MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 14

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

$m$ and

$n$ meters 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%
$\dfrac{m}{n}$
0%
None of these

Explanation

$In\quad \triangle ABD,\\ \tan { \theta =\dfrac { AB }{ BD } =\dfrac { h }{ m } } \quad \quad \quad ...(1)\\ In\quad \triangle ABC,\\ \tan { \left( 90-\theta \right) =\dfrac { AB }{ BC } } =\dfrac { h }{ n } \quad \quad \quad \quad \quad \\ \cot { \theta =\dfrac { h }{ n } } \\ \tan { \theta =\dfrac { n }{ h } } \\ Using\quad equation\quad (1),\\ \dfrac { h }{ m } =\dfrac { n }{ h } \\ { h }^{ 2 }=mn\\ h=\sqrt { mn }$

At the foot of the mountain the elevation of its summit is

${ 45 }^{ 0 }$ ; after ascending

$1000m$ towards the mountain up a slope of

${ 30 }^{ 0 }$ inclination, the elevation is found to be

${ 60 }^{ 0 }$ . The height of the mountain is

0%
$\displaystyle \frac { \sqrt { 3 } +1 }{ 2 } m$
0%
$\displaystyle \frac { \sqrt { 3 } -1 }{ 2 } m$
0%
$\displaystyle \frac { \sqrt { 3 } +1 }{ 2\sqrt { 3 } }$
0%
none of these

Explanation

Let

$P$ be the summit of the mountain and

$Q$ be the foot.

Let

$A$ be the first position and

$B$ the second position of observation.

$BN$ and

$BM$ are

$\bot$ s to

$PQ$ and

$AQ$ respectively.

Then

$AB=1000m=1km$ ,

$\angle MAB={ 30 }^{ 0 },\angle MAP={ 45 }^{ 0 },\angle NBP={ 60 }^{ 0 }$

Now,

$\angle BAP=\angle MAP-\angle MAB={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$

$\angle APB=\angle APN-\angle BPN={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$ .........

$[ \angle BNP = 90^{\circ}, \angle BPN = 90- 60=30^{\circ}]$

$\therefore \triangle ABP$ is isosceles and

$\therefore AB=BP$

But

$AB=1$ kilometer,

$\therefore BP=1$ kilometer

Now

$PQ=PN+NQ=PN+BM$

$=BP\sin { { 60 }^{ 0 } } +AB\sin { { 30 }^{ 0 } }$ .........

$\because sin(\Theta ) = \dfrac {Opposite}{Hypotenuse}$ for right

$\angle \triangle$

$\displaystyle =1.\frac { \sqrt { 3 } }{ 2 } +1.\frac { 1 }{ 2 } =\frac { \sqrt { 3 } +1 }{ 2 } m$

The height of the mountain is

$\dfrac { \sqrt { 3 } +1 }{ 2 } m$ .

$6$ ft-tall man finds that the angle of elevation of the top of a

$24$ ft-high pillar and the angle of depression of its base are complementary angles.The distance of the man from the pillar is

0%
$\displaystyle 2\sqrt{3}\ ft$
0%
$\displaystyle 8\sqrt{3}\ ft$
0%
$\displaystyle 6\sqrt{3}\ ft$
0%
none of these

Explanation

Let the distance between the tower and the man be x ft

$\Rightarrow AP=CQ=x\ ft$
In rt

$\Delta BCQ$

$\tan \alpha=\displaystyle \dfrac{18}{x}$ ....(1)

In rt

$\Delta ACQ$

$\tan (90-\alpha)=\displaystyle \dfrac{6}{x}$

$\Rightarrow \cot \alpha=\displaystyle \dfrac{6}{x}$ ....(2)

Multiplying (1) and (2), we get

$\Rightarrow tan(\alpha)\times cot(\alpha)=\dfrac{18\times 6}{x\times x}$

$\Rightarrow 1=\dfrac{18\times 6}{x^2}$

$\Rightarrow x^2=36\times 3$

$\Rightarrow x=\sqrt{36\times 3}$

$\Rightarrow x=6\sqrt{3}\ ft$

A piece of paper in the shape of a sector of a circle of radius 10cm and of angle

$\displaystyle 216^{\circ}$ just covers the lateral surface of a right circular cone of vertical angle

$\displaystyle 2\theta$ .Then

$\displaystyle \sin\: \theta$ is

0%
$\displaystyle \frac{3}{5}$
0%
$\displaystyle \frac{4}{5}$
0%
$\displaystyle \frac{3}{4}$
0%
none of these

As observed from the top of a 60 m high lighthouse from the sea-level, the angles of depression of two ships are

${30}^{o}$ and

${45}^{o}$ . If one ship is exactly behind the other on the same side of the lighthouse. Find the distance between the two ships.

