MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 3

$A$ man standing on a level plane observes the elevation of the top of a pole to be

$\alpha$ . He then walks a distance equal to double the height of the pole and then finds that the elevation is now

$2\alpha$ . Then

$\alpha=$

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$30^{\circ}$
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$15^{\circ}$
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$60^{\circ}$
0%
$45^{\circ}$

Explanation

Let,

$OP$ be the height of tower

$h$ and

$\angle OAP=\alpha,\angle OBP=2\alpha$

$AB=2h$ (Given) and

$AB=OA-OB\dots 1$

$\triangle OBP$ :

$\dfrac{OB}{OP}=\cot 2\alpha$

$OB=OP \cot 2\alpha$

$OB=h\cot2\alpha$

$\triangle OAP$ :

$\dfrac{OA}{OP}=\cot \alpha$

$OA=OP \cot \alpha$

$OA=h\cot \alpha$

Substituting the values of

$AB$ ,

$OA$ and

$OB$ in equation

$1$ :

$2h=h\cot { \alpha } -h\cot { 2\alpha }$

$2=\dfrac{\cos \alpha}{\sin \alpha}-\dfrac{\cos 2\alpha}{\sin 2\alpha}$

$\cos { \alpha } \sin { 2\alpha } -\sin { \alpha } \cos { 2\alpha } =2\sin { \alpha } \sin { 2\alpha }$

$\sin { \alpha } =2\sin { \alpha } \sin { 2\alpha }$

$\displaystyle \sin { 2\alpha } =\frac { 1 }{ 2 } \quad \quad \quad \left( \because \sin { \alpha } \neq 0 \right)$

$\displaystyle 2\alpha =\frac { \pi }{ 6 } \Rightarrow \alpha =\frac { \pi }{ 12 } ={ 15 }^{ \circ}$

A person walking along a straight road towards a hill observes at two points distance

$\sqrt{3}$ km, the angle of elevation of the hill to be

$30^{0}$ and

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

0%
$\dfrac{3}{2}$ km
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$\sqrt{\dfrac{2{1}}{3{2}}}$
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$\displaystyle \frac{(\sqrt{3}+1)}{2}$ km
0%
$\sqrt{3}$ km

Explanation

$\tan 60= \dfrac{h}{d}$

$\tan 30= \dfrac{h}{ \sqrt{3}+d }$

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

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

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

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

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

$A$ man observes a tower

$AB$ of height

$h$ from a point

$P$ on the ground. He moves a distance

$d$ towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance

$\dfrac {3d}{4}$ in the same direction and the angle of elevation is three times that at

$P$ . Then

$\displaystyle \frac{h^{2}}{d^{2}}=$

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$\dfrac {35}{9}$
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$\dfrac {35}{36}$
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$\dfrac {36}{5}$
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$\dfrac {36}{35}$

Explanation

$\triangle BQR$ apply sine formula

$\displaystyle\frac { d }{ \sin { \left( \pi -3\alpha \right) } } =\frac { \dfrac { 3 }{ 4 } d }{ \sin { \alpha } } =4\sin { \alpha } -3\sin { 3\alpha }$

$\Rightarrow 4\sin { \alpha } =3\left( 3\sin { \alpha } -4\sin ^{ 3 }{ \alpha } \right)$
Since

$\displaystyle\sin { \alpha } \neq 0,4=9-12\sin ^{ 2 }{ \alpha } \Rightarrow \sin ^{ 2 }{ \alpha } =\frac { 5 }{ 12 }$
From

$\triangle BQA$

$h=BQ\sin ^{ 2 }{ \alpha } =2d\sin { \alpha }\cos { \alpha } =2d\sin { d }\sqrt { 1-\sin ^{ 2 }{ \alpha } }$

$\displaystyle=2d\sqrt { \frac { 5 }{ 12 } } \sqrt { 1-\frac { 5 }{ 12 } } =\frac { 2d }{ 12 } \sqrt { 35 }$

