MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 11

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

$30^{\circ}$ . On moving a distance of 20 meters towards the foot of the tower to a point B, the angle of elevation increases to

$60^{\circ}$ . The height of the tower is:

0%
$\sqrt{3}m$
0%
$5\sqrt{3}m$
0%
$10\sqrt{3}m$
0%
$20\sqrt{3}m$

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

$80$ m wide. From the points between them on the road, the elevation of the top of the poles are

${60^ \circ }$ and

${30^ \circ }$ respectively. Find the height of the poles.

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

Explanation

In $\Delta DPC$

$\tan p = {{CD} \over {CP}}$

$\tan {30^0}$

${1 \over {\sqrt 3 }} = {{CD} \over {CP}}$

$CP = {{CP} \over {\sqrt 3 }}$ (1)

IN $\Delta APB$

$\tan p = {{AB} \over {PB}}$

$\tan {60^0} = {{AB} \over {BP}}$

$\sqrt 3 BP = AB$

$CD = \sqrt 3 BP$ (2)

FROM (1) & (2)

${{CP} \over {\sqrt 3 }} = \sqrt 3 BP$

$CP = 3BP$

NOW , $BC = BP + CP$

$80 = BP + 3BP$

$BP = 20m$

$CP = 60m$

$CD = \sqrt 3 BP$

$= 20\sqrt 3$

A ladder rests against a wall at an angle

$\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance

$a$ slides a distance

$b$ down the wall making an angle

$\beta$ with the horizontal. Choose the correct option-

0%
$a=b\tan { \left( \dfrac { \alpha +\beta }{ 2 } \right) }$
0%
$b=a\tan { \dfrac { \alpha +\beta }{ 2 } }$
0%
$a=b\sin { \dfrac { \alpha +\beta }{ 2 } }$
0%
$b=a\sin { \dfrac { \alpha +\beta }{ 2 } }$

Explanation

$\displaystyle In \, \triangle OQS , \cos \beta = \frac{OS}{QS } \Rightarrow OS = \ell \cos \beta$

$\displaystyle In \,\triangle OPR , \cos \propto = \frac{OR}{PR} \Rightarrow OR = \ell \cos \propto$

$\displaystyle a = OS - OR = \ell (\cos \beta - \cos \propto)$

$\displaystyle In \,\triangle OPR , \sin \propto = \frac{OP}{PR} \Rightarrow OP = \ell \sin \propto$

$\displaystyle \triangle OQS , \sin \beta = \frac{OQ}{QS} \Rightarrow OQ = \ell \sin \beta$

$\displaystyle b= OP - OQ = \ell (\sin \propto - \sin \beta)$

$\displaystyle \dfrac{a}{b} = \dfrac{\cos \beta - \cos \propto}{\sin \propto - \sin \beta} = \dfrac{2 \sin \left (\dfrac{\propto + \beta}{2}\right)\sin \left (\dfrac{\propto - \beta}{2}\right)}{2 \cos\left (\dfrac{\propto + \beta}{2}\right) \sin \left(\dfrac{\propto -\beta}{2}\right)}$

$\displaystyle \frac{a}{b} = \tan [\frac{\propto + \beta}{2}]$

The angle of elevation of jet fighter from

$A$ point on the ground is

$60^0$ . After a fight of

$15$ seconds angle of elevation changes to

$30^0$ . If the jet is flying at a speed of

$720$ km/hr. Find the constant height at which jet is flying. (Take

$\sqrt3= 1.732$ ).

0%
$2.598km$
0%
$2.598m$
0%
$280m$
0%
$2.97m$

Explanation

speed=

$720\ km/h$

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

distance travelled in

$16 sec$

$200\times15$

now

$\tan30^{\circ}=\cfrac{h}{x+3000}=\cfrac{1}{\sqrt{3}}$

$\tan60^{\circ}=\cfrac{h}{x}={\sqrt{3}}$

$=\cfrac { \cfrac { h }{ x } }{ \cfrac { h }{ x+3000 } } =\cfrac { \sqrt { 3 } }{ \cfrac { 1 }{ \sqrt { 3 } } } \\ =\cfrac { x+3000 }{ x } =3\\ =x+3000=3x=2x=3000\\ =x=1500$

The constant height

$=1500\sqrt{3}$

$=1500\times1.732$

$=2598m=2.598km$

1025845_1050385_ans_7378a105a83f44898c5375fa7c641432.png

The height of the tower is

$50$ m when angle of elevation changes from

$45^0$ to

$30^0$ , the shadow of tower becomes

$x$ meters more, find the value of

$x$ .