0%
$20\ m$
0%
$46.5\ m$
0%
$54.9\ m$
0%
$60.1\ m$

Explanation

From the given figure

$tan\ 45^{o}=\dfrac{75}{y}$
Hence

$y=75\ m$
And

$tan\ 30^{o}=\dfrac{75}{75+x}$

$75+x=75\sqrt{3}$

$x=75(\sqrt{3}-1)\ m=54.9 \ m$ =Distance between the ships.

Which of the following statements is true?

$I$ . The angle of elevation of the top of a hill at the foot of the tower is

$60^o$ and the angle of elevation of the top of the tower from the foot of the hill is

$30^o$ . If the tower is

$50\ m$ high, then height of the hill is

$150\ m$ .

$II$ . An aeroplane flying horizontally

$1\ km$ above the ground is observed at an angle

$60^o$ . After

$10\ \text{seconds}$ its elevation changes to

$30^o$ . Then the speed of the aeroplane is

$527.04\ \text{km/h}$ .

$III$ . A man in a boat rowing away from light house

$100\ m$ high takes

$2\ \text{minutes}$ to change the angle of elevation of the top of the light house from

$60^o$ to

$30^o$ . Then the speed of the boat is

$40\ \text{m/minute}$ .

0%
$I$
0%
$II$
0%
$III$
0%
$I \ \&\ III$

Explanation

$I$

Let the height of the hill be

$'h'm$

$\&$ distance between the foot of hill and tower be

$'x'm.$

Now, In

$\triangle ACB,$

$\Rightarrow \tan { 60° } =\dfrac { h }{ x }$

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

$\Rightarrow x=\dfrac { h }{ \sqrt{3} }\longrightarrow (2)$

$\triangle DBC,$

$\Rightarrow \tan { 30° } =\dfrac { 50 }{ x }$

$\Rightarrow \dfrac{1}{\sqrt {3}}=\dfrac { 50 }{ x }$

$\Rightarrow x=50\sqrt {3}$

From

$\left( 1 \right) \;\xi\;\left( 2 \right),$ we get

$\Rightarrow \dfrac { h }{ \sqrt{3} }=50\sqrt {3}$

$\Rightarrow h=150m$

$II$

Let the distance travelled in

$2$ min be

$'y'm.$

$'x'$ be the distance initially.

Now, In

$\triangle EDC,$

$\Rightarrow \tan { 60° } =\dfrac { EC }{ DC }=\dfrac { 1000 }{ x }$

$\Rightarrow \sqrt {3} =\dfrac { 1000 }{ x }$

$\Rightarrow x=\dfrac { 1000 }{ \sqrt {3} }m \longrightarrow (1)$

Now, In

$\triangle ADB,$

$\Rightarrow \tan { 30° } =\dfrac { AB }{ DB }=\dfrac { 1000 }{ x+y }$

$\Rightarrow \dfrac { 1 }{ \sqrt {3} }=\dfrac { 1000 }{ \dfrac{1000}{\sqrt{3}}+y }$

$[\;From\;(1)\;]$

$1000 \sqrt {3}=\dfrac{1000}{\sqrt{3}}+y$

$\Rightarrow y=1000 \sqrt {3}-\dfrac{1000}{\sqrt{3}}$

$y=\dfrac{3000-1000}{\sqrt{3}}=\dfrac{2000}{\sqrt{3}}=\dfrac{2000\sqrt{3}}{3}m$

Now,

$y$ is the distance travelled in

$2\;min.$

$\therefore$ Speed of plane

$=\dfrac{Distance}{Time}=\dfrac{2000\sqrt{3}}{6}m/min.$

$=\dfrac{2000\sqrt{3}\times 60}{6\times 1000}km/h$

$=20\sqrt{3}km/h$

$III$

Let

$'y'$ be the initial distance of boat from light house.

$'x'$ be the distance travelled in

$2\;min.$

Now, In

$\triangle ACB,$

$\Rightarrow \tan { 60° } =\dfrac { AB }{ BC }=\dfrac { 100 }{ y }$

$\Rightarrow \sqrt {3} =\dfrac { 1000 }{ y }$

$\Rightarrow y=\dfrac { 100 }{ \sqrt {3} }m \longrightarrow (1)$

Now, In

$\triangle ADB,$

$\Rightarrow \tan { 30° } =\dfrac { AB }{ BD }=\dfrac { 100 }{ x+y }$

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

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

$[\;From\;(1)\;]$

$\Rightarrow x=100 \sqrt {3}-\dfrac{100}{\sqrt{3}}=\dfrac{300-100}{\sqrt{3}}=\dfrac{200}{\sqrt{3}}$

$\Rightarrow x=\dfrac{200\sqrt{3}}{3}m$

Now, Speed of the boat

$=\dfrac{Distance\;'x\;'}{Time}=\dfrac{200\sqrt{3}}{2\times 3}=\dfrac{100\sqrt{3}}{3}m/min.$

Hence the answer is only

$I$ statement is true.

