Rightarrow \beta+90^o-\alpha+\angle PAC+90^o=180^o$

$\Rightarrow \beta-\alpha+\angle PAC+180^o=180^o$

$\Rightarrow \angle PAC=\alpha-\beta$

In right angle triangle

$PBQ,$

$\dfrac {x}{BQ} = \tan \alpha ..... (i)$

And, in right angle triangle

$PCA,$

$\dfrac {x - h}{AC} = \tan (\alpha - \beta)$

$\Rightarrow \dfrac {x - h}{BQ} = \tan (\alpha - \beta) .... (ii)$

Now divide equation

$(ii)$ by

$(i),$

$\dfrac {x - h}{x} = \dfrac {\tan (\alpha - \beta)}{\tan \alpha}$

$\Rightarrow 1 -\dfrac {h}{x} = \dfrac {\cot \alpha}{\cot (\alpha - \beta)}$

$\Rightarrow x = \dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) - \cot \alpha}$

Hence, option

$A$ is correct.

The angle of elevation of a jet fighter from a point

$A$ on the ground is

$60$ . After a flight of

$15$ seconds, the angle of elevation changes to

$30$ . If the jet is flying at a speed of

$720km/hr$ , find the constant height in m.

0%
$2598 m$
0%
$2690 m$
0%
$2355 m$
0%
$2405 m$

Explanation

Let

$A$ be the point of observation on the ground and

$B$ and

$C$ be two positions of the jet.

Let

$BL=CM=h$ metres.

Let

$AL=x$ metres.

Speed of the jet

$=720$ km\hr

$=720 \times \cfrac{5}{18}=200$ m\s.

Time taken to cover the distance

$BC$ is

$15$ sec.

Distance

$BC=LM=200 \times 15$ m=

$3000$ m

$\triangle ALB$ ,

$\cot 60^{\circ}=\cfrac{AL}{BL}=\cfrac{1}{\sqrt 3}=\cfrac{x}{h}$

$\therefore x=\cfrac{h}{\sqrt 3}$ -------------------------1

$\triangle AMC$ ,

$\cot 30^{\circ}=\cfrac{AM}{MC}$

$\Rightarrow \sqrt 3=\cfrac{AL+LM}{MC}$

$\Rightarrow \sqrt 3=\cfrac{x+3000}{h}$

$\Rightarrow h\sqrt 3-3000=x$ --------------------------2

From Equation 1 & 2,

$\cfrac{h}{\sqrt 3}=h\sqrt 3-3000$

$\Rightarrow h=3h-\sqrt 3\times 3000$

$\Rightarrow 2h=5196$

$\Rightarrow h=2598m$

The angles of depression of the top and the bottom of a 7 m tall building from the top of a tower are

$45^0$ and

$60^0$ respectively.Find the height of the tower in metres.

0%
$7(3+\sqrt{3})$
0%
$\dfrac{7}{2}(3-\sqrt{3})$
0%
$\dfrac{7}{2}(3+\sqrt{3})$
0%
$7(3-\sqrt{3})$

Explanation

Height of the building

$AB=7\:m$

Let height of the tower be

$PC$ and the distance between the building and the tower be

$AC$ .

Angle of depression to top of building is

$\angle {QPB}=45^o$

Angle of depression to bottom of building is

$\angle {QPA}=60^o$

We have to find the height of the tower

$i.e., PC$ .

Draw

$BD$ parallel to

$AC \: and\: PQ$

Since

$PQ || BD, \angle{PBD}=\angle{QPB}\Rightarrow \angle{PBD}=45^o$

Since

$PQ || AC, \angle{PAC}=\angle{QPA}\Rightarrow \angle{PAC}=60^o$

Now,

$AC,BD$ are parallel lines

$\Rightarrow AC=BD$

Also,

$AB,CD$ are parallel lines

$\Rightarrow AB=CD$ .

