MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 4

A tower stands vertically on the ground. From a point on the ground which is

$25\ m$ away from the foot of the tower, the angle of elevation of the top of the tower is found to be

$45^{\circ}$ . Then the height

$(in\ meters)$ of the tower is:

0%
$25 \sqrt{2}\ m$
0%
$25 \sqrt{3}\ m$
0%
$25\ m$
0%
$12.5\ m$

Explanation

Let the height of the tower be

$'h'$

$m.$

Now, in

$\triangle ACB,$

$\Rightarrow \tan { 45^{\circ}} =\dfrac { AB }{ BC } =\dfrac { h }{ 25 }$

$\Rightarrow 1=\dfrac { h }{ 25 }$

$\Rightarrow h=25\ m$

Hence, the height of tower is

$25\ m.$

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground, making an angle of

$30^o$ with the horizontal. The distance from the foot of the tree to the point where the top touches the ground is

$10$ m. The height of the tree is

0%
$10\left ( \sqrt{3}+1 \right )$ m
0%
$10\sqrt{3}$ m
0%
$10\left ( \sqrt{3}-1 \right )$ m
0%
$\dfrac{30}{\sqrt{3}}$ m

Explanation

The broken part touches the ground, making a triangle with the ground and foot of the tree as shown in the figure.

Now,

$\tan 30=\dfrac{b}{10}$

$b=10\tan 30$

$b=10\times\dfrac{1}{\sqrt3}$

$\cos 30=\dfrac{10}{l}$

$\dfrac{\sqrt3}{2}=\dfrac{10}{l}$

$l=\dfrac{20}{\sqrt3}$
Therefore, length of the tree is

$l+b=\dfrac{10+20}{\sqrt3}=\dfrac{30}{\sqrt3}$

If the distance between the objects is

$p$ metres, then the height

$h$ of the tower is:

0%
$\dfrac{p\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$
0%
$\dfrac{\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$
0%
$\dfrac{p(\tan \alpha -\tan \beta )}{\tan \alpha \tan \beta }$
0%
none of the above

Explanation

$Height \: of \: the \: tower \: (AB)=h\:m$

$distance \: (CD) =p \: m$

$Let \:distance (BC) =xm$

$\angle ACB=\alpha \: and\: \angle ADB=\beta$
In right

$\triangle ABD$ ,

$\frac{AB}{BC}=\tan \alpha$

$\Rightarrow \frac{h}{x}=\tan \alpha$

$\Rightarrow {h}={x}\tan \alpha$ .....( i)
In right

$\triangle ABD$ ,

$\frac{AB}{BD}=\tan \beta$

$\Rightarrow \frac{h}{BC+CD}=\tan \beta$

$\Rightarrow h=(x+p)\tan \beta$ ..... (ii)
From (1), we get

$x=\frac{h}{\tan \alpha }$
Thus,

$h = (\frac{h}{\tan\alpha} + p) \tan \beta$

$h = (\frac{h + p \tan \alpha}{\tan \alpha}) \tan \beta$

$h \tan \alpha = h \tan \beta + p \tan \alpha \tan \beta$

$h (\tan \alpha - \tan \beta) = p \tan \alpha \tan \beta$

$h = \frac{p \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}$
Hence proved.

Based on the above figure, which of the following is/are correct?

0%
$\theta _1$ is the angle of elevation.
0%
$\theta _2$ is the angle of depression.
0%
The angle of elevation or depression is always measured from horizontal line through the point of observation.
0%
$\theta _1$ and $\theta _2$ are always equal.

An aeroplane flying horizontally 1 km. above the ground is observed at an elevation of

$60^o$ and after

$10$ seconds the elevation is observed to be

$30^o$ . The uniform speed of the aeroplane in

$km/h$ is

0%
$240$
0%
$240\sqrt{3}$
0%
$60\sqrt{3}$
0%
None of these

Explanation

In triangle with angle of elevation

$30^{\circ}$

$\tan 30 = \dfrac{1}{d_a}$

$d_a = \cot 30 = \sqrt{3}$

In triangle with angle of elevation

$60^{\circ}$

$\tan 60 = \dfrac{1}{d_b}$

$d_b = \cot 60 =\dfrac{1}{\sqrt{3}}$
Thus,

$d = d_a - d_b = \sqrt{3} - \dfrac{1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}$ km
Now