The angles of depression of two boats as observed from the mast-head of a ship

$50m$ high are

$45^0$ and

$30^0.$ Tho distance between the boats, if they are on the same side of mast head in line with it, is

0%
$50 \sqrt{3}m$
0%
$50(\sqrt{3}+1)$ m
0%
$50(\sqrt{3}-1)$ m
0%
$5(1-\dfrac{1}{\sqrt{3}})$ m

Explanation

Consider the above given figure
Let

$AB$ be the mast of the ship and

$DC$ be the distance between the boats
Consider triangle

$ABC$

$\angle ACB=45^0$
Therefore,

$\angle ABC=45^0$
Hence triangle

$ABC$ is a an isosceles triangle.
Therefore

$AB=AC=50$ m
Consider triangle

$ABD$ ,

$\tan 30^0=\dfrac{1}{\sqrt{3}}$

$=\dfrac{AB}{AD}=\dfrac{50}{50+CD}$

$\Rightarrow 50+CD=50\sqrt{3}$

$\Rightarrow CD=50(\sqrt{3}-1)$ m
Hence, the answer is option C.

A vertical lamp-post, 6 m high, stands at a distance of 2 m from a wall, 4 m high. A 1.5-m-tall man starts to walk away from the wall on the other side of the wall, in line with the lamp-post.The maximum distance to which the man can walk remaining in the shadow is

0%
$\displaystyle \displaystyle \frac{5}{2}m$
0%
$\displaystyle \frac{3}{2}m$
0%
$\displaystyle 4m$
0%
none of these

Explanation

From similar

$\triangle ABQ$ and

$\triangle CDQ$ , we get

$\cfrac { BR }{ DR } =\cfrac { AB }{ CD } =\cfrac { 4 }{ \cfrac { 3 }{ 2 } } =\cfrac { 8 }{ 3 }$ ...(1)
Now, from similar

$\triangle PQR$ and

$\triangle ABR$ , we get

$\cfrac { PQ }{ AB } =\cfrac { QR }{ BR } \Rightarrow \cfrac { 6 }{ 4 } =\cfrac { BQ+BR }{ BR } =\cfrac { 2+BR }{ BR }$

$\Rightarrow BR=4$ ...(2)
Substituting (2) in (1), we get

$\cfrac { 4 }{ DR } =\cfrac { 8 }{ 3 } \Rightarrow DR=\cfrac { 3 }{ 2 }$ ...(3)
From (2) and (3)

$BD=BR-DR=4-\cfrac { 3 }{ 2 } =\cfrac { 5 }{ 2 }$

The angle of elevation

$\theta$ of a vertical tower from a point

$A$ on the ground is such that its tangent is

$\displaystyle\frac{5}{12}$ . On walking

$192\ m$ towards tower in the same straight line, the tangent of the angle of elevation

$\phi$ formed to be

$\displaystyle\frac{3}{4}$ . The height of the tower is

0%
$810$ $m$
0%
$108$ $m$
0%
$180$ $m$
0%
$801$ $m$

Explanation

Let the height of the tower be

$h$ .
Let the distance between the foot of the tower and the point till which the man walks

$192 m$ be

$y$ .
Hence

$\tan\theta=\dfrac{h}{192+y}=\dfrac{5}{12}$

$12h=960+5y$ ..(i)

$\tan(\phi)=\dfrac{h}{y}=\dfrac{3}{4}$

$4h=3y$ ...(ii)
Substituting in i, we get

$9y=960+5y$

$4y=960$

$y=240\ m$
Hence

$h=\dfrac{3}{4}\times240$

$=180\ m$

At the foot of the mountain the elevation of its summit is

${45}^{0}$ ; after ascending

$1000\ m$ towards the mountain up a slope of

${30}^{0}$ inclination, the elevation is found to be

${60}^{0}$ . The height of the mountain is

0%
$\displaystyle \frac { \sqrt { 3 } +1 }{ 2 }$ $km$
0%
$\displaystyle \frac { \sqrt { 3 } -1 }{ 2 }$ $km$
0%
$\displaystyle \frac { \sqrt { 3 } +1 }{ 2\sqrt { 3 } }$ $km$
0%
None of these

Explanation

Let

$P$ be the summit of the mountain and

$Q$ be the foot.

Let

$A$ be the first position and

$B$ the second position of observation.

$BN$ and

$BM$ are perpendiculars from

$B$ to

$PQ$ and

$AQ$ respectively.