$\therefore CD=7\:m$

Since

$PC \:perpendicular \:to\:AC$ ,

$\angle{PDB}=\angle{PCA}=90^o$

In right triangle

$PBD, tan\:B=\dfrac{PD}{BD}$

$\Rightarrow Cot\:45=\dfrac{BD}{PD}$

$\Rightarrow 1=\dfrac{BD}{PD}$

$\Rightarrow BD=PD \quad ...(1)$

In right triangle

$PAC, tan\:A=\dfrac{PC}{AC}$

$\Rightarrow cot\:60=\dfrac{AC}{PC}$

$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{AC}{PC}$

$\Rightarrow AC=\dfrac{PC}{\sqrt{3}}$

$\Rightarrow BD=\dfrac{PC}{\sqrt{3}} (\because AC=BD) \quad ...(2)$

From

$(1), (2)$ we get,

$PD=\dfrac{PC}{\sqrt{3}}$

$\Rightarrow PD=\dfrac{PD+DC}{\sqrt{3}}$

$\Rightarrow \sqrt{3}PD=PD+DC$

$\Rightarrow (\sqrt{3}-1)PD=7$

$\Rightarrow PD=\dfrac{7}{\sqrt{3}-1}$

Rationalising we get

$PD=\dfrac{7(\sqrt{3}+1)}{3-1}=\dfrac{7(\sqrt{3}+1)}{2}$

Now,

$PC=PD+DC$

$=\dfrac{7(\sqrt{3}+1)}{2}+7$

$=\dfrac{7\sqrt{3}+7+14}{2}$

$=\dfrac{7\sqrt{3}+21}{2}$

$PC=\dfrac{7(\sqrt{3}+3)}{2}$

Hence the height of the tower is

$\dfrac{7(\sqrt{3}+3)}{2}\: m$

A spherical balloon of radius

$r$ subtends an angle

$\angle$ at the eye of an observer, while the angle of elevation of its centre is

$\beta$ . What is the height of the centre of the balloon (neglecting the height of the observer)?

0%
$\cfrac { r\sin { \beta } }{ \sin { \left( \cfrac { \alpha }{ 2 } \right) } }$
0%
$\cfrac { r\sin { \beta } }{ \sin { \left( \cfrac { \alpha }{ 4 } \right) } }$
0%
$\quad \cfrac { r\sin { \left( \cfrac { \beta }{ 2 } \right) } }{ r\sin { \alpha } }$
0%
$\cfrac { r\sin { \alpha } }{ \sin { \left( \cfrac { \beta }{ 2 } \right) } }$

Explanation

$\rightarrow$ Let

$'O'$ be the centre of the balloon,

$P$ be the eye of the observer, and

$\angle APB$ be the angle subtended by the balloon at the eye of the observer.

$\therefore \angle APB = \alpha$

$\angle APO = \angle BPO = \dfrac{\alpha} { 2}$

$\triangle OAP \ \sin \dfrac {\alpha}{2} = \dfrac {OA}{OP}$

$\rightarrow \sin \dfrac {\alpha}{2} = \dfrac {r}{OP}$ or

$OP = r \text{cosec} \dfrac {\alpha}{2}$

$OPL$

$\sin \beta = \dfrac {OL}{OP}$ or

$PL = OP\sin \beta$

$\therefore OL = r\text{cosec} \dfrac {\alpha}{2} \sin \beta$

$= \dfrac {r\sin \beta}{\sin \left (\dfrac {\alpha}{2}\right )}$ .

A man observes two objects in a straight line in the West. On walking a distance c to the North, the objects subtend an angle

$\alpha$ , in front of him and on walking a further distance c to the North, they subtend an angle

$\beta$ . The distance between the objects is

$\dfrac{3c}{2 \cot \beta - \cot \alpha}$

0%
True
0%
False

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of

$30^\circ$ , which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be

$60^\circ$ . Find the time taken by the car to reach the foot of the tower from this point.

0%
$3\text{ sec}$
0%
$4\text{ sec}$
0%
$2\text{ sec}$
0%
$6\text{ sec}$

Explanation

Let, the distance traveled by car in six seconds be

$x$ and the height of the tower be

$h$ .