0%
$50$
0%
$50\sqrt{30}$
0%
$50(\ 1-\sqrt3)$
0%
$25$

Explanation

let

$BC=y$

$\triangle ABC$ ,

$\tan { { 30 }^{ \circ } }$

$=\cfrac { 50 }{ y }$

$\cfrac { 1 }{ \sqrt { 3 } } =\cfrac { 50 }{ y }$

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

$\triangle ABD$ ,

$\tan { { 45 }^{ \circ } } =\cfrac { 50 }{ y+x }$

$\Rightarrow 1=\cfrac { 50 }{ y+x }$

$y+x=50 \ \Rightarrow x=50-y$

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

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

1026804_1052171_ans_ce28626b519f43b5bdd206c5ae7d96e9.png

The shadow of a vertical tower on a level ground increases by

$10$ when the altitude of the sun changes from

$45^0$ to

$30^0$ . Find the height of the tower, correct to two decimal places.

0%
$14$ m
0%
$13.69$ m
0%
$3.69$ m
0%
$14.69$ m

Explanation

$\triangle ABC$

$\tan45^o=\cfrac {x}{y}$

$\Rightarrow x=y\longrightarrow (1)$

$\triangle ABD$

$\tan 30^o=\cfrac {x}{y+10}$

$\Rightarrow \cfrac {1}{\sqrt {3}}=\cfrac {x}{x+10}$ {from equation (1)}

$\Rightarrow x+10=\sqrt {3}x$

$\Rightarrow x=\cfrac {10}{\sqrt {3}-1}$

$\because \sqrt{3}=1.732$

$\Rightarrow x= 13.69m$

1006195_1053983_ans_38fd6994c3494f019a69d752a56d65a5.PNG

An aeroplane is moving one kilometer high from West to East horizontally. From a point on the ground the angle of elevation of the aeroplane is

${60}^{o}$ , and after

$10$ seconds, the angle of elevation of the aeroplane is observed as

${30}^{o}$ , then the speed of the aeroplane in

$km/hour$ is.

0%
$120\sqrt {3}$
0%
$240\sqrt {3}$
0%
$120$
0%
$240$

From the top of a cliff

$25 \ m$ height, the angle of the elevation of the the top of tower is found to be equal to the angle of depresion of the foot of the tower. The height of the tower is

0%
$25m$
0%
$50\ m$
0%
$\dfrac{25}{\sqrt{2}} m$
0%
$25 \sqrt{2} m$

Explanation

R.E.F image

Consider

$\Delta DCB$ right angled at c,

$tan x^{\circ} = \dfrac{BC}{CD} = \dfrac{25}{CD}...(1)$

Now, consider

$\Delta DCE$ right angled atc,

$tan x^{\circ} = \dfrac{CE}{CD} = ...(2)$

equating eq

$^{n}$ (1) & (2),

$\Rightarrow \dfrac{25}{CD} = \dfrac{CE}{CD}$

$\Rightarrow CE = 25$

$\therefore$ Total ht. of tower is (25+25) m

(

$\because$ BE is ht.of tower such that BC+CE & BC = AD = 25

$\Rightarrow 25+CE$ )

$= 50 m$ Ans

So Option b is correct

1120312_1140126_ans_b6b3af7d59074819912f0763d09d040b.png

The lower window of a house is at

$2\ m$ above the ground and its upper window is

$4\ m$ vertically above the lower window. At certain instant the angle of elevation of balloon from these windows are observed to be