Then

$AB=1000 m=1 km$

$\angle MAB={ 30 }^{ 0 };\angle MAP={ 45 }^{ 0 };\angle NBP={ 60 }^{ 0 }$

Now,

$\angle BAP=\angle MAP-\angle MAB={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$

$\angle APB=\angle APN-\angle BPN={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$

$\therefore \triangle ABP$ is isosceles and

$\therefore AB=BP$

But

$AB=1$ kilometer,

$\therefore BP=1$ kilometer

Now,

$PQ=PN+NQ=PN+BM$

$=BP\sin { { 60 }^{ 0 } } +AB\sin { { 30 }^{ 0 } }$

$\displaystyle =1.\frac { \sqrt { 3 } }{ 2 } +1.\frac { 1 }{ 2 } =\frac { \sqrt { 3 } +1 }{ 2 }$

$km$

From a window (

$h$ metres high above the ground) of a house in a street, the angle if elevation and depression of the top and the foot of another house on the opposite side of the street are

$\theta$ and

$\phi$ respectively.
Find the height of opposite house if

$\theta={60}^{o}, \phi={45}^{o}, h=120$ metres.

0%
$317.84m$
0%
$366.97m$
0%
$301.44m$
0%
None of these

Explanation

Let

$L$ be the window of a house

$LM$ . Let

$PQ$ be the height of the opposite house such that,

$PQ=$ height of the opposite house

$=H$ metres. Then

$LM=h$ metres. Through

$L$ draw

$LN\bot PQ$ , meeting

$PQ$ at

$N$ . As per condition of the problem,

$\angle PLN=\theta, \angle NLQ=\phi$ , Let

$PQ=H$ metres then

$PN=(H-h)$ metres. Let

$LN=MQ=x$ metres
(

$\because NQ=LM=h$ metres)
From the right

$\triangle LNP$ ,

$\cfrac{PN}{LN}=\tan {\theta}$

$\cfrac{H-h}{x}=\tan{\theta}$

$x=\cfrac{H-h}{\tan{\theta}}.......(1)$
From the right

$\triangle LNQ$

$\cfrac{NQ}{LN}=\tan{\phi}$

$\cfrac{h}{x}=\tan{\phi}$

$\cfrac {h}{\tan{\phi}}=x.........(2)$
From

$(1)$ and

$(2)$ , we get:

$\cfrac{H-h}{\tan{\theta}}=\cfrac{h}{\tan{\phi}}$

$\Rightarrow$

$H\tan{\phi}-h\tan{\phi}=h\tan{\theta}$

$\Rightarrow$

$H\tan{\phi}=h(\tan{\phi}+\tan{\theta})$
Required height of opposite house.

$=H=h(1+\tan{\theta}\cot{\phi})$
For

$\theta={60}^{o}, \phi={45}^{o},h=120m$ ,
The height of the opposite house

$H=120(1+\tan{{60}^{o}}\cot{{45}^{o}}=120(1+\sqrt 3\times 1)$

$H=120(1+1.732)=120\times 2.732=327.84m$

A tree

$10(2+\sqrt 3)$ metres high is broken by the wind at a height

$10\sqrt 3$ metres from its root in such a way that top struck the ground at certain angle and horizontal distance from the root of the tree to the point where the top meets the ground is

$10m$ . Find the angle of elevation made by top of the tree with the ground.

0%
${80}^{o}$
0%
${60}^{o}$
0%
${40}^{o}$
0%
${20}^{o}$

Explanation

Consider the given figure,

$AC$

$=10(2+\sqrt{3})-10\sqrt{3}$

$=20m$

$=Hypotenuse.$
Hence

$\sin(\angle C)$

$=\dfrac{AB}{AC}$

$=\dfrac{10\sqrt{3}}{20}$

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

$=\sin60^{0}$
Hence

$\angle C= 60^{0}$
Hence the angle of elevation made by top of the tree with the ground is

$60^{0}$ .

Two housed are collinear with the base of a tower and are at distances 3 m and 12 m ( on the same side ) from the base of the tower. The angles of elevation form these two houses of the top of the tower are complementary. what is the height of the tower?

0%
4 m
0%
6 m
0%
7.5 m
0%
36 m

Explanation

Let the height of tower =

$h$

Angle of elevation at 3 m distance =

$\alpha$

Angle of elevation at 12 m distance =

$90 - \alpha$

Now, The line of sight, the ground and the tower will form a right angled triangle.