$\triangle BCD,$

$\tan 60 ^{\circ} = \dfrac{CD}{BC}$

$\Rightarrow \sqrt{3} = \dfrac{h}{d}$

$\Rightarrow h = \sqrt{3}d \quad\quad\quad\dots(i)$

$\triangle ACD,$

$\tan 30^{\circ} = \dfrac{CD}{AC}$

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{x+d}$

$\Rightarrow h = \dfrac{x+d}{\sqrt{3}}\quad\quad\quad\dots(ii)$

On comparing

$(i)$ in

$(ii)$ we get,

$\sqrt{3}d = \dfrac{x+d}{\sqrt{3}}$

$\Rightarrow 3d = x+d$

$\Rightarrow x = 2d$

$\Rightarrow d = x/2$

Time required to travel

$x$ distance is equal to

$6$ seconds. So, distance

$\dfrac{x}{2}$ will be covered in time

$3$ seconds.

1148985_875155_ans_79912d24c92646c49938380426629a63.png

The foot of the ladder leaning against a wall of length 5 metre rest on a level ground

$5\sqrt{3}$ metre from the base of the wall. The angle of inclination of the ladder with the ground is:

0%
$90^{\circ}$
0%
$50^{\circ}$
0%
$40^{\circ}$
0%
$30^{\circ}$

The angles of elevation of the top of a mountain from two points A and B on a horizontal line are

${ 15 }^{ \circ }$ and

${ 75 }^{ \circ }.$ If AB=650 mt, then the height of the mountain is ______________.

0%
$\dfrac { 325\sqrt { 3 } }{ 3 } mt$
0%
$\dfrac { 325\sqrt { 3 } }{ 2 } mt$
0%
$\dfrac { 324\sqrt { 3 } }{ 3 } mt$
0%
$\dfrac { 324\sqrt { 3 } }{ 2 } mt$

The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 m from its base is

$45^{\circ}$ . If the angle of elevation of the top of the complete pillar at the same point is to be

$60^{\circ}$ , then the height of the incomplete pillar is to be increased by :

0%
$50\sqrt{2}$
0%
$100$
0%
$100(\sqrt{3}-1)$
0%
$100(\sqrt{3}+1)$

Explanation

Let

$D$ be the fixed point,

$AC$ be the pillar

$\Rightarrow CD=100$ m

$AB$ is the height to be increased.

$\triangle{BCD}, \tan{{45}^{\circ}}=\dfrac{BC}{DC}$

$\Rightarrow 1=\dfrac{BC}{DC}=\dfrac{BC}{100}$

$\triangle{ACD},\tan{{60}^{\circ}}=\dfrac{AC}{DC}$

$\Rightarrow \sqrt{3}=\dfrac{AC}{100}$

$\Rightarrow AC=100\sqrt{3}$ m

Height to be increased

$=AB=AC-BC=100\sqrt{3}-100=100\left(\sqrt{3}-1\right)$ m

$\therefore$ to complete the pillar

$100\left(\sqrt{3}-1\right)$ m has to be increased.

1098427_1227620_ans_bc3bda0e067d4650a4bf75bd9e3026c3.png

There are two statins

$P.Q$ due north, due south of a tower of height

$15$ meters. The angle of depression of

$P$ and

$Q$ as seen from top a tower are

$\cot^{-1}\dfrac{12}{5},\sin^{-1}\dfrac{3}{5}$ . The distance between

$P$ and

$Q$ is ..

0%
$48$
0%
$56$
0%
$65$
0%
$25$

The angular elevation of tower CD at a point A due south of it is

${ 60 }^{ \circ }$ and at a point B due west of A, The elevation is

${ 30 }^{ \circ }$ If AB= 3 km, The height of lower is

0%
$2\sqrt { 3 } km$
0%
$2\sqrt { 6 } km$
0%
$\dfrac { 3\sqrt { 3 } }{ 2 } km$
0%
$\dfrac { 3\sqrt { 6 } }{ 4 } km$

A ladder rest against 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 distance 'b' down the wall making an angle

$\beta$ with the horizontal , then

$tan\dfrac{\alpha + \beta}{2}$ =

0%
b/a
0%
a/b
0%
2/ab
0%
2a/b

A balloon of radius

$r$ subtends an angle

$\alpha$ at the eye of an observer and the elevation of the centre of the balloon from the eye is

$\beta$ , the height

$h$ of the centre of the balloon is given by

0%
$\dfrac{r\sin\beta}{\sin\alpha}$
0%
$r\sin\beta\sin\alpha$
0%
$\dfrac{r\sin\beta}{\sin(\alpha/2)}$
0%
$\dfrac{r\sin\beta}{\sin(\beta/2)}$