$60^0$ and

$30^0$ respectively. Then the height of the balloon above ground is

0%
$12\ m$
0%
$14\ m$
0%
$8\ m$
0%
$16\ m$

Explanation

Let the height of the first window be

$EG=2\ m$

and, the height of the second window be

$CG=(2+4)\ m=6\ m$

Let,

$AB$ be

$x$ and

$FG$ be

$y$

From

$\Delta ABC$

$\implies \tan 30°=\cfrac {AB} {BC}$

$\implies \cfrac 1 {\sqrt3}= \cfrac x y$

$\implies y=\sqrt{3}x$

From

$\Delta ADE$

$\implies \tan 60°=\cfrac {AD} {DE}$

$\implies \sqrt{3}= \cfrac {x+4} y$

$\implies \sqrt{3}y=x+4$

$\implies \sqrt{3}\times\sqrt{3}x=x+4$

$\implies 3x=x+4$

$\implies 2x=4$

$\implies x=2$

Hence, the height of the balloon from the ground is

$=(2+4+2)\ m=8\ m$

2105518_1057156_ans_077174e90a2d4691abaff3ff8d03ee8f.png

From the top of a building

$15\text{ m}$ high the angle of elevation of the top a tower is found to be

$30^\circ$ . From the bottom of the same building, the angle of elevation of the tower id found to be

$60^\circ$ . Find the height of the tower and the distance between the tower and the building.

0%
$h= 22.5\text{ m} , D=38.97\text{ m}$
0%
$h=22.5\text{ m}, D=12.97\text{ m}$
0%
$h= 38.97\text{ m}, D=22.5\text{ m}$
0%
$h=12.97\text{ m}, D= 22.5\text{ m}$

Explanation

Let the height of the tower be

$h$ and the distance between tower and building be

$d.$

This means the height from the top of the building to the top of the tower is

$h-15 \text{ m}.$

Now, from the top of building,

$\tan 30^\circ$

$=\dfrac{h-15}{d}\quad\quad\quad\dots(i)$

And, from the bottom of the building,

$\tan 60^\circ$

$=\dfrac{h}{d}\quad\quad\quad\dots(ii)$

Dividing

$(i)$ by

$(ii)$ ,

$\dfrac{h-15}{h}$

$=\dfrac{\tan 30^\circ}{\tan 60^\circ}$

$\Rightarrow$

$h$

$= 22.5 \text{ m}$

Now, using equation

$(ii)$ ,

$\tan 60^\circ$

$=\dfrac{h}{d}$

$\Rightarrow$

$d=\dfrac{h}{\tan 60^0}$

$\Rightarrow$

$d$

$=12.97\text{ m}$

So. the correct option is

$B$ .

1016252_1060447_ans_4f949ebe68974aa593cd3777e45a9655.png

The angle of elevation of the top of a tower at any point on the ground is

$\dfrac{\pi}{6}$ and after moving

$20$ meters towards the tower it becomes

${\pi}{3}$ .The height of the tower is equal to :

0%
$10 meters$
0%
${10}{\sqrt3} meters$
0%
$\dfrac{10}{\sqrt3} meters$
0%
$5\sqrt3 meters$

Explanation

We can see from the diagram that

$CD=20\ m$

Let

$BC=x\\AB=h$

Now, in

$\triangle ABC$

$\tan60^O=\dfrac hx\Rightarrow x\sqrt3=h........................(1)\\\tan30^o=\dfrac h{x+20}\Rightarrow x+20=\sqrt3h\Rightarrow x+20=\sqrt3*x\sqrt3\Rightarrow x+20=3x\Rightarrow x=10$

from (1)

$h=x\sqrt3\Rightarrow h=10\sqrt3$

1110877_1108486_ans_2c711b77e8f8481990e048d3b7e0cf83.png

The angle of elevation of the top of a tower from a point

$20m$ away from its base is

${60^ \circ }$ . The height of the tower is-

0%
$2\sqrt{3}\,m$
0%
$-20\sqrt{3}\,m$
0%
$\sqrt{3}\,m$
0%
$20\sqrt{3}\,m$

Explanation

Let the distance between point of observation and foot of tower $BC=20\,m\,$

Height of tower $AB=H$

Angle of elevation of top of tower $\theta ={{60}^{0}}$

Now from figure $\Delta ABC$

We know that, $\Delta ABC$ is a right angle triangle.

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

$\tan {{60}^{0}}=\dfrac{h}{20}$

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

$\Rightarrow h=20\sqrt{3}\,\,m.$

Therefore, height of tower

$=20\sqrt{3}\,m$
1038047_1140119_ans_8b9a85ceb9ec411a94237edf07f3d9bd.jpg

If the angles of elevation of the top of a tower from the top and foot of a pole height

$10mt$ are

$30^{o}$ and

$60^{o}$ then the height of the tower is

0%
$12\ mt$
0%
$15\ mt$
0%
$20\ mt$
0%
$36\ mt$

Explanation

REF.Image.