Thus,

$\tan \angle$ of elevation =

$\dfrac{height}{distance}$

Thus, considering the 3 m distance,

$\tan \alpha = \dfrac{h}{3}$ (1)

Considering the 12 m distance,

$\tan (90 - \alpha) = \dfrac{h}{12}$ (2)

Multiplying the two equations,

$\tan \alpha \tan (90 - \alpha) = \dfrac{h}{3}. \dfrac{h}{12}$

$\tan \alpha \cot \alpha = \dfrac{h}{3}. \dfrac{h}{12}$

$1 = \dfrac{h^2}{36}$

$h = 6$ m

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

${30}^{o}$ , 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

${60}^{o}$ . Find the time taken by the car to reach the foot of the tower.

0%
3 seconds
0%
18 seconds
0%
2 minutes
0%
5 minutes

Explanation

Let

$PQ=h$ meters be the height of the tower.

$P$ is the top of the tower.

$PX$ is horizontal line through

$P$ .

The first and second positions of the car are at

$A$ and

$B$ respectively.

(Angle of depression of the car when observed at

$B$ ) and

$\angle BPX={60}^{o}$

Let the speed of the car be

$x \ m/second$ ,

Time required to travel A to B is 6 second

Then distance

$AB=6 x$ meters. .....

$[Distance = speed \times time]$

Let time taken from

$B$ to

$Q$ be

$n$ second.
Then distance

$BQ=n \ x$ meters.

We know,

$tan(\Theta ) = \dfrac {Opposite}{Adjacent}$

In right

$\triangle PAQ$ ,

$\cfrac{h}{6x+nx}=\tan {{30}^{o}}=\cfrac{1}{\sqrt 3}$

$\Rightarrow$

$h=\cfrac {(n+6)x}{\sqrt 3}........(1)$

In right

$\triangle PBQ$

$\cfrac{h}{nx}=\tan {{60}^{o}}=\sqrt 3$

$\Rightarrow$

$h= nx\sqrt 3........(2)$

From

$(1)$ and

$(2)$ we have

$\cfrac {(n+6)x}{\sqrt 3}=nx\sqrt 3$

$\Rightarrow n+6=n\sqrt 3\times \sqrt 3$

$\Rightarrow 3n=n+6$

$\Rightarrow 2n=6$

$\Rightarrow n=3$

Hence, the time from

$B$ to

$Q=3$ seconds

A boy standing on a horizontal plane finds a bird flying at a distance of

$100\ m$ from him at an elevation of

${30}^{o}$ . A girl standing on the roof of

$20\ m$ high building, finds the angle of elevation of the same bird to be

${45}^{o}$ . Both the boy and the girl are on opposite sides of the bird. Find the distance of the bird from the girls.

0%
$44.62\ m$
0%
$52.35\ m$
0%
$36.92\ m$
0%
$42.42\ m$

Explanation

The position of the boy is at point B of elevation

$30^{o}$ and that of the girl is at point G of elevation

$45^{o}$ .

$\triangle PQB$

$sin\ 30^{o}=\dfrac{PQ}{PB}$

$=>\dfrac{1}{2}=\dfrac{PQ}{100}$

$=>PQ=50\ m$

Now,

$PS=PQ-SQ$

$=PQ-GR$

$=(50-20)\ m$

$=30\ m$

$\triangle PSG$

$\sin 45^{o}=\dfrac{PS}{PG}$

$=>\dfrac{1}{\sqrt 2}=\dfrac{30}{PG}$

$=>PG=30\sqrt 2\ m$

$=>PG=3\times 1.41429$ [since

$\sqrt 2=1.41429$ ]

$=>PG=42.42$

The distance of the bird from the girls is

$42.42\ m$

$1.2m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of

$88.2m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is

${60}^{o}$ . After some time, the angle of elevation reduces to

${30}^{o}$ . Find the distance travelled by the balloon during the interval.

0%
$17\sqrt 2m$
0%
$34m$
0%
$58\sqrt 3m$
0%
$67m$

Explanation

From the figure, we have

$\angle POQ={30}^{o}$ is the angle of elevation for the first position of the balloon.

Let

$OQ=y \ m$ .

We are given that

$AC=88.2m; AB=88.2=1.2=87m$

For the second position of the balloon, we have

$\angle POQ={30}^{o}$ .

Let

$OB=xm$ .

We have to find

$D=BQ=(y-x)$

We know,

$tan(\Theta ) = \dfrac {Opposite}{Adjacent}$

$\cfrac{AB}{OB}=\tan {60}^{o}$ and

$\cfrac {PQ}{OQ}=\tan {30}^{o}$

$\Rightarrow$

$\cfrac{87}{x}=\sqrt 3$ and

$\cfrac{87}{y}=\cfrac{1}{\sqrt 3}$

$\Rightarrow$

$x=\cfrac{87}{\sqrt 3}m$ and

$y=87\sqrt 3 m$

Then

$D=y-x={87\sqrt 3-\cfrac{87}{\sqrt 3}}m$

$=87\times \left(\sqrt 3-\cfrac {1}{\sqrt 3}\right)m$

$=87\times \cfrac{2}{\sqrt 3}m=87\times \cfrac{2}{3}\times \sqrt 3m$

$=\cfrac {174}{3}\sqrt 3m=58\sqrt 3m$

The distance traveled by the balloon is

$58\sqrt 3m$ .