Explanation

$\begin{array}{l} \angle APO=\angle BPO=\frac { \alpha }{ 2 } \\ \Rightarrow In\, \, \Delta APO,\, \, \sin \frac { \alpha }{ 2 } =\frac { { OA } }{ { OP } } ,\, \, OP=\left( { \gamma \cos ec\frac { \alpha }{ 2 } } \right) \\ and\, \, In\, \, \Delta OPG,\, \, \sin \beta =\frac { { OG } }{ { OP } } ,\, \, OG=OP,\, \, \, \sin \beta \\ \Rightarrow OG=\left( { \gamma \cos ec\frac { \alpha }{ 2 } .\sin \beta } \right) \\ \Rightarrow Thus,\, \, height\, \, of\, \, the\, \, centre\, \, of\, \, the\, \, balloon\, \, is\left( { \gamma \sin \beta .\cos ec\frac { \alpha }{ 2 } } \right) \end{array}$
1202014_1314823_ans_9a71fc453b52401aa454df2dd99ce50a.png

1202014_1314823_ans_9a71fc453b52401aa454df2dd99ce50a.png

The angle of elevation of the top of the tower observed from each of three points A, B, C on the horizontal ground, forming a triangle ABC, is observed to be

$'\theta '$ . If 'R' be the circumradius of the triangle, then height of tower is :

0%
$R\tan { \theta }$
0%
$R\cot { \theta }$
0%
$R\sin { \theta }$
0%
$R\cos { \theta }$

Explanation

Here, Height of tower

$=$ OD

Circumradius

$=OA=OB=OC=R$

We know

$:\tan\theta =\dfrac{opposite}{adjacent}$ , so

In triangle OAD, we get

$\tan(\alpha OAD)=\dfrac{OD}{OA}$

$\Rightarrow \tan x=\dfrac{OD}{R}$

$\Rightarrow OD=R\tan x$

$\therefore$ Height of tower

$=$ R tan x.

1226440_1316479_ans_e2b704fb2923493fad6c457786c995fe.jpg

A man standing on a level plain observes the elevation of the top of the pole of height

$h$ to be

$\dfrac {\pi}{12}$ . He then walks a distance

$x$ towards the pole and finds that the elevation is now

$\dfrac {\pi}{6}$ . If

$h=33\ m$ then

$x^ {2}$

0%
$4356$
0%
$4355$
0%
$4354$
0%
$4353$

A tower

${ T }_{ 1 }$ of higher 60 m is located exactly opposite to a tower

${ T }_{ 2 }$ of height 80 m on a straight road. From the top of

${ T }_{ 1 }$ , if the ngle of depression of the foot of

${ T }_{ 2 }$ is twice the angle of elevation of the top of

${ T }_{ 2 }$ , then the width (in m) of the road between the feet of the towers

${ T }_{ 1 }$ and

${ T }_{ 2 }$ is :

0%
$10\sqrt { 2 }$
0%
$10\sqrt { 3 }$
0%
$20\sqrt { 3 }$
0%
$20\sqrt { 2 }$

Over a tower AB of height 10 mt there is a flag staff BC if AB and BC makes equal angles at a point of distance 12 mt from foot A of the tower , them height of flag staff BC is:-

0%
$\frac { 510 }{ 11 } mt$
0%
$\frac { 610 }{ 11 } mt$
0%
$\frac { 710 }{ 11 } mt$
0%
none of these

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 distance 'd' down wall, finally making an angle

$\beta$ with the horizontal then

$tan\left( \dfrac { \alpha +\beta }{ 2 } \right) =$

0%
$\dfrac { a }{ b }$
0%
$\dfrac { b }{ a }$
0%
$\dfrac { 2ab }{ { a }^{ 2 }-{ b }^{ 2 } }$
0%
$\dfrac { 2ab }{ b^{ 2 }-{ a }^{ 2 } }$

The shadow of the tower standing on a level ground is found to be 60 meters longer when the sun altitude is

${ 30 }^{ \circ }$ than when it is

${ 45 }^{ \circ }$ . The height of the tower is

0%
60 m
0%
30 m
0%
$60\sqrt { 3 } m$
0%
$30\left( \sqrt { 3 } +1 \right) m$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 15 - 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 15 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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