$speed = \dfrac{distance}{time}$

$speed = \frac{\dfrac{2}{\sqrt{3}}}{10} \times 60 \times 60$

$speed = 240 \sqrt{3}$ km/hr

If the length of the shadow of a tower is

$\sqrt{3}$ times that of its height, then the angle of elevation of the sun is

0%
$15^o$
0%
$30^o$
0%
$45^o$
0%
$60^o$

Explanation

Let height of tower (AB) be

$h\ \text{metres}$

Then length of its shadow will be

$BC = \sqrt{3}\:h\:\text{metres}$
Let angle of elevation be

$\theta$
Now, In

$\triangle ABC$

$\tan \theta = \dfrac{P}{B} = \dfrac{AB}{BC}$

$\tan \theta = \dfrac{h}{\sqrt{3}h}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \theta =30^o$

A Lamp Post,

$5\sqrt{3}m$ high, cast a shadow

$5\ m$ long on the ground. The suns elevation of this point is:

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

Explanation

Let the angle of elevation be

$'\theta'.$

In,

$\triangle ACB,$

$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 5\sqrt { 3 } }{ 5 }$

$\Rightarrow \tan { \theta } =\sqrt { 3 }$

$\Rightarrow$ Or

$\theta =60^{\circ}$

Hence, the answer is

$60^{\circ}.$

If the angle of depression of an object from a

$75\ m$ high tower is

$30^o$ , then the distance of the object from the tower is

0%
$25\sqrt{3}\: m$
0%
$50\sqrt{3}\: m$
0%
$75\sqrt{3}\: m$
0%
$150\ m$

Explanation

Considering the

$\Delta AOB$
Angle of depression =

$\angle AOB$ (Alternate angles are equal)
Since,

$\Delta AOB$ is a right angled triangle with

$\angle B = 90^o$ ,

$\tan 30^o = \dfrac{AB}{OB}$

$\therefore \dfrac{1}{\sqrt{3}} = \dfrac{75}{OB}$

$\therefore OB = 75 \sqrt{3}\ m$

Hence, option C.

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

$Let\ height\ of\ the\ pole\ be\ h\ m.\\ Then,\ length\ of\ shadow\ of\ the\ pole=h\ m\\ Now,\ In\ \triangle ABC,\\ \tan { { \theta }^{ \circ }=\frac { AB }{ BC } } =\frac { AB }{ AB } =1\\ { \theta }^{ \circ }=45$

A tree breaks due to storm and the broken part bends so that the top of the tree just touches the ground, making an angle of

$30^o$ with the horizontal. The distance from the foot of the tree to the point where the top touches the ground is 10 m. The height of the tree is

0%
$10(\sqrt 3+1)\ m$
0%
$10\sqrt 3\ m$
0%
$10(\sqrt 3-1)\ m$
0%
$\frac {10}{\sqrt 3}\ m$

Explanation

Let BD be the tree broken at point C such that the broken part CD takes the position CA and strikes the ground at A. It is given that

$AB=10\ \text{m}$ and

$\angle{BAC=30^{\circ}}$

Let

$BC=x\ \text{m}$ and

$CD=CA=y\ \text{m}$

In right angled triangle ABC, we have

$\tan30^{\circ}=\dfrac{BC}{AB}$

$\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{x}{10}$

$\Rightarrow x=\dfrac{10}{\sqrt3}=\dfrac{10\sqrt3}{3}$

And

$\cos30^{\circ}=\dfrac{AB}{AC}$

$\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{10}{y}$

$\Rightarrow y=\dfrac{20}{\sqrt3}=\dfrac{20\sqrt3}{3}$

Now,

Height of the tree

$=x+y$

$=\dfrac{10\sqrt3}{3}+\dfrac{20\sqrt3}{3}$

$=\dfrac{30\sqrt3}{3}$

$=10\sqrt3\ \text{m}$

Hence, the height of the tree is

$10\sqrt3\ \text{m}$ .