The angle of elevation of the top of the tower is

${ 45 }^{ \circ }$ on walking up a slope inclined at an angle of

${ 30 }^{ \circ }$ the horizontal a distance 20 my, the angle of elevation of top of tower is observed to be

${ 60 }^{ \circ }$ . Then height of the tower

0%
$10\left( \sqrt { 3 } +1 \right) mt$
0%
$20\left( \sqrt { 3 } +1 \right) mt$
0%
$100\sqrt { 3 } mt$
0%
$50\left( 3+\sqrt { 3 } \right) mt$

A 10 meters high tower is standing at the centre of an equilateral triangle and each side of the triangle subtends an angle of

${60^o}$ at the top of the tower. Then the length of each side of the triangle is

0%
$5m$
0%
$5\sqrt 6 m$
0%
$4\sqrt 6$
0%
$4m$

$1.5m$ tall boy is standing at some distance from a

$30m$ tall building. The angles of elevation from his eyes to the top of the building increases from

$30$ to

${60}^{o}$ as he walks towards the building. The distance he walked towards the building is:

0%
$19\sqrt 3m$
0%
$57\sqrt 3m$
0%
$38\sqrt 3m$
0%
$18\sqrt 3m$

Explanation

Consider the given figure:

$\tan30^{0}=\dfrac{28.5m}{AB+BC}$

Hence,

$AB+BC=28.5\sqrt{3}m$ ...(i)

And,

$\tan60^{0}$

$=\dfrac{28.5m}{BC}$

$=\sqrt{3}$

Hence,

$BC=\dfrac{28.5}{\sqrt{3}}$

Hence,

$AB=28.5\sqrt{3}-BC$

$=28.5\sqrt{3}-\dfrac{28.5}{\sqrt{3}}$

$=28.5\left[\dfrac{3-1}{\sqrt{3}}\right]$

$=28.5\left[\dfrac{2}{\sqrt{3}}\right]$

$=\dfrac{57}{\sqrt{3}}$

$=19\sqrt{3}m$

A straight highway leads to the foot of a tower of height

$50m$ . From the top of the tower, the angles of depression of two cars standing on the highway are

${30}^{o}$ and

${60}^{o}$ . What is the distance between the two cars and how far is each car from the tower?

0%
$22.31m$ ; $28.86m$ ; $86.60m$
0%
$522.31m$ ; $35.24m$ ; $86.60m$
0%
$57.73m$ ; $35.24m$ ; $40.19m$
0%
None of these

Explanation

Let the first car be x distance away from the foot of the tower.
Let distance between the first and second car be y.
Hence applying trigonometric ratios we get

$\tan60^{0}=\sqrt{3}$

$=\dfrac{50}{x}$
Hence

$x=\dfrac{50}{\sqrt{3}}=28.86m$
And

$\tan30^{0}=\dfrac{1}{\sqrt{3}}=\dfrac{50}{x+y}$
Hence

$50\sqrt{3}=x+y$
Or

$50\sqrt{3}=\dfrac{50}{\sqrt{3}}+y$

$y=50(\sqrt{3}-\dfrac{1}{\sqrt{3}})$

$y=50(\dfrac{2}{\sqrt{3}}$

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

$=57.73m$
Hence the distance of the second car from the foot of the tower is

$=57.73+28.86$

$=86.59m$

A tree is broken at certain height and its upper part

$9\sqrt2m$ long not completely separated meet the ground at an angle of

${45}^{o}$ . Find the height of the tree before it was broken and also find the distance from the root of the tree to the point where the top of the tree meets the ground.

0%
$8\sqrt 2$ ; $11m$
0%
$7(\sqrt 2-1)$ ; $15m$
0%
$9(\sqrt 2+1)$ ; $9m$
0%
None of these

Explanation

Consider the below triangle

$h=9\sqrt{2}sin45^{0}$

$=\dfrac{9\sqrt{2}}{\sqrt{2}}$

$=9m$
Hence the total height will be

$9\sqrt{2}+9$

$=9(\sqrt{2}+1)m$
Now

$\tan45^{0}=1$

$=\dfrac{h}{BC}$
Hence

$h=BC=9m$
=The distance from the root of the tree to the point where the top of the tree meets the ground.