$\triangle ABE$

$tan 60 ^{\circ} = \dfrac{EB}{AB}$

$\sqrt{3} = \dfrac{10+h}{x} ..(1)$

and In

$\triangle CDE$

$tan30^{\circ} =\dfrac{h}{x}$

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

$x = \sqrt{3}h ...(2)$

substituting x value in (1)

$\Rightarrow \sqrt{3} = \dfrac{10+h}{\sqrt{3}h}$

$3h = 10+h$

$2h = 10, h = 5$

Tower's Height = BE = 10+h = 10+5 = 15

$\therefore$ Height of Tower = 15 m

option B is correct.

The shadow of the tower standing on a level ground is found to be

$60$ meters longer when the suns altitude is

$30^{o}$ than when it is

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

0%
$60m$
0%
$30m$
0%
$60\sqrt{3}m$
0%
$20(\sqrt{3}+1)m$

Explanation

REF.Image

$\triangle XYZ$

$tan 45^{\circ}=\dfrac{XY}{YZ}=\dfrac{XY}{x}$

$I=\dfrac{XY}{x}$

$XY=x$ cm

$\triangle XYA$

$tan 30^{\circ}= \dfrac{XY}{AY}$

$\dfrac{1}{\sqrt{3}}= \dfrac{x}{60+x}$

$60+x=\sqrt{3}x$

$\dfrac{60}{(\sqrt{3}-1)}=x$

$x= 81.96m$

or Height of Tower = 82 m

$\dfrac{60}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$

$= \dfrac{60(\sqrt{3}+1)}{(\sqrt{3})^{2}-(1)^{2}}$

$= \dfrac{60(\sqrt{3}+1)}{3-1}$

$= 20(\sqrt{3}+1)$

1155604_1144323_ans_20bfaded69284e32b8049f1d09a4c822.jpeg

A vertical to stands on a horizontal plane and is surmounted by a vertical flag staff of height 6 meters. At point on the plane angle of elevation of the bottom and the top of the flag staff are respectively

${ 30 }^{ \circ }and60^{ \circ }$ . Find the height of tower.

0%
5 m
0%
3m
0%
10 m
0%
7.5 m

The angle of elevation of the sun, when the length of the shadow of a tree is

$\sqrt{3}$ times the height of the tree, is

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

Explanation

REF. Image

Let AB be the tree and AC be its

shadow

Let

$\angle ACB =\theta$

$\Rightarrow \frac{AC}{AB}=\sqrt{3}$

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

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

1225006_1250705_ans_48d2cd5dee0c4e38ab04942ad00ec20b.PNG

The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun's altitude is 30 than when it was 45.

then height of tower

$x ( \sqrt { 3 } + 2 )$ . that is true /false

0%
True
0%
False

Explanation

Let height of the tower

$AB = h$ meters

Let distance

$(BC) = y$ meters

$\triangle ABC$

$\Rightarrow \tan 45^{\circ} = \dfrac{AB}{BC} \Rightarrow 1 = \dfrac{h}{y}$

$h = y$
In

$\triangle ABD$

$\tan 30^{\circ} = \dfrac{AB}{BD}$

$\dfrac{1}{\sqrt{3}} = \dfrac{h}{2x + y} \Rightarrow \sqrt{3} h = 2x + y$

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

$h (\sqrt{3} - 1) = 2x \Rightarrow h = \dfrac{2x}{\sqrt{3} - 1}$

$h = \dfrac{2x (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} = \dfrac{2x (\sqrt{3} + 1)}{3 - 1}$

$h = \dfrac{2x (\sqrt{3} + 1)}{2} = (\sqrt{3} + 1)x\ m$

$\therefore$ Height of tower

$= (\sqrt{3} + 1)x\ m$

2095079_1138236_ans_a1292d2e086349eeb5f34620f9092c74.png

From the bottom of a pole of height

$h$ the angle of elevation the top of a tower is

$\alpha$ and the pole subtends an angle

$\beta$ and top of the tower. The height of the tower is:

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

Explanation

${\textbf{Step - 1: Find tan }}\alpha$

${\text{H = height of the tower}}$

$\alpha {\text{ = from bottom of a pole the angle of elevation the top of a tower}}$

$\beta {\text{ = subtends angle}}$

${\text{Let AO = BM = a and AB = OM = h}}{\text{.}}$

$\therefore {\text{PM = H - h}}$

${\text{In }} \Delta {\text{AOP}}$

${\text{tan}}\alpha {\text{ = }}\dfrac{{{\text{PO}}}}{{{\text{AO}}}}$

$\Rightarrow {\text{AO = Hcot}}\alpha$

$\Rightarrow {\text{BM = Hcot}}\alpha \;\;......{\text{(i)}}$

${\textbf{Step -2: Find the height of the tower}}{\text{.}}$

${\text{In }}\Delta {\text{PBM}},{\text{tan}}(90 - \alpha + \beta ) = \dfrac{{{\text{BM}}}}{{{\text{PM}}}}$

$\Rightarrow {\text{BM = PM tan}}(90 - \alpha + \beta )\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\;$

$= {\text{(H - h)cot}}(\alpha - \beta ) \ldots ..{\text{(ii) by using [tan}}(90 - \theta ){\text{ = cot}}\theta ]$

${\text{From (i) and (ii),}}$

${\text{Hcot}}\alpha = {\text{(H - h)cot}}(\alpha - \beta )$

$\Rightarrow {\text{hcot}}(\alpha - \beta ){\text{ = H[cot}}(\alpha - \beta ) - {\text{cot}}\alpha ]$

$\Rightarrow {\text{H = }}\dfrac{{{\text{hcot}}(\alpha - \beta )}}{{{\text{cot}}(\alpha - \beta ) - {\text{cot}}\alpha }}\;\;\;.....\;{\text{[Proved]}}$

${\textbf{Hence the correct answer is option B}}$

A person , standing on the bank of a river observer that the angle subtended by tree on the opposite bank is, when he retreats 40 meters from the bank, he finds the angle to be . the height of the tree and the breadth of the river are :

0%
$10\sqrt 3 \,\,meters\,,\,10\,meters$
0%
$20\sqrt 3 \,\,meters\,,\,10\,meters$
0%
$\,20\sqrt 3 \,\,meters\,,\,20\,\,meters$
0%
none of these

Explanation

Here

$h$ is height of the tree, and

$AB$ be the breath of the river

$A$ be the river position of the man

Now in

$\triangle ABC \tan^{o}=\dfrac{h}{x}=\sqrt{3}$

$h=x\sqrt{3}$

$\triangle ABC = \tan30^{o}=\dfrac{h}{40+x}=\dfrac{1}{\sqrt{3}}$

$=\dfrac{x\sqrt{3}}{40+x}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow 3x=40+x$

$\Rightarrow x=20 \ metres$

Height

$= x\sqrt{3}=20\sqrt{3}\ metres$

1373584_1162421_ans_b99d74c4e97e434aa2ebfd9400605233.png

From a point at a height of 27 metres above a lake, the angle of elevation of the top of a tree on opposite side is

$30^{o}$ and the angle of depression of the image is

$45^{o}$ . The height of the tree from water level is (in metres)

0%
$10(2+\sqrt{3})$
0%
$10(2-\sqrt{3})$
0%
$27(2-\sqrt{3})$
0%
$27(2+\sqrt{3})$

A fire at a building is reported by a telephone of two fire stations

${F_1}\,\& \,{F_2}$ which are

$10$ km apart from each on a straight road.

${F_1}\,\& \,{F_2}$ observe that the fire is at an angle of

${60^0}\,\& \,{45^0}$ to the road. which station should send its team & how much it have to travel.

0%
${F_1},7.32m$
0%
${F_1},17.32m$
0%
${F_2},7.32m$
0%
${F_2},17.32m$

A bird is sitting on the top of a tree, which is

$80\,m$ high. The angle of the elevation of the bird , from a point on the ground is

$45^\circ$ . The bird flies away from the point of observation horizontally and remains at a constant height. After

$2$ seconds, the angle of elevation of the bird from the point of observation becomes

$30^\circ$ . Find the speed of the bird.