The angles of elevation of the top of a tower from two points at distance m and n metres are complementary. If the two points and the base of the tower are on the same straight line, then the height of the tower is

0%
$\sqrt {mn}$
0%
mn
0%
$\frac {m}{n}$
0%
None of these

Explanation

From

$\Delta ADB,$

$\dfrac {h}{m}=tan\theta$

From

$\Delta ABC,$

$\dfrac {h}{n}=tan(90-\theta)$

$=cot\theta$

$\therefore \dfrac {h}{m}\times \dfrac {h}{n}=tan\theta \times cot\theta$

$=1$

$\Rightarrow h=\sqrt {mn}$

The angle of elevation of the sun when the length of the shadow of a pole is

$\sqrt 3$ times the height of the pole is

0%
$30^0$
0%
$45^0$
0%
$60^0$
0%
$75^0$

Explanation

$tan\angle MOP=\frac {MP}{MO}=\frac {1}{\sqrt 3}=tan 30^0$

$\therefore\ \angle MOP=30^0$

The length of the ladder making an angle of

$\displaystyle 45^{\circ}$ with the ground and whose foot is

$7$ m away from the wall is

0%
$\displaystyle \frac{7\sqrt{2}}{2}m$
0%
$\displaystyle 7\sqrt{2}m$
0%
$\displaystyle 14\sqrt{2}m$
0%
$14 m$

Explanation

$AC=\text{lenght of ladder}=x,\quad AB=\text{height of wall}$
Given,

$BC=7m$

Let

$AC$ be

$x$

$\cos 45^{0}=\cfrac{BC}{AC}$

$\therefore \dfrac{1}{\sqrt 2}=\cfrac{7}{x}$

$\therefore x =7\sqrt 2$ m

A ladder is inclined to a wall making and angle of

$30^0$ with it. A man is ascending the ladder at the rate of

$2\ \text{m/s}$ . How fast is he approaching the wall?

0%
$2\ \text{m/s}$
0%
$1.5\ \text{m/s}$
0%
$1\ \text{m/s}$
0%
$0.5\ \text{m/s}$

Explanation

According to the figure, let AC represent the magnitude of velocity vector from point C on ground to point A on the wall, that is AC= 2 m per s and

$\angle BAC =30^o$ degree.

Then, by trigonometry,

$sin 30^{o}=\frac{BC}{AC}$

where BC is the magnitude of velocity vector from C to wall.

Then

$sin 30^{o}=\dfrac{BC}{AC}$

$\Rightarrow \dfrac{1}{2}=\dfrac{BC}{2m/s}$

$\Rightarrow BC=1 m/s$

Then he approaching the wall

A tree

$6\ m$ tall casts a

$4\ m$ long shadow. At the same time, a flag pole casts a shadow

$50\ m$ long. How long is the flag pole?

0%
$75\ m$
0%
$100\ m$
0%
$150\ m$
0%
$50\ m$

Explanation

Let AB is the length of tree and Pq is the lenght of flag

We know that the angle of evaluation at same time on a point is same

In triangle ABC

The

$tan \theta =\frac{AB}{BC}=\frac{6}{4}$

In triangle PQR

The

$tan \theta =\frac{PQ}{QR}=\frac{PQ}{50}$

$\frac{6}{4}=\frac{PQ}{50}\Rightarrow PQ =\frac{6\times 50}{4}=75$ m

From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be

$\alpha$ and

$\beta$ where

$\alpha >\beta.$ The height of the tower is

$130\ m,$

$\alpha =60^\circ$ and

$\beta =30^\circ.$ The distance of the extreme object from the top of the tower is ______.

0%
$65\ m$
0%
$130\ m$
0%
$260\ m$
0%
None of the above

Explanation

Let

$AB$ be the tower

Given: Height of the tower

$h=130$ m

and

$\alpha=60^\circ,\ \beta=30^\circ$

The extreme object from the top of the tower has low angle of depression.