A man stands on the ground at a point A, Which is on the same horizontal plane as B, the foot of a vertical Pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground He observes the angle of clevation at C, the top of the pole as

$\displaystyle x^{\circ}$ ,
where tan

$\displaystyle x^{\circ}=\frac{2}{5}.$ The distance AB ( in meters ) is

0%
15 m
0%
18 m
0%
20 m
0%
16 m

Explanation

Height of the pole = 10 m
height of the man's eye above the ground = 2 m
Let the man's eye be at point A' and similar point on the pole be B'
Now, In

$\triangle A'B'C$

$\tan x = \dfrac{B'C}{A'B'}$

$\dfrac{2}{5} = \dfrac{8}{A'B'}$

$A'B' = 20$ m

Now,

$AB = A'B' = 20$ m

The elevation of the top of a hill at each of the three angular points X, Y, Z of a horizontal

$\displaystyle \Delta XYZ$ is

$\displaystyle \alpha$ , then the height of the hill is

0%
$\displaystyle \frac{YZ.\tan \alpha. \mathrm{cosec} \:X}{2}$
0%
$\displaystyle X Z. \tan \alpha .\mathrm{cosec}\: Y$
0%
$\displaystyle \frac{XY.\cot \alpha }{2\sin Z}$
0%
$\displaystyle \frac{XY.\tan \alpha }{2\sin Y}$

Horizontal distance between two pillars of different height is 60 m. it was observed that the angular elevation form form the top of the shorter pillar to the top of the taller pillar is

$\displaystyle 45^{\circ}$ if the height of taller pillar is 130 m, the height of the shorter pillar

0%
45 m
0%
70 m
0%
80 m
0%
60 m

Explanation

Height of bigger pilar = 130 m
Horizontal distance between the pillars =

$60$ m
Angle of elevation from top of smaller pillar to top of bigger pillar =

$\alpha$ =

$45^{\circ}$
Thus,

$\tan \alpha = \dfrac{difference \quad of \quad height}{horizontal \quad distance}$

$\tan 45 = \dfrac{difference \quad of \quad height}{60}$

$difference \quad of \quad height = 60$ m

Thus, height of smaller pillar = height of larger pillar - difference in heights
=

$130 - 60 = 70$ m

From the top of a hill, the angles of depression of two consecutive kilometer stones due east are found to be

$30^o$ and

$45^o$ . Find the height of the hill

0%
$1.365$ km
0%
$1.5$ km
0%
$1.7$ km
0%
$1.1$ km

Explanation

Let the distance between the nearer kilometre stone and the hill be ' $x$ ' km.

So, the distance between the farther kilometre stone and the hill is ' $1+x$ ' km since both are on the same side of the hill.

In triangle APB,

$\tan 45 ^0=\dfrac hx$

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

$\Rightarrow$ $h=x$

In triangle AQB,

$\tan 30 ^0=\dfrac {h}{1+x}$

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

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

From equation 1,

$1+h=\sqrt { 3 } h \Rightarrow$ $1=\sqrt { 3 } h-h$

$\Rightarrow$ $h=\dfrac { 1 }{ \sqrt { 3 } -1 }$

$\Rightarrow$ $h=1.365km$

Hence option A is correct.

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is

$\dfrac{5}{12}$ . On walking

$160$ metres towards the tower, the tangent of the angle of elevation is

$\dfrac{3}{4}$ . The height of the tower is equal to

0%
$150$ m
0%
$180$ m
0%
$200$ m
0%
None

Explanation

Let

$h$ be the height of the tower.

Given, at 1st point on the ground,

$\tan { \alpha } =\dfrac { 5 }{ 12 }$ ...given

$\therefore \tan { \alpha } =\dfrac { h }{ x+160 }$

Now we can write

$\dfrac { 5 }{ 12 } =\dfrac { h }{ x+160 }$

$12h=5x+800$ ...(1)
At second point on the ground,

$\tan { \beta } =\dfrac { 3 }{ 4 }$ ...given

$\tan { \beta } =\dfrac { h }{ x }$

where,

$x$ is the distance of 2nd point on the ground from tower
Now we can write

$\dfrac { h }{ x } =\dfrac { 3 }{ 4 }$

$\therefore x=\dfrac { 4h }{ 3 }$ ...(2)
Put the value of

$x$ from eq 2 in eq 1, we get

$h=150m$

On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are

$45^o$ and

$60^o$ . If the height of the tower is

$50\sqrt 3$ , then the distance between the objects is

0%
$\displaystyle\frac{50}{\sqrt 3}$ m
0%
$150$ m
0%
$50(\sqrt 3-1)$ m
0%
$25(\sqrt 3-1)$ m

Explanation

Let AB be the tower which is $50\sqrt { 3 } m$ high and C and D be the positions of the two towers such that

