MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 5

The angle of elevation of the top of a lamp-post as observed from a point

$40 m$ distant from the foot of the post, is

$30^0.$ The height of the lamp post is

0%
$40\sqrt{3}$ m
0%
$\dfrac{40\sqrt{3}}{3}$ m
0%
$20 m$
0%
$\dfrac{40\sqrt{3}}{2}$

Explanation

Consider the given triangle

Let AB be the height of the lamp post denoted by h

Let BC bet the horizontal distance of the point from the foot of the lamp post

Given that angle

$C=30^0$

Therefore

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

$=\dfrac{AB}{BC}$

$=\dfrac{h}{40}$

$h=\dfrac{40}{\sqrt3}$

Rationalising the denominator we get

$h=\dfrac{40\sqrt{3}}{3}$

Hence answer is B

A light-house

$100$ m high observe that two ships are approaching it from West and South respectively. If the angles of depression of the two ships are

$30$

$^{\circ}$ and

$45$

$^{\circ}$ respectively, then the distance between the two ships is -

0%
$100(\sqrt{3}+1)$ m
0%
$100(\sqrt{3}+1)^2$ m
0%
$200$ m
0%
$400$ m

Explanation

we can observe that in

$\triangle ABC$

$\tan45^o=\dfrac{100}{a}\Rightarrow a=100$

Similarly, in

$\triangle ACD$

$\tan30^o=\dfrac{100}{b}\Rightarrow b=100\sqrt3$

So, in

$\triangle BCD$ (Top View)

$a^2+b^2=d^2\Rightarrow d^2=10000+30000=40000\Rightarrow d=200$

An electric pole

$10$ m high is tied to three steel wires which are fixed to the ground at points which are at equal distances from the foot of the pole. These points, if joined form an equilateral triangle. If the steel wires are each inclined to the ground at

$60$

$^{\circ}$ , then each side of the equilateral triangle measures

0%
$10$ m
0%
$\dfrac {10}{\sqrt 3}$ m
0%
$10$ $\sqrt{3}$ m
0%
None of the above

Explanation

Let C, D, F be the points at which are steel wires are fixed to the ground. B is the foot of the pole.

From fig. (ii),

$\cot 60^{\circ} = \cfrac{CB}{AB} \therefore \cfrac{1}{\sqrt{3}} = \cfrac{CB}{10 }$ or

$CB = \cfrac{10}{\sqrt{3}}$

From fig. (i),

$\cfrac{CE}{CB} = \cos 30^{\circ}$

$\therefore CE = \cfrac{CB \times \sqrt{3}}{2} = \cfrac{10}{2} = 5, CD = 10$

Calculate the height AD.

0%
12.12
0%
8.96
0%
5.14
0%
18.24

Explanation

$tan 75^{\circ} = \frac{AD}{2.4}$

$tan 75^{\circ} = tan (45^{\circ} + 30^{\circ})$

$\displaystyle =\frac{tan 45^{\circ} + tan 30^{\circ}}{1 - tan 45^{\circ} tan 30^{\circ}} = \frac{1+\frac{1}{\sqrt{3}}}{1- \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$

$=\displaystyle \frac{(\sqrt{3} + 1)^2}{3-1} = \frac{1+3 + 2 \sqrt{3}}{2} = 2 + \sqrt{3} = 3.732$

$AD = 2.4 \times tan 75^{\circ} = 8.96$

A man standing between two vertical posts finds that the angle subtended at his eyes by the tops of the posts is a right angle. If the height of the two posts are two times and four times the height of the man and the distance between the poles is equal to the height of the largest pole, then the ratio of the distance of the man from the shorter and the longer post is

0%
$3 :1$
0%
$2 : 3$
0%
$3 : 2$
0%
$1 : 3$

Explanation

Let

$AB$ be the men of height

$h\ m$ ,

$CD$ and

$EF$ be the two towers of height

$2h\ m$ and

$4h\ m$ respectively.

$\therefore \angle CAE={ 90 }^{ o }$

Draw

$PQ\parallel DF$ passing through

$A$ .