$\therefore$ The angle of elevation of the point

$=30^{\circ}$

Let the distance of the point from the top of the tower

$=p$

Thus,

$\sin 30^\circ = \dfrac{h}{p}$

$\Rightarrow \dfrac{1}{2} = \dfrac{130}{p}$

$\Rightarrow p = 260$ m

The angles of elevation of an artificial satellite measured from two earth stations are

$30^0$ and

$60^0$ respectively. If the distance between the earth stations is 4000 km, then the height of the satellite is

0%
2000 km
0%
6000 km
0%
3464 km
0%
2828 km

The Qutub Minar casts a shadow

$150~ \text{m}$ long, at the same time the Vikas Minar casts a shadow

$120 ~\text{m}$ long on the ground. If the height of the Vikas Minar is

$80~ \text{m}$ , find the height of the Qutub Minar.

0%
$180~\text{m}$
0%
$100~\text{m}$
0%
$150~\text{m}$
0%
$120~\text{m}$

Explanation

$\text{Since both the Qutub Minar and the Vikas Minar cast the shadow at the same time,the elevation } \alpha \text{ is same for both so}\\ \text{far as the shadow cast by the sun is concerned}.\\ \text{Given V=80 m, height of Vikas Minar}\\ \quad \quad \quad\text{ R=120 m, shadow of Vikas Minar}\\ \quad \quad \quad \text{S=150 m, shadow of Qutub Minar}\\ \quad \quad \quad \text{Q=?, height of Qutub Minar}\\ \text{now},\dfrac { \text{V} }{ \text{R} } =\tan { \alpha } \quad \quad \\ \therefore \tan { \alpha } =\dfrac { 80 }{ 120 } =\dfrac { 2 }{ 3 } (\text{for Vikas Minar})\\ \therefore \text{for Qutab Minar}, \frac { \text{Q} }{ \text{S} } =\tan { \alpha } =\dfrac { 2 }{ 3\\ } \\ \implies \text{Q}= \dfrac { 2 }{ 3 } \times \text{S}\\ \implies \text{Q}=\dfrac{2}{3}\times 150=100~\text{m}\\$

The distance between the tops of two trees

$20\ m$ and

$28\ m$ high is

$17\ m$ . The horizontal distance between the two tree is

0%
$9\ m$
0%
$11\ m$
0%
$15\ m$
0%
$31\ m$

Explanation

In right

$\triangle ABE$

$AB^2=BE^2+AE^2$

$AE^2=AB^2-BE^2$

$={17}^2-8^2$

$=289-64$

$=225$

$\therefore AE=\sqrt {225}$

$\Rightarrow AE=15\Rightarrow CD=15\ m$

One side of a parallelogram is 12 cm and its area is

$60 cm^2$ . If the angle between the adjacent sides is

$30^0$ , then its other side is

0%
$10$ cm
0%
$8$ cm
0%
$6$ cm
0%
$4$ cm

Explanation

Given that, in the parallelogram

$ABCD$ ,

$AD=12\ cm, \angle DAE=30^{o}$ and

$A(ABCD)=60\ cm^2$

Draw a line from

$D$ perpendicular to

$AB$ which intersects it at

$E$ .

Now,

$\triangle AED=90^{o}$ is a right-angled triangle

$\implies \sin 30^{o}=\dfrac{DE}{AD}$

$\implies \dfrac{1}{2}=\dfrac{DE}{12}$

$\implies DE=6\ cm$

$A(Parallelogram)=b \times h$

Now, we know that

$A(ABCD)=AB\times DE$

$\implies 60=AB\times 6$

$\implies AB=10\ cm$

Then the other side of a parallelogram is

$10\ cm$ .

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

$30^0.$ At a point on the horizontal line passing through the foot of the tower and

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

$60^0,$ then height of the tower is

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

Explanation

Consider the triangle where

$AB=h=$ height of the tower

$AD=x$ ,

$CD=100$ m
In triangle

$ABD$ , we have

$\tan60^0=\sqrt{3}=\dfrac{h}{x}$

$\therefore h=\sqrt{3}x$ ...(i)
Consider

$\triangle ABC$

$\tan30^0=\dfrac{1}{\sqrt{3}}=\dfrac{h}{x+100}$

$\therefore x+100=\sqrt{3}h$ ...(ii)
Substituting the value of

$h$ from equation (i), we get

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

$x=50$ m
Therefore from (i), we get

$h=50\sqrt{3}$ metre
Hence, the answer is option A.