$\angle ADB=45 ^0$ and $\angle ACB=60 ^0$

Now, in triangle ABC,

$\tan {60 ^0}=\dfrac {AB}{BC}$

$\Rightarrow$ $\sqrt { 3 } =\dfrac { 50\sqrt { 3 } }{ BC }$

$\Rightarrow$ $BC=50m$

Also, in triangle ABD,

$\tan 45 ^0=\dfrac {AB}{BD}$

$\Rightarrow \tan 45 ^0=\dfrac {AB}{BC+CD}$

$\Rightarrow$ $1=\dfrac { 50\sqrt { 3 } }{ 50+CD }$

$\Rightarrow$ $50+CD=50\sqrt { 3 }$

$\Rightarrow$ $CD=50\sqrt { 3 } -50$

$\Rightarrow$ $CD=50\left( \sqrt { 3 } -1 \right) m$ .

Hence option C is correct.

The upper part of a tree is broken over by the wind makes an angle of

$\displaystyle 30^{\circ}$ with the ground and the distance from the root to the point where the top of the tree meets the ground is 15 m The height of the broken part is

0%
$\displaystyle 15\sin 30^{\circ}m$
0%
$\displaystyle 15\cos 30^{\circ}m$
0%
$\displaystyle 15\tan 30^{\circ}m$
0%
$\displaystyle 15\sec 30^{\circ}m$

Explanation

Let the length of the broken part =

$l$
Angle made by the broken part with the ground =

$30^{\circ}$
the distance on the ground =

$15$ m
The ground, the broken part and the unbroken part forms a right angled triangle.
Hence,

$\cos 30 = \dfrac{15}{l}$

$l = \dfrac{15}{\cos 30^{\circ}}$

$l = 15 \sec 30^{\circ}$ m

The angle of elevation of an aeroplane from a point on the ground is

$\displaystyle 45^{\circ}$ . After a flight of

$15$ sec the elevation changes to

$\displaystyle 30^{\circ}$ . If the aeroplane is flying at a height of

$3000$ metres, find the speed of the aeroplane.

0%
$304$ km/hr
0%
$457$ km/hr
0%
$321.37$ km/hr
0%
$527.04$ km/hr

Explanation

Let the point in the ground is

$E$ which is y metres from point

$B$ and let after

$15$ sec flight, it covers

$x$ metres distance
In

$\displaystyle \Delta AEB$

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

$\displaystyle \therefore 1=\frac{3000}{y}$

$\displaystyle \therefore y=3000m$ .......(i)

$\displaystyle \Delta CED$

$\displaystyle \tan 30^{\circ}=\frac{CD}{ED}$

$\displaystyle \therefore \frac{1}{\sqrt{3}}=\frac{3000}{x+y}$ (

$\displaystyle \because$

$AB = CD$ )

$\displaystyle \therefore x+y=3000\sqrt{3}$ .......(ii)

From equation (i) and (ii)

$\displaystyle x+3000=3000\sqrt{3}$

$\displaystyle \therefore x=3000\sqrt{3}-3000$

$\displaystyle \therefore x=3000\left ( \sqrt{3}-1 \right )$

$\displaystyle \therefore$

$x=3000(1.732 - 1)$

$\displaystyle \therefore$

$x=3000(0.732)$

$\displaystyle \therefore$

$x=2196 m$

Speed of Aeroplane =

$\displaystyle \frac{\text{Distance covered}}{\text{Time taken}}$

$=\displaystyle \frac{2196}{15}$ m/sec = 146.4 m/sec

$=\displaystyle \frac{2196}{15}\times \frac{18}{5}$ km/hr

$=527.04$ km/hr
Hence the speed of aeroplane is

$527.04$ km/hr

If the angle of elevation of a cloud from a point 200 metres above a lake is

$\displaystyle 30^{\circ}$ and the angle of depression of its reflection inthe lake is

$\displaystyle 60^{\circ}$ then the height of the cloud (n metres ) above the lake is

0%
200
0%
300
0%
500
0%
None of these

Explanation

Assumptions:

Let the dotted line represent the lake and pt A be the position of the cloud.

Pt. C is a point 200m above the lake.

To find:

Height of cloud from the dotted line (lake).

Solution:

$\displaystyle \Delta ABC$

$\displaystyle \tan 30^{\circ}=\frac{h}{x}$
x =

$\displaystyle h\sqrt{3}$ .......(i)
In

$\displaystyle \Delta BDC$

$\displaystyle \tan 60^{\circ}=\frac{h+400}{x}$

$\displaystyle \sqrt{3}x=h+400$ .......(ii)
from (i) and (ii)

$\displaystyle h\sqrt{3}.\sqrt{3}=h+400$
3h - h = 400
h =

$\displaystyle \frac{400}{2}=200$
h = 200
So, height of object above the lake = h + 200 = 200 + 200 = 400m

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 14 - 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 14 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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