$\Rightarrow CP=PD=h\ m,QF=h\ m$

$\therefore EQ=3h\ m$

Let

$DB$ be

$d\ m$

$\therefore DB=PA=d\ m$ and

$BF=AQ=\left( 4h-d \right) m$

PAQ is a straight line.

$\therefore \angle PAC+\angle CAE+\angle EAQ={ 180 }^{ o }$

$\angle PAC+{ 90 }^{ o }+\angle EAQ={ 180 }^{ o }$

$\therefore \angle PAC+\angle EAQ={ 90 }^{ o }$

Let

$\angle PAC$ be

$x$ .

$\therefore \angle EAQ={ 90 }^{ o }-x$ ...(1)

$\triangle CPA,\angle PAC=x$

$\Rightarrow \angle CPA={ 90 }^{ o }-x$ ...(2)

$\triangle CPA$ and

$\triangle AQE$

$\angle PCE=\angle EAQ$

$\left[ each{ 90 }^{ o }-x \right]$

$\angle CPA=\angle EAQ$

$\left[ each{ 90 }^{ o } \right]$

$\therefore \triangle CPA\sim \triangle AQE$

$\displaystyle\Rightarrow \frac { CP }{ AQ } =\frac { PA }{ QE } =\frac { CA }{ AE }$ (ESST)

$\displaystyle\therefore \frac { h }{ 4h-d } =\frac { d }{ 3h }$

$\Rightarrow 3{ h }^{ 2 }=d\left( 4h-d \right)$

$\Rightarrow 3{ h }^{ 2 }=4dh-{ d }^{ 2 }$

$\left( d-3h \right) \left( d-h \right) =0$

$d-3h=0$ or

$d=h$

when

$d=3h$

$\displaystyle\frac { DB }{ BF } =\frac { 3h }{ h } =\frac { 3 }{ 1 } =3:1$

when

$d=h$

$\displaystyle\frac { DB }{ BF } =\frac { h }{ 3h } =\frac { 1 }{ 3 } =1:3$

Henc,e option

$D$ is correct.

A vertical lamp-post of height 9 meters stands at the corner of a rectangular field.The angle of elevation of its top from the farthest corner is

$\displaystyle 30^{\circ}$ ,while from another corner it is

$\displaystyle 45^{\circ}.$ The area of the field is

0%
$\displaystyle 9\sqrt{2}\: meter\:^{2}$
0%
$\displaystyle 81\sqrt{3} \:meter\:^{2}$
0%
$\displaystyle 81\sqrt{2}\: meter\:^{2}$
0%
$\displaystyle 9\sqrt{3}\: meter\:^{2}$

Explanation

Consider rectangle

$ABCD$ be the rectangular field. Let

$PD$ be the height of vertical lamp post.

it is given that Height of the Pole

$=PD=9$ m

Now in

$\triangle PDA$

$tan45^o=\dfrac{PD}{DA}=\dfrac{9}{DA}\Rightarrow DA=9$

Now in

$\triangle PDB$

$tan30^o=\dfrac{9}{DB}\Rightarrow DB=9\sqrt3$

Also, we can see that

$\triangle DAB$ is Right Angled Triangle

$\therefore DA^2+AB^2=DB^2\Rightarrow 9^2+AB^2=(9\sqrt3)^2\Rightarrow AB^2=81\times2\Rightarrow AB=9\sqrt2$

Now, Area of Rectangle

$=DA\times AB=9\sqrt2\times 9=81\sqrt2$

$m^2$

Therefore, option (C) is the correct answer.

When a eucalyptus tree is broken by strong wind, its top strikes the ground at an angle of

$30^{\circ}$ and at a distance of

$15$ m from the foot. What is the height of the tree?

0%
$15$ $\sqrt{3}$
0%
$10$ $\sqrt{3}$ m
0%
$20$ m
0%
$10$ m

Explanation

Let the height of the tree be

$A'C.$

$'h'$ is the height of the tree from the broken point to the bottom.