A man observes the elevation of a tower to be

$30^0.$ After advancing 11 cm towards it, he finds that the elevation is

$45^0.$ The height of the tower to the nearest metro is

0%
10
0%
15
0%
20
0%
22

Explanation

Let AB the height of the tower

$\angle ACB=45^0$ and

$\angle D=30^0$ ...(given)
Consider triangle ABC

$tan45^0=1=\frac{AB}{BC}$
Therefore

$AB=BC$ ...(i)
Consider Triangle ABD

$tan30^0=\frac{1}{\sqrt{3}}=\frac{AB}{BC+CD}$

$BC+CD=\sqrt{3}AB$

$AB+CD=\sqrt{3}AB$ ...from i

$11=AB(\sqrt{3}-1)$

$AB=\frac{11(\sqrt{3}+1)}{2}$ (after rationalization)

$AB=15.026$
Hence answer is B

A rope of length

$5$ metres is tightly tied with one end at the top of a vertical pole and other end to the horizontal ground. If the rope makes an angle

$30^0$ to the horizontal, then the height of the pole is

0%
$\dfrac{5}{2}$ m
0%
$\dfrac{5}{\sqrt{2}}$ m
0%
$5\sqrt{2}$ m
0%
$5$ m

Explanation

Consider the above given triangle
Let

$OP=h$ be the height of the vertical pole
Let

$HP$ be the length of the rope
angle

$PHO=30^0$ .......(given)
Therefore,

$\sin 30^0=\dfrac{OP}{HP}=\dfrac{h}{5}$

$\Rightarrow \dfrac{1}{2}=\dfrac{h}{5}$

$\Rightarrow h=\dfrac{5}{2}$
Answer is

$\dfrac{5}{2}$ metre.

A round balloon of radius r subtends an angle

$\alpha$ at the eye of the observer while the angle of elevation of its center is

$\beta$ . Then the height of the balloon is

0%
$r \cos \dfrac{\alpha}{2}\sin \beta$
0%
$\dfrac{r \cos \alpha}{2 \sin \beta}$
0%
$\dfrac{r \cos \alpha}{2 \cos \beta}$
0%
$r \sin \alpha \sin \beta$

Explanation

Let

$O$ be the center of the balloon,

$P$ be the eye of the observer, and let

$\angle APB=\alpha$

OL will be the height of the ballon if we draw a

$\bot$ from point

$O$ .

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

$(\because \triangle AOP\cong \triangle BOP)$

We know,

$sin(\Theta ) = Opposite / Hypotenuse$

$\triangle AOP$

$\sin\dfrac{\alpha}{2}=\dfrac{OA}{OP}=\dfrac{r}{OP}$

$OP=r \cos\dfrac{\alpha}{2} \rightarrow(1)$

$\triangle OPL$

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

$\Rightarrow OL=OP\sin\beta$

$\Rightarrow OL=r\cos\dfrac{\alpha}{2}\sin\beta$ (from(1))

$\therefore \ A$ is the answer.

The angle of elevation of a cloud from a point

$h$ meters above the surface of a lake is

$300$ and the angle of depression of its reflection is

$600$ . Then the height of the cloud above the surface of the lake is

0%
$\frac{h}{\sqrt{3}}$
0%
$h$
0%
$\sqrt{2}h$
0%
$2 h$

Explanation

Consider the following diagram with the cloud at the point A
Let the perpendicular distance of the cloud from the the elevation be

$x=AC$
Angle ABC

$=30^0$
angle DBC

$=60^0$
Therefore
Consider triangle ABC

$tan30^0=\frac{x}{BC}=\frac{1}{\sqrt{3}}$

$BC=\sqrt{3}x$ ...(i)
Similarly consider triangle BDC

$tan60^0=\frac{h+h+x}{BC}=\sqrt{3}$

$2h+x=\sqrt{3}BC$ ...(ii)
Substituting the value of BC from equation i we get

$2h+x=\sqrt{3}\sqrt{3}x$

$3x=2h+x$

$h=x$
Therefore the height of the cloud above the surface of the water

$=h+x$

$=2h$
Hence answer is D

The shadow of a stick of height 1 meter, when the angle of elevation of the Sun is