$\Rightarrow 36{ h }^{ 2 }=35{ d }^{ 2 }$

$\Rightarrow \dfrac{h^2}{d^2}=\dfrac{35}{36}$

An observer finds that the angular elevation of a tower is

$\theta$ . On advancing

$a$ metres towards the tower, the elevation is

$45^{0}$ and on advancing

$b$ metres nearer the elevation is

$90^{0}-\theta$ , then the height of the tower (in metres) is

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$\displaystyle \frac{ab}{a+b}$
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$\displaystyle \frac{ab}{a-b}$
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$\displaystyle \frac{2ab}{a+b}$
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$\displaystyle \frac{2ab}{a-b}$

Explanation

Let

$b$ be the initial point and

$AE$ be the height of the tower.
In,

$\triangle ADE,$

$\Rightarrow \tan { \left( 90°-\theta \right) } =\dfrac { h }{ x }$

$\Rightarrow \cot { \theta } =\dfrac { h }{ x }$

$\Rightarrow x=h\tan { \theta } \longrightarrow (1)$

$\Rightarrow \tan { \theta } =\dfrac { x }{ h }$

In,

$\triangle ACE,$

$\Rightarrow \tan { 45° } =\dfrac { h }{ b+x }$

$\Rightarrow b+x=h$

$\Rightarrow b+h\tan { \theta } =h$ [From

$(1)$ ]

$\Rightarrow b=h\left( 1-\tan { \theta } \right) \longrightarrow (2)$

In,

$\triangle ABE,$

$\Rightarrow \tan { \theta } =\dfrac { h }{ a+b+x }$

$\Rightarrow \tan { \theta } =\dfrac { h }{ a+h }$

$\Rightarrow \dfrac { x }{ h } =\dfrac { h }{ a+h }$

$\Rightarrow \dfrac { h-b }{ h } =\dfrac { h }{ a+h }$

$\Rightarrow \left( h-b \right) \left( a+h \right) ={ h }^{ 2 }$

$\Rightarrow ah-ab+{ h }^{ 2 }-hb={ h }^{ 2 }$

$\Rightarrow h\left( a-b \right) =ab$

$\Rightarrow h=\dfrac { ab }{ a-b }$

Hence, the answer is

$\dfrac { ab }{ a-b }.$

A pole of height

$h$ stands at one corner of a park in the shape of an equilateral triangle. If

$\alpha$ is the angle which the pole subtends at the midpoint of the opposite side, the length of each side of the park is:

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$\displaystyle \frac{\sqrt{3}}{2}h\cot\alpha$
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$\displaystyle \frac{2}{\sqrt{3}}h\cot\alpha$
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$\displaystyle \frac{\sqrt{3}}{2}h\tan\alpha$
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$\displaystyle \frac{2}{\sqrt{3}}h\tan\alpha$

C is the mid point of the line joining two pionts A,B on the ground. A tower at C slightly leans towards B. If the angles of elevation of the top of the tower from A and B are

$30^{0},60^{0}$ respectvely, the angle made by the tower with the horizontal is

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$45^{0}$
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$60^{0}$
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$75^{0}$
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$30^{0}$

$A$ tower standing at point

$A$ leans towards west making an angle

$\alpha$ with the vertical. The angular elevation of

$B$ , the top most point of the tower is

$\beta$ as observed from a point

$C$ due east of

$A$ at a distance

$d$ from

$A$ . lf the angular elevation of

$B$ from a point due east of

$C$ at a distance

$2d$ from

$C$ is

$\gamma$ , then

$2\tan\alpha$ can be written as

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3 $\cot\beta-2\cot\gamma$
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$\cot\gamma -\cot\beta$
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$3 \cot\beta-\cot\gamma$
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$\cot\beta-3\cot\gamma$