Now, In

$\triangle BAC,$

$\Rightarrow \tan { 30° } =\dfrac { AC }{ BC } =\dfrac { h }{ 15 }$

$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ 15 }$

$\Rightarrow h=\dfrac { 15 }{ \sqrt { 3 } } m$

Again,

$\cos { 30° } =\dfrac { BC }{ BA } =\dfrac { 15 }{ BA }$

$\Rightarrow \dfrac { \sqrt { 3 } }{ 2 } =\dfrac { 15 }{ BA }$

$\Rightarrow BA=\dfrac { 30 }{ \sqrt { 3 } } m$

Now, the height of tree

$=AB+AC$

$=\dfrac { 30 }{ \sqrt { 3 } } +\dfrac { 15 }{ \sqrt { 3 } }$

$=\dfrac { 45 }{ \sqrt { 3 } }$

$=\dfrac { 45\sqrt { 3 } }{ 3 }$

$=15\sqrt { 3 } m$

Hence, the answer is

$15\sqrt { 3 } m.$

STATEMENT -

$1$ : The angle of elevation of a point viewed is the angle formed by the line of sight with the horizontal when the point being view is above the horizontal level.
STATEMENT -

$2$ : Then the angle of depression of a point view is the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level.

0%
Statement - $1$ is True, Statement - $2$ is True, Statement - $2$ is a correct explanation for Statement - $1$
0%
Statement - $1$ is True, Statement - $2$ is True : Statement $2$ is NOT a correct explanation for statement - $1$
0%
Statement - $1$ is True, Statement - $2$ is False
0%
Statement - $1$ is False, Statement - $2$ is True

Explanation

Both the statements are true but the second statement is not a correct explanation of the first statement.

Hence, the answer is option

$B.$

A person, standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is

$60^{\circ}$ when he retreats

$20$ m from the bank, he finds the angle to be

$30^{\circ}$ . The height of the tree and the breadth of the river.

0%
$10$ $\sqrt{3}$ m, $10$ m
0%
$10$ ; $10$ $\sqrt{3}$ m
0%
$20$ m, $30$ m
0%
None of these

Explanation

$\textbf{Step-1: Draw the appropriate diagram & label it.}$

$\text{Let AB be the breadth of the river and BC be the height of the}$

$\text{tree which makes a}$

$\angle$

$\text{of 60}$

$^{\circ}$

$\text{at a point A on the opposite bank.}$

$\text{Let D be the position of the person after retreating 20 m from the bank.}$

$\text{Let AB}$

$= x$

$\text{metres and BC}$

$= h$

$\text{metres.}$

$\text{From right angled}$

$\triangle$

$\text{ABC and DBC,}$

$\text{we have,}$

$\tan 60^{\circ} = \cfrac{BC}{AB}$

$\text{and}$

$\tan 30^{\circ} = \cfrac{h}{20+x}$

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

$\text{and}$

$\cfrac{1}{\sqrt{3}} = \cfrac{h}{x+20}$

$\textbf{Step-2: Use the above results & get the required unknown.}$

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

$\text{and}$

$h = \cfrac{x+20}{\sqrt{3}}$

$\Rightarrow x \sqrt{3} = \cfrac{x+20}{\sqrt{3}} \Rightarrow 3x = x+20 \Rightarrow x=10 m$

$\text{Putting}$

$x=10$

$\text{in}$

$h = \sqrt{3}x$ ,

$\text{we get,}$

$h= 10 \sqrt{3} = 17.32 m$

$\textbf{Hence, the height of the tree = 17.32 m and the breadth of the river =10 m.}$

2164625_107879_ans_cf2ba335827a4565b211ba417e448083.png

A tree is broken by the wind. The top struck the ground at an angle of

$30^{\circ}$ and at a distance of 30 m from the root, then the whole height of the tree.