$60^{\circ}$ , will be

0%
$\dfrac{1}{\sqrt{3}}$ metre
0%
$\dfrac{1}{3}$ metre
0%
$\sqrt{3}$ metre
0%
3 metre

Explanation

Let AB

$=$ 1 m be the stick.
Let AC be the shadow of length

$x.$
From right angled

$\Delta ACB,$

$\tan 60^0=\dfrac{1}{x}$

or

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

or

$x=\dfrac{1}{\sqrt{3}}$ m

A man looks from the top of a vertical tower 30 metres high at a marked point upon the horizontal plane on which the tower stands. The angle of depression of this point is

$30^0.$ The distance of the marked point from the foot of the tower is

0%
$\frac{30}{\sqrt{3}}$ m
0%
$\frac{30\sqrt{3}}{2}$ m
0%
30 m
0%
$30 \sqrt{3}$ m

Explanation

Consider the above diagram
In triangle POM
Let

$OM$ be the height of the tower
Let

$OP$ be the horizontal distance of the marked point from the foot of the tower
It is given that angle of depression

$=P=30^0$
Therefore

$\frac{MO}{PO}$

$=\frac{30m}{PO}$

$=tan30^0=\frac{1}{\sqrt{3}}$
Therefore

$PO=30\sqrt{3}$ metres
Hence answer is D

Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of

$30^0$ with the ground. If the length of the broken part is 20 metros, then the remaining part of the tree is of length

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

Explanation

$The\quad tree\quad breaks\quad at\quad P.\\ \therefore \quad PO=20m,\quad \angle POB={ 30 }^{ \circ }\\ \therefore \quad \frac { PB }{ PO } =sin{ 30 }^{ \circ }\\ or\quad \frac { PB }{ 20 } =\frac { 1 }{ 2 } \\ or\quad PB=\frac { 1 }{ 2 } \times 20=10m\quad \quad (Ans)$ .

The angle of elevation of the top of a tower as observed from a point on the horizontal ground is x. If we move a distance d towards the foot of the tower, the angle of elevation increases to y, then the height of the tower is

0%
$\dfrac{d tan x tan y}{tan y - tan x}$
0%
$d (tan y + tan x)$
0%
$d (tan y - tan x )$
0%
$\dfrac{d tan x tan y}{tan y + tan x}$

Explanation

Consider the above given triangle
Let AB be the height of the tower and

$\angle ACB=x$ and

$\angle ADB=y$ ...(given)
Consider triangle ADB

$tany=\frac{AB}{BD}$

$BD={AB}coty$ ...(i)
Consider triangle ACD

$tanx=\frac{AB}{BD+CD}$

$BD+CD=ABcotx$

${AB}coty+CD=ABcotx$ ...from(i)

$d=AB(cotx-coty)$

$AB=\frac{d}{cotx-coty}$
Upon simplifying we get

$AB=\frac{d(tanx tany)}{tany-tanx}$
Hence answer is A

A man observes the elevation of a balloon to be

$30^0.$ He then walks 1 km. towards the balloon and finds the angle of elevation now is

$60^0.$ The height (in km) of the balloon is

0%
$\dfrac{\sqrt{3}}{2}$
0%
$\sqrt{3}+1$
0%
$\sqrt{3}-1$
0%
$\dfrac{2}{\sqrt{3}}$

Explanation

Consider triangle ABCD

Let AB=h be the height of the balloon

Let CD be the distance walked by the man

It is given that

$\angle D=30^0$ and

$\angle C=60^0$ which are the angles of elevation before and after the man travelled a distance of 1Km

Consider triangle ABC

$\tan60^0=\dfrac{AB}{BC}=\sqrt{3}$

$\dfrac{AB}{\sqrt{3}}=BC$ ...(i)

Consider triangle ABD

$\tan30^0=\dfrac{1}{\sqrt{3}}$

$=\dfrac{AB}{1+BC}$

$1+BC=\sqrt{3}AB$

$1=\sqrt{3}AB-\dfrac{AB}{\sqrt{3}}$ ... from i

Therefore

$\dfrac{2AB}{\sqrt{3}}=1$

$AB=\dfrac{\sqrt{3}}{2}$

Hence answer is A

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 4 - 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 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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