Explanation

Let

$AB,$ be the tower leaning towards west making an angle

$\alpha$ with vertical at

$C$ , angle of elevation of

$B$ is

$\beta$

$abd$ at

$D$ us

$\gamma$

$CA=AD=d$ in

$\triangle ABH$

$\tan { \alpha } =\cfrac { AH }{ h }$

$\Rightarrow AH=h\tan { \alpha } ......... 1$

$\triangle BCH,\tan { \beta } =\cfrac { h }{ CH }$

$\Rightarrow CH=h\cot { \beta }$

$d-AH=h\cot { \beta } ,$ put the value of AH from

$1$

$d-h\tan { \alpha } =h\cot { \beta } \Rightarrow d=h\left( \cot { \beta } +\tan { \alpha } \right) ....... 2$

$\triangle BDH,\tan { \gamma } =\cfrac { BH }{ HD } =\cfrac { h }{ AH+d } =AH+d=h\cot { \gamma }$

$\Rightarrow h\tan { \alpha } +d=h\cot { \gamma } \Rightarrow d=h\left( \cot { \gamma } -\tan { \alpha } \right) \rightarrow 3$

From

$2$ and

$3$ we get

$h\left( \cot { \beta } +\tan { \alpha } \right) =h\left( \cot { \gamma } -\tan { \alpha } \right)$

$2\tan { \alpha } =\cot { \gamma } -\cot { \beta }$

Assertion (

$A$ ): ladder rests against a wall at an angle

$30^{0}$ to the horizontal. Its foot is pulled away through a distance

$x$ ' so that it slides a distance

$y$ ' down the wall finally making an angle

$60^{0}$ with the horizontal then

$x=y$ .
Reason (

$R$ ):

$A$ ladder rests against a wall at angle

$\alpha$ to the horizontal. Its foot is pulled a way through a distence

$a$ ' so that it slides a distence

$b$ ' down the wall, finally making an angle

$\beta$ with the horizonal then

$\displaystyle \tan(\frac{\alpha+\beta}{2})=b/a$

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Both $A$ and $R$ are ture and ` $R$ ' is the correct

explanation of $A$
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Both $A$ and $R$ are true and ` $R$ ' is not correct

explanation of $A$
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$A$ is true but ` $R$ ' is false
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$A$ ' is false but $R$ ' is true.

Explanation

$AB=A^{\prime}B^{\prime}=l$

$\sin \alpha =\dfrac {AC}l$

$\sin \beta =\dfrac {A^{\prime}C}l$

$\sin \alpha -\sin \beta =\dfrac bl$

$\cos \beta -\cos \alpha =\dfrac al$

$\dfrac {2\sin \left(\dfrac {\alpha -\beta}2\right)\cos \left( \dfrac {\alpha +\beta}2\right)}{2\sin \left( \dfrac {\alpha +\beta}2\sin \left(\dfrac {\alpha -\beta}2\right)\right)}=\dfrac ba$

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

From the top of a tree a man observes the angle of depression of a moving car is

$30^{o}$ and after

$3$ minutes he finds the angle of depression is

$60^{o}$ . How much time will the car take to reach the tree?

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$4$ minutes
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$3$ minutes
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$1.5$ minutes
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$2$ minutes

On one side of a road of width

$d$ metres there is a point of observation

$P$ at a height

$h$ metres from the ground. If a tree on the other side of the road, makes a right angle at

$P$ , height of the tree in metres is:

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$\displaystyle \frac{h^{2}-d^{2}}{h}$
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$\displaystyle \frac{h^{2}+d^{2}}{h}$
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$\displaystyle \frac{d^{2}-h^{2}}{h}$
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$\displaystyle \frac{2d^{2}+h^{2}}{h}$