$(\sqrt3 =1.73)$ is 51.9 m
If true then enter

$1$ and if false then enter

$0$

0%
TRUE
0%
FALSE
0%
can't say
0%
NONE

Explanation

Angle made by the broken part with ground =

$30^{\circ}$
broken tree is at distance (d) = 30 m from the roots
Let broken part be of length = b
and unbroken part of length = u
Now, the broken part, ground and the unbroken part of the tree forms a right angled triangle
Thus,

$\tan 30 = \dfrac{u}{d}$

$u = \dfrac{30}{\sqrt{3}}$

$u = 10\sqrt{3}$

$\sin 30 = \dfrac{u}{b}$ $

$\dfrac{1}{2} = \dfrac{10\sqrt{3}}{b}$

$b = 20\sqrt{3}$
Thus, length of tree =

$u + b$ =

$10\sqrt{3} + 20\sqrt{3} = 30\sqrt{3}= 51.96$ m

A kite is attached to a

$100\ m$ long string. Find the greatest height reached by the kite when its string makes an angle of

$\displaystyle 60^{\circ}$ with the level ground.

0%
$86.6$
0%
$45.7$
0%
$63.8$
0%
$72.0$

Explanation

The highest height the kite reaches be P
The length of string is L = 100 m
The angle made by the string with the ground =

$60^{\circ}$

now,

$\sin 60 = \dfrac{P}{L}$

$P = \dfrac{100 \sqrt{3}}{2}$

$P = 50 \sqrt{3}$

$P = 86.6$ m

Choose the correct option for the following.

A kite is flying at a height of

$60$ m above the ground. The string attached to the kite is temporarily tied to the ground. The inclination of the string with the ground is

$60^{\circ}$ , then the length of the string, assuming that there is no slack in the string is

$69.2$ m.

$(\sqrt3 = 1.73)$

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Height of the kite =

$60$ m

Angle made by the string with the ground, =

$60^{\circ}$ .

The kite, string, and the ground form a right-angled triangle

Thus,

$\sin 60 = \dfrac{60}{string}$

$\dfrac{\sqrt{3}}{2} = \dfrac{60}{string}$

$string= \dfrac{120}{\sqrt{3}}$

Length of string

$= 69.2$ m

Choose the correct option for the following.

From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be

$60^{\circ}$ . If the height of the lighthouse is 90 meters then that ship is 51.9 m from the lighthouse.

$\displaystyle (\sqrt3 = 1.73)$ .

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Given, height of lighthouse =

$90$ m

Angle of depression =

$60^{\circ}$

The angle of elevation from the ship to the top of lighthouse = angle of depression =

$60^{\circ}$

Now,

$\tan 60 = \dfrac{height}{distance}$

$\sqrt{3} = \dfrac{90}{distance}$

$distance = 51.9$ m

Two buildings are in front of each other on either side of a road of width

$10$ metres. From the top of the first building, which is

$30$ metres high, the angle of elevation of the top of the second is

$45^{\circ}$ . What is the height of the second building (in metres)?

0%
$10$
0%
$30$
0%
$40$
0%
$25$

Explanation

$CD=EF=10$

we can see that

$CDFE$ is a rectangle,

$\therefore CE=DF=30\\ \& \ CD=EF=10$

Now, in

$\triangle ACD$

$\tan45^o=\dfrac{AC}{10}\Rightarrow AC=10$

Therefore, height of the second Building

$=AC+CE=10+30=40$

Therefore, Answer is

$40$

A ladder is placed against a vertical tower. If the ladder makes an angle of

$\displaystyle 30^{\circ}$ with the ground and reaches up to a height of

$15\ m$ of the tower; find length of the ladder in cm.