Explanation

$\triangle PBA,{ AP }^{ 2 }={ h }^{ 2 }+{ d }^{ 2 }$

$\triangle PCD,{ PC }^{ 2 }={ (H-h) }^{ 2 }+{ d }^{ 2 }$

$\triangle ADC,{ AC }^{ 2 }={ AP }^{ 2 }+{ PC }^{ 2 }$

${ H }^{ 2 }={ h }^{ 2 }+{ d }^{ 2 }+{ (H-h) }^{ 2 }+{ d }^{ 2 }$

${ H }^{ 2 }={ h }^{ 2 }+2{ d }^{ 2 }+{ H }^{ 2 }+{ h }^{ 2 }-2Hh$

$2Hh=2{ h }^{ 2 }+2{ d }^{ 2 }\Rightarrow H=\cfrac { { h }^{ 2 }+{ d }^{ 2 } }{ h }$

Flag-staff of length

$d$ stands on a tower of height

$h$ . lf at a point on the ground the angles of elevation of the tower and the top of the flag-staff be

$\alpha,\ \beta$ respectively, then

$h=$

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$\displaystyle \dfrac{d\cot\beta}{\cot\beta-\cot\alpha}$
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$\displaystyle \dfrac{d\tan\beta}{\tan\alpha-\tan\beta}$
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$\displaystyle d[\frac{\tan\alpha+\tan\beta}{\cot\alpha-\cot\beta}]$
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$\displaystyle \frac{d\tan\alpha}{\tan\beta}$

Explanation

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

$\displaystyle x = h \cot \alpha$
where

$x$ is the horizontal distance from base of the tower

$\displaystyle \tan \beta= \dfrac{d+h}{x}$
Substituting for

$x$

$\displaystyle \tan \beta= \dfrac{d+h}{ h \cot \alpha}$

$\displaystyle h(\cot \alpha \tan \beta-1)=d$

$\displaystyle \therefore h=\dfrac{d \cot \beta}{\cot \beta- \cot\alpha}$

The height of a hill is

$3,300$ metres. From the point

$D$ on the ground the angle of elevation of the top of the hill is

$60^{0}$ . A balloon is moving with constant speed vertically upwards from

$D.$ After

$5$ minutes of its movement a person sitting in it observes the angle of elevation of the top of the hill as

$30^{0}$ . The speed of the balloon is

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$2.64$ km/hr
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$26.4$ km/hr
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$22.4$ km/hr
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$2.24$ km/hr

Explanation

We can see that,

$BCDE$ is a Rectangle

So,

$BE=CD$ and

$BC=ED$

$\triangle AED$

$\tan60^o=\dfrac{AE}{ED}\Rightarrow \sqrt3=\dfrac{3300}{ED}\Rightarrow ED=\dfrac{3300}{\sqrt3}=1100\sqrt3$

$ED=BC=1100\sqrt3$

$\triangle ABC$

$tan30^o=\dfrac{AB}{BC}\Rightarrow AB=BC\tan30^o\Rightarrow 1100$

Distance Covered by Balloon is

$CD=BE=AE-AB=3300-1100=2200\ m=2.2 \ km$

So, Speed will be

$=\dfrac{2.2\ km}{\dfrac{5}{60}\ hr}=26.4 \ km/hr$

An aeroplane flying horizontally

$1$ km above the ground is observed at an elevation of

$60^{0}$ . If after

$10$ secs the elevation is observed to be

$30^{0}$ . Then the uniform speed per hour the aeroplane is

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$20\sqrt{3}\ km$
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$240\sqrt{3}\ km$
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$256\sqrt{3}\ km$
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$250\sqrt{3}\ km$

Explanation

$\triangle AED$

$tan60^o=\dfrac{1}{ED}\Rightarrow ED=\dfrac{1}{\sqrt3}$

Similarly, in

$\triangle ABC$

$tan30^o=\dfrac{1}{EC}\Rightarrow EC=\sqrt3$

Distance travelled by aeroplane in 10 seconds

$=EC-ED=\sqrt3-\dfrac{1}{\sqrt3}=\dfrac{2}{\sqrt3}$

$Speed =\dfrac{Distance}{Time}=\dfrac{\dfrac{2}{\sqrt3}}{\dfrac{10}{3600}}=240\sqrt3$

Therefore, Answer is

$240\sqrt3$

A pole 6 m high casts a shadow

$2\sqrt{3}$ m long on the ground, then the Sun's elevation is

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

Explanation

The sun's elevation angle will be opposite to the side which depicts the height of the pole, and base will be the length of the shadow.
Base=

$2\sqrt{3}\ m$
height=

$6\ m$

$\tan(\theta)=\dfrac{6}{2\sqrt3}$

$=\sqrt{3}$

$\theta=\tan^{-1}(\sqrt{3})$

$=60^0$

$\theta$ being the angle of elevation.