0%
2311
0%
3000
0%
1688
0%
1200

Explanation

Consider the ladder and wall system as the right angled triangle, such that length of wall (W) till the ladder reaches be the perpendicular= 15 cm and angle (G) made by ladder (L) with the ground is

$30^{\circ}$ ,
and length of ladder be the Hypotenuse.
Thus,

$\sin G = \dfrac{P}{H}$

$\sin 30 = \dfrac{W}{L}$

$\dfrac{1}{2} = \dfrac{15}{L}$

$L = 30\ m=3000\ cm$

Two poles of height 18 metres and 7 metres are erected on the ground. A wire of length

$22$ metres tied to the top of the poles. Find the angle made by the wire with the horizontal.(in degree)

0%
$30$
0%
$10$
0%
$20$
0%
$15$

Explanation

Angle of elevation from smaller pole to bigger =

$x$
Height of smaller building

$= 7$ m
Height of bigger pole

$= 18$ m
Length of the rope

$= 22$ m
Difference between the two buildings

$= 11$ m

Now,

$\sin x = \cfrac{opposite\ side}{hypotaneous}$

$\sin x = \cfrac{11}{22} = \dfrac{1}{2}$

$\sin x = \sin 30$

$x = 30^{\circ}$

Determine

$x$ .

0%
$x=4m$
0%
$x=5m$
0%
$x=6m$
0%
$x=7m$

Explanation

Let

$AB$ be the tree broken by the wind at the point

$C$ . Its top

$B$ strike the ground at the point

$D$ such that

$\angle CDA=\theta$ and

$CB$ takes the position of

$CD$ i.e.,

$CD=CB=y$ metres. But

$AC=x$ metres then the height of the tree

$=AB=(x+y)=15m, CD=y=(15-x)$ metres

$ADC$ is rt. angles

$\triangle$ at

$A$

${CD}^{2}={AC}^{2}+{AD}^{2}$

$\Rightarrow$

${(15-x)}^{2}={x}^{2}={(5\sqrt 3)}^{2}={x}^{2}+75$ (Pythagoras theorem)

$\Rightarrow$

$225+{x}^{2}-30x={x}^{2}+75$

$\Rightarrow$

$30x=150m$

$\Rightarrow$

$x=5m$

$\therefore$ The tree is broken at a height of

$5m$ from the bottom or ground.

A man

$1.8m$ tall stands at distance of

$3.6m$ from a lamp post and casts a shadow of

$5.4m$ on the ground. Find the height of the lamp post.

0%
$2m$
0%
$5m$
0%
$3m$
0%
$7m$

Explanation

Let

$\theta$ be the unknown angle of elevation.
Let DF be the man and AB be the lamp post.
Considering the given figure.

$\tan\theta$

$=\dfrac{1.8}{5.4}$

$=\dfrac{x}{3.6}$
Hence

$x=\dfrac{(3.6).(1.8)}{5.4}$

$=1.2m$
Hence
The height of the lamp post is

$1.2m+1.8m$

$=3m$

A lamp post

$\displaystyle 5\sqrt{3}$ m high casts a shadow 5 long on the ground. The Sun's elevation at this point is

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

Explanation

Given,

Height of the post is

$5\sqrt 3$ and base is

$5$ m

$\tan θ=\dfrac{AB}{BC}$

$\therefore \tan θ=\dfrac{5\sqrt 3}{5}$

$=\sqrt 3$

∴

$\therefore θ=60^{0}$ ............ [since

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

find the height of the tower from the cliff.

0%
$30$ metres
0%
$60$ metres
0%
$80$ metres
0%
$40$ metres

Explanation

Let

$PM$ be the height of the cliff whose height is

$60$ metres.

$P$ be its top such that

$PM=$ height of cliff

$=60m$ . Let

$BA$ be the tower. Through

$P$ draw

$PC$ horizontally then by the conditions of th e questions.

$\angle CPA={30}^{o}$ and

$\angle CPB={60}^{o}$
Let the required height of the tower

$AB$ or

$BA$

$=h$ metres.
From

$A$ draw

$AD\bot PM$ , then:

$\angle CPA=alt. \angle DAP$

$\Rightarrow$

$\angle DAP={30}^{o}$

$\angle CPB=alt. \angle MBP$

$\Rightarrow$

$\angle MBP={60}^{o}$
Now

$AD=BM....(1)$ [Note this step carefully]

$DP=MP-MD=MP-BA=(60-h)$ metres
From the right

$\triangle ADP$ ,

$\cfrac{AD}{DP}=\cot {{30}^{o}}=\sqrt 3$

$AD=DP.\sqrt 3$

$AD=(60-h)\sqrt 3........(2)$
And from the right

$\triangle BMP$

$\cfrac{BM}{MP}=\cot {{60}^{o}}=\cfrac{1}{\sqrt 3}$

$BM=\cfrac{MP}{\sqrt 3}=\cfrac{60}{\sqrt 3}$

$BM=20\sqrt 3...........(3)$
Using

$(2)$ and

$(3)$ in

$(1)$ , we get:

$AD=BM$

$\Rightarrow$

$(60-h)\sqrt 3=20\sqrt 3$

$\Rightarrow$

$h=40m$

Two poles of equal heights are standing opposite to each other on either side of a road, which is

$80$ metres wide. From a point between them on the road, the angles of elevation of their top are

${30}^{o}$ and

${60}^{o}$ . Find the position of the point and also the height of the poles.

0%
$50m$
0%
$35m$
0%
$20m$
0%
$10m$

Explanation

Let

$AB$ and

$CD$ be two poles of equal height standing opposite to each other on either side of the road

$BD$

$\Rightarrow$

$AB=CD=h$ metres.
Let

$P$ be the observation point on the road

$BD$ . The angles of elevation of their top are

${30}^{o}$ and

${60}^{o}$ .

$\angle APB={30}^{o}$ ,

$\angle CPD={60}^{o}$
The width of the road

$=BD=80m$ , let

$PD=x$ metres
Then

$BP=(80-x)$ metres consider right

$\triangle CDP$ , we have:

$\cfrac{CD}{PD}=\tan{{60}^{o}}$

$\Rightarrow$

$\cfrac{h}{x}=\sqrt 3 \Rightarrow h=\sqrt 3x.........(1)$
In right

$\triangle ABP$ we ave:

$\cfrac{AB}{BP}=\tan{{30}^{o}} \Rightarrow \cfrac{h}{80-x}=\cfrac{1}{\sqrt 3} \Rightarrow h=\cfrac{80-x}{\sqrt 3}......(2)$
From

$(1)$ and

$(2)$ we get

$h=\sqrt 3x$

$h=\cfrac{80-x}{\sqrt 3}$

$\Rightarrow$

$(80-x)=3x \Rightarrow 4x=80 \Rightarrow x=20$
Height of each pole

$=AB=CD=\sqrt 3.x=20\sqrt 3=20(1.732)=34.64$ metres
Position of point

$P$ is

$20$ from the first and

$60m$ from the second pole.
i.e., the position of the point

$P$ is

$20m$ from either of the poles.

the height of the post.

0%
$5\sqrt2m$
0%
$15m$
0%
$7\sqrt3m$
0%
$19m$

Explanation

Let

$AB$ be the vertical post its shadow is

$21m$ when the altitude of the sun is

${30}^{o}$

$BC=21m,\angle ACB={30}^{o},AB=h$ metres (figure)

$ABC$ is rt.

$\triangle,\cfrac{AB}{BC}=\tan{{30}^{o}}$

$\Rightarrow$

$\cfrac{h}{21}=\cfrac{1}{\sqrt 3}$

$\Rightarrow$

$h=\cfrac{21}{\sqrt 3}=\cfrac{7\times \sqrt 3\times \sqrt 3}{\sqrt 3}=7\sqrt 3m$

$\Rightarrow$

$AB=h$ , Height of the pole

$=7\sqrt 3m$

A boy

$1.7$ m tall, is

$25$ m away from a tower and observes the angle of elevation of the top of the tower to be

${60}^{o}$ . Find the height of the tower.