The angle of elevation of a tower from a point

$A$ due south of it, is

$x$ , from a point

$B$ due east of

$A$ , is

$y$ . If

$AB=l,$ then the height

$h$ of the tower is given by

0%
$\displaystyle \frac { l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } }$
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$\displaystyle \frac { l }{ \sqrt { \tan ^{ 2 }{ y } -\tan ^{ 2 }{ x } } }$
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$\displaystyle \frac { 2l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } }$
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None of these

Explanation

Let

$OP$ be the tower of height

$h$ .

In right angles

$\triangle OAP$

$\displaystyle \angle OAP=x\Rightarrow \frac { OA }{ h } =\cot { x } \Rightarrow OA=h\cot { x }$ ...(1)

In right angled

$\triangle OBP$

$\displaystyle \angle OBP=y\Rightarrow \frac { OB }{ h } =\cot { y } \Rightarrow OB=h\cot { y }$ ...(2)

In right angles

$\triangle OAB$

${ AB }^{ 2 }+{ OA }^{ 2 }={ OB }^{ 2 }\Rightarrow { l }^{ 2 }+{ h }^{ 2 }\cot ^{ 2 }{ x } ={ h }^{ 2 }\cot ^{ 2 }{ y }$

$\displaystyle \Rightarrow { h }^{ 2 }\left( \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } \right) ={ l }^{ 2 }\Rightarrow h=\frac { l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x } } }$

The angle of elevation a vertical tower standing inside a triangular at the vertices of the field are each equal to

$\theta$ . If the length of the sides of the field are

$30\ m,\ 50\ m$ and

$70\ m$ , the height of the tower is:

0%
$70\sqrt{3}\tan\theta\ m$
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$\displaystyle \frac{70}{\sqrt{3}}\tan\theta\ m$
0%
$\displaystyle \frac{50}{\sqrt{3}}\tan\theta\ m$
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$75\sqrt{3}\tan\theta\ m$

Explanation

$tan \theta = \dfrac hR$

$R= \dfrac{abc}{40}$

$\Delta = \sqrt{\left ( \dfrac{3+5+7}{2} \right )\left ( \dfrac{3+5+7}{2} \right )\left ( \dfrac{7+3-5}{2} \right )\left ( \dfrac{7+5-3}{2} \right )\times 10^4} = \dfrac{15\times 1\times 5\times 9\times 10^4}{10}$

$\Delta = 25\times 5\times 3\sqrt{3}$

$R= \dfrac{30\times 50\times 70}{4\times 25\times 5\times 3\sqrt{3}}$

$R= \dfrac{70}{\sqrt{3}}$

$h= \dfrac {70}{\sqrt{3}} tan \theta$

A vertical tower of height

$50$ meters high stands on a sloping ground. The bottom of the tower is at the same level as the middle point of a vertical flagpole. From the top of the tower, the angles of depression of the top and bottom of the flagpole are

$15^{0}$ and

$45^{0}$ respectively. The height of the flagpole is

0%
$\displaystyle \frac{50}{\sqrt{3}}m$
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$50\sqrt{3}m$
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$\displaystyle \frac{100}{\sqrt{3}}m$
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$100\sqrt{3}m$