0%
$55$ m
0%
$45$ m
0%
$20$ m
0%
$35$ m

Explanation

Let

$AA'$ be the boy and the tower be

$B'C$
Therefore,

$AA'=1.7$ m

$=BB'$
and

$A'B'=25\ m=AB$
Now, angle of elevation is

$\angle CAB=\theta$ .
Given,

$\theta=60^o$

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

$\Rightarrow 25\tan 60^o=BC$

$\Rightarrow BC=25\sqrt3$
Now, height of tower

$=BC+BB'$

$=25\sqrt3+1.7$

$=45$ m

A captain of an aeroplane flying at an altitude of

$1000$ metres sights two ships as shown in the figure. If the angle of depression are

${60}^{o}$ and

${30}^{o}$ . Find the distance between the ships.

0%
$2210$ m
0%
$2196$ m
0%
$2219.7$ m
0%
$2309.3$ m

Explanation

Let

$A$ be the position of the captain of an aeroplane flying at the altitude of

$1000$ metres from the ground.

$AB=$ the altitude of the aeroplane from the ground

$=1000$ m

$P$ and

$Q$ be the position of two ships.
Let

$PB=x$ metres and

$BQ=y$ metres
Required:

$PQ=$ Distance between the ships

$=(x+y)$ metres

$ABP$ is right angled

$\triangle$ at

$B$

$\cfrac{AB}{PB}=\tan {{60}^{o}}$

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

$x=\cfrac{1000(1.732)}{3}=577.3$

$m$

$ABQ$ is right angled

$\triangle$ at

$B$

$\cfrac{AB}{BQ}=\tan {{30}^{o}}$

$\cfrac{1000}{y}=\cfrac{1}{\sqrt 3} \Rightarrow y=1000\sqrt 3$

$y=1000(1.732)=1732$ m
Required distance between the ships

$=(x+y)$ metres

$=(577.3+1732)$ m

$=2309.3$ m

The length of a string between a kite and a point on the roof of a building

$10m$ high is

$180m$ . If the string makes an angle

$\theta$ with the level ground such that

$\tan {\theta}=\dfrac43$ , how high is the kite from the ground?

0%
$154\ m$
0%
$176\ m$
0%
$198\ m$
0%
$214\ m$

Explanation

$\tan\theta=\dfrac{4}{3}$
Hence

$\sin\theta=\dfrac{4}{5}$ and

$\cos\theta=\dfrac{3}{5}$
Hence the perpendicular height of the kite from the rooftop of the building is

$=180\sin\theta$

$=180.\dfrac{4}{5}$

$=36(4)=144m$
Hence the height of the kite from the ground will be

$=$ Height of the building

$+$ height of the kite from the rooftop of the building.

$=144+10$

$=154\ m$

The Shadow of a tower when the angle of elevation of the sun is

$\displaystyle 30^{\circ}$ is found to be

$5 m$ longer than when it was

$\displaystyle 45^{\circ}$ then the height of tower in meter is

0%
$\displaystyle \frac{5}{\sqrt{3}+1}$
0%
$\displaystyle \frac{5}{2}(\sqrt{3}-1)$
0%
$\displaystyle \frac{5}{2}(\sqrt{3}+1)$
0%
None of these

Explanation

Let the height of the tower be

$h$

$\displaystyle \therefore$ In

$\displaystyle \Delta ABC$

$\displaystyle \tan 45^{\circ}=\frac{h}{x}$

$\displaystyle \therefore x=h$
In

$\displaystyle \Delta ABD$

$\displaystyle \tan 30^{\circ}=\frac{h}{x+5}$

$\therefore \displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x+5}$

$\therefore h + 5 = \displaystyle h\sqrt{3}$

$\therefore \displaystyle h\left ( \sqrt{3}-1 \right )=5$

$\therefore h = \displaystyle \frac{5\left ( \sqrt{3}+1 \right )}{\sqrt{3}-1\left ( \sqrt{3}-1 \right )}=\frac{5}{2}\left ( \sqrt{3}+1 \right )m$

The angle of elevation of a cloud from a point

$200$ metres above a lake is

${30}^{o}$ and the angle of depression of the reflection of the cloud in the lake is

${60}^{o}$ . Find the height of the cloud.

0%
$=400m$
0%
The height of the cloud $=280m$
0%
The height of the cloud $=340m$
0%
None of these

Explanation

Let

$QR$ be the surface of the lake and

$P$ be the point of observation which is

$200m$ above

$Q$
i.e.,

$PQ=200m$ . Let

$C$ be the position of the cloud.
Through

$C$ draw

$CR$ perpendicular to th esurface of the lake and

$D$ the reflection of the cloud in the lake then

$CR=RD=h$ metres. (Assume) (laws of relection)
Through

$P$ draw

$PQ\bot$ on the surface of lake and

$PM\bot CD$ .

$\angle CPM={30}^{o},\angle MPD={60}^{o}$

$MR=PQ=200m$ , Let

$PM=x$ metres

$CM=CR-MR=(h-200m)$ , (

$\because CR=h$ metres)

$DM=DR+RM=(h+200m)$ , (

$\because DR=h$ metres)
In rt.

$\triangle CMP$

$\cfrac{CM}{PM}=\tan {{30}^{o}}$

$\cfrac{h-200}{x}=\cfrac{1}{\sqrt 3}$

$x=\sqrt 3(h-200)........(1)$
From rt.

$\triangle PMD$ ,

$\cfrac{MD}{PM}=\tan {{60}^{o}}$

$\cfrac{h+200}{x}=\sqrt 3$

$x=\cfrac{h+200}{\sqrt 3}........(2)$
From

$(1)$ and

$(2)$ , we get

$\sqrt (h-200)=\cfrac{h+200}{\sqrt 3}$

$\Rightarrow$

$3(h-200)=h+200$

$\Rightarrow$

$3h-h=600+200=800$

$\Rightarrow$

$2h=800$

$\Rightarrow$

$h=\cfrac{800}{2}=400$
Required height of the cloud

$=h=CR=400m$

What is the angle of elevation of a vertical flagstaff of height

$100\sqrt 3m$ from a point

$100m$ from its foot.

0%
${18}^{o}$
0%
${30}^{o}$
0%
${48}^{o}$
0%
${60}^{o}$

Explanation

Height of flagstaff (H) =

$100 \sqrt{3}$
Distance between the point and the flagstaff (D) = 100 m
Let the angle of elevation be

$\theta$
Then,

$\tan \theta = \dfrac{D}{H}$

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

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

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

$\theta = 60^{\circ}$

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

$\displaystyle 30^{\circ}$ with the ground if the length of the broken part is 20 metres then remaining part of the tree is of length

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

Explanation

$\triangle ABC$

$\sin30^o=\dfrac{x}{20}\Rightarrow x=10 \ meters$

The angle of elevation of the top of a tower as seen from two points

$A$ &

$B$ situated the same line and at distance '

$p$ ' and '

$q$ ' respectively from the foot of the tower are complementary, then height of the tower is

0%
$pq$
0%
$\displaystyle \frac{p}{q}$
0%
$\displaystyle \sqrt{pq}$
0%
none of these

Explanation

Let

$OT$ be the tower of height

$h$
Let

$O$ be the base of the tower
Let

$A$ and

$B$ be two points on the through the base such that

$OA = p$ and

$OB = q$
In

$\displaystyle \Delta AOT$

$\displaystyle \tan \alpha =\frac{OT}{OA}=\frac{h}{p}$ .....(i)
In

$\displaystyle \Delta BOT$

$\displaystyle \Rightarrow \tan \left ( 90-\alpha \right )=\frac{OT}{OB}=\frac{h}{q}$ or

$\cot \alpha =\dfrac{h}{q}$ ......(ii)
Multiplying (i) and (ii), we have

$\displaystyle \Rightarrow \tan \alpha \cot \alpha =\frac{h}{p}\times \frac{h}{q}$

$\displaystyle \Rightarrow \tan \alpha* \frac{1}{\tan \alpha }=\frac{h^{2}}{pq}$

$\displaystyle \Rightarrow 1=\frac{h^{2}}{pq}$

$\displaystyle \Rightarrow h^{2}=pq$

$\displaystyle \Rightarrow h=\sqrt{pq}$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 5 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 5 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.