Explanation

In,

$\triangle ABC,$

$\Rightarrow \tan { 15° } =\dfrac { 50-x }{ a }$

$\Rightarrow a=\dfrac { 50-x }{ \tan { 15° } }$

In,

$\triangle ADE,$

$\Rightarrow \tan { 45° } =\frac { 50+x }{ a }$

$\Rightarrow a=50+x$

$\Rightarrow \dfrac { \left( 50-x \right) }{ \tan { 15° } } =50+x$

$\Rightarrow 50-x=\tan { 15° } \left( 50+x \right)$

$\Rightarrow 50-x=\dfrac { \left( \sqrt { 3 } -1 \right) }{ \left( \sqrt { 3 } +1 \right) } \left( 50+x \right)$

$\Rightarrow \left( 50-x \right) \left( \sqrt { 3 } +1 \right) =\left( \sqrt { 3 } -1 \right) \left( 50+x \right)$

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

$\Rightarrow 2\sqrt { 3 } x=100$

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

$\therefore$ Height of flagpole

$=2x=\dfrac { 100 }{ \sqrt { 3 } } m$

Hence, the answer is

$\dfrac { 100 }{ \sqrt { 3 } } m.$

Two poles of height

$a$ and

$b$ stand at the centres of two circular plots which touch each other externally at a point and the two poles subtends angles of

$30^{o}$ and

$60^{o}$ respectively at this point. Then the distance between the centres of these plots is:

0%
$a+b$
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$\displaystyle \frac{3a+b}{\sqrt{3}}$
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$\displaystyle \frac{a+3b}{\sqrt{3}}$
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${a}\sqrt{3}+b$

Explanation

$\triangle ABC,\tan { 30° } =\cfrac { a }{ BC }$

$BC=\sqrt { 3 } a.......(1)$

$\triangle ECD,\tan { 60° } =\cfrac { b }{ CD }$

$CD=\cfrac { b }{ \sqrt { 3 } } .......(2)$

Required distance is equal to

$BC+CD$

$BC+CD=a\sqrt { 3 } +\cfrac { b }{ \sqrt { 3 } } =\cfrac { 3a+b }{ \sqrt { 3 } }$

Distance between poles

$=\cfrac { 3a+b }{ \sqrt { 3 } }$

Each side of square subtends an angle of

$60^{o}$ at the top of a tower of

$h$ meter height standing in the centre of the square. If

$a$ is the length of each side of the square then which of the following is/are correct?

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$2a^{2}=h^{2}$
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$2h^{2}=a^{2}$
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$3a^{2}=2h^{2}$
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$2h^{2}=3a^{2}$

Explanation

As per question, the tower is standing on the center of square

Then the tower stands the point of intersection of two diagonal of square

Then the base of square of triangle be

$\dfrac{1}{2}a$ and the diagonal of square

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

And its perpendicular to be

$h$ meters,

So in this triangle

$\therefore \tan 60^o=\dfrac{h}{\frac{a}{\sqrt{2}}}$

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

$\Rightarrow \sqrt{3}a=\sqrt{2}h$

Squaring both side we get

$3a^{2}=2h^{2}$

If the altitude of the sun is

$60^{\circ}$ , the height of a tower which casts a shadow of length 30 m is :

0%
$30\sqrt{3}m$
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15m
0%
$\dfrac{30}{\sqrt{3}}m$
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$15\sqrt{2}m$

Explanation

$\tan { 60° } =\dfrac { h }{ x }$

$\Rightarrow \sqrt { 3 } =\dfrac { h }{ 30 }$

$\Rightarrow h=30\sqrt { 3 } m$

Hence, the answer is

$30\sqrt { 3 } m.$

The ratio of the length of a pole and its shadow is 1:

$\sqrt{3}$ . The angle of elevation of the sun is :

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

Explanation

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

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

$\Rightarrow \tan { \theta } =\tan { 30° }$

$\Rightarrow \theta =30°$

Hence, the answer is

$=30°.$

If the ratio of height of a tower and the length of its shadow on the ground is

$\sqrt{3}:1$ , then the angle of elevation of the sun is

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

A pole

$10\ m$ high cast a shadow

$10\ m$ long on the ground, then the sun's elevation is:

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

Explanation

$\triangle BAC,$

$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 10 }{ 10 } =1$

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

Hence, the answer is

$45^{\circ}.$

A ladder reaches a point on a wall which is 20 m above the ground and its foot is

$\displaystyle 20\sqrt{3}$ m away from the ground. The angle made by the ladder with the wall is

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

Explanation

Let the angle made by the ladder be

$\theta$

$tan θ=\dfrac{BC}{AB}$

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

$=\sqrt 3$

$\tan θ=\sqrt 3$

∴

$θ=60^{0}$ [since

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

The measure of angle of elevation of top of tower

$75\sqrt{3}$ m high from a point at a distance of 75 m from foot of tower in a horizontal plane is :

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

Explanation

Let

$AB$ be the tower and

$C$ be the point from which angle of elevation is observed.
Let the angle of elevation be

$\theta.$

$\triangle ACB,$

$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 75\sqrt { 3 } }{ 75 }$

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

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

Hence, the answer is

$60^{\circ}.$

The given figure, shows the observation of point

$C$ from point

$A$ . The angle of depression from

$A$ is:

0%
$60^{\circ}$
0%
$30^{\circ}$
0%
$45^{\circ}$
0%
$75^{\circ}$

Explanation

Let angle of depression be

$\theta$

Now,

$\angle ACB=\theta$ {Alternate angles}

$\triangle ACB,$

$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 2 }{ 2\sqrt { 3 } } =\dfrac { 1 }{ \sqrt { 3 } }$

$\Rightarrow \tan { \theta } =\tan { 30^{\circ} }$

$\Rightarrow \theta =30^{\circ}$

Hence, the answer is

$30^{\circ}.$

The angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level is called:

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Vertical Angle
0%
Angle of Depression
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Angle of Elevation
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Obtuse Angle

The angle of elevation of the sun, when the length of the shadow of a pole is equal to its height, is

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

Explanation

Let the height of the pole be

$AB=x\ m$

Then the length of the shadow of the pole

$AB$ will be

$OB=x\ m$

Let the angle of elevation be

$\theta$ i.e.,

$\angle AOB=\theta$

In the right-angled triangle

$OAB,$ we have

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

$=\dfrac{x}{x}$

$=1$

$\Rightarrow \theta=\tan^{-1} 45^\circ$

Therefore angle of elevation is

$45^\circ$

When the angle of elevation of the sun is

$30^{\circ}$ , the length of the shadow cast by 50 m high building is

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$\dfrac{50}{\sqrt{3}}m$
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$50\sqrt{3}m$
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$25\sqrt{3}m$
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$100\sqrt{3}m$

Explanation

Let the length of shadow be

$'x'\ m$

$ACB,$

$\Rightarrow \tan { 30^{\circ} } =\dfrac { AB }{ BC } =\dfrac { 50 }{ x }$

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

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

Hence, the answer is

$50\sqrt { 3 } m.$

An aeroplane at a height of

$600$ m passes vertically above another aeroplane at an instant where their angles of elevation at the same observing point are

$60^o$ and

$45^o$ respectively. How many metres higher is the one from the other?

0%
$286.53$ m
0%
$274.53$ m
0%
$253.58$ m
0%
$263.83$ m

Explanation

Let the aeroplanes are at point

$A$ and

$D$ respectively. Aeroplane

$A$ is flying

$600$ m above the ground.
So,

$AB=600$

$\angle ACB=60^o,\angle DCB=45^o$
From

$\triangle ABC$ , we have

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

$\Rightarrow BC=\cfrac{600}{\sqrt{3}}=200\sqrt{3}$
From

$\triangle DCB,$

$\cfrac{DB}{BC}=\tan 45^o\Rightarrow DB=200\sqrt{3}$
Since, the distance

$AD = AB -BD = 600-200\sqrt{3}$

$AD =200(3-\sqrt{3})=200(3-1.7321)=253.58$ m
Hence, the distance between the two aeroplanes is

$253.58$ m.

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