0%
$45(\sqrt3 -1)$
0%
$40(\sqrt4 -1)$
0%
$40(\sqrt3 -1)$
0%
$30(\sqrt3 -1)$

Explanation

REF.Image

Let AC be the tree.

$tan 45^{\circ}= \dfrac{AC}{BC}$

$1= \dfrac{80}{BC}$

$BC= 80m$

$tan 30^{\circ}= \dfrac{DE}{BC+CE}$

$\dfrac{1}{\sqrt{3}}=\dfrac{80}{80+x}$

$80+x= 80\sqrt{3}$

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

$x= 80(\sqrt{3}-1)cm$

The bird travelled

$80(\sqrt{3}-1)m$

Speed of bird =

$\dfrac{Distance}{time}$

$=\dfrac{80(\sqrt{3}-1)}{2}$

$= AO(\sqrt{3}-1)m/s$

$\therefore$ Speed =

$40(\sqrt{3}-1)m/s$

1152313_1165987_ans_29e673040c2a43a09593bd13b1c4fad7.png

If the angles of the elevation of a tower from two points distant

$a$ and

$b\ (a > b)$ from its foot and in the same straight line from it are

$30^0$ and

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

0%
$\sqrt{a+b}$
0%
$\sqrt{ab}$
0%
$\sqrt{a-b}$
0%
$\sqrt{\dfrac{a}{b}}$

Explanation

Let AB be the height of the tower and B be its foot

$AB=h$

$BD=b$

$BC=a$

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

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

$\triangle{ABC}$

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

$\dfrac{1}{\sqrt3}=\dfrac{h}{a}$

$h=\dfrac{a}{\sqrt3}.....(1)$

$\triangle{ADB}$

$\tan60=\dfrac{h}{b}$

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

$h=b\sqrt3.....(2)$

Multiplying (1) and (2)

$h^2=\dfrac{a}{\sqrt3}\times b\sqrt3$

$h^2=ab$

$h=\sqrt{ab}$

1103763_1200557_ans_3d0a04fa881b480da7fe0319b45afcc4.png

$P$ and

$Q$ are two points observed from the top of a building

$10\sqrt{3}\ \text{ m}$ high if the angles of depression of the points are complementary and

$PQ=20\ \text{m}$ , then the distance of

$P$ from the building is : (P being the farther point)

0%
$30\ \text{m}$
0%
$75\ \text{m}$
0%
$45\ \text{m}$
0%
$40\ \text{m}$

Explanation

Angle of depressions are

$x$ and

$90-x$ as shown in the figure

From,

$\triangle POR$

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

$\implies PR=\dfrac{h}{\tan x}$ ......(1)

From,

$\triangle ROQ$

$\tan (90-x)=\dfrac {h}{QR}$

$\implies QR=\dfrac{h}{\tan (90-x)}$

$\implies QR=\dfrac{h}{\cot x}$

$[\because \tan(90-\theta)=\cot \theta]$

$\implies QR= h\tan x$ .....(2)

$[\because \dfrac{1}{\cot x}=\tan x ]$

$\because PQ=PR-QR$

$\implies 20=\dfrac{h}{\tan x}-h\tan x$

$\implies 20 \tan x=h-h\tan^2{x}$

$\implies 20 \tan x=10\sqrt{3}-10\sqrt{3}\tan^2{x}$

$[\because h=10\sqrt{3}]$

$\implies \tan^2 x+\dfrac{2}{\sqrt{3}} \tan x-1=0$

$\implies \tan x=\dfrac{-\dfrac{2}{\sqrt{3}} \pm \sqrt{\left(\dfrac{2}{\sqrt{3}}\right)^2+4}}{2}$ (using quadratic formula)

$\implies \tan x=\dfrac{-\dfrac{2}{\sqrt{3}} \pm \dfrac{4}{\sqrt{3}}}{2}$

$\implies \tan x=-\dfrac{1}{\sqrt{3}} \pm \dfrac{2}{\sqrt{3}}$

$\implies \tan x=\dfrac{1}{\sqrt{3}}$ or

$\tan x=-\sqrt{3}$

Taking only,

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

$\implies x=30^{\circ}$

From (1)

$PR=\dfrac{10\sqrt{3}}{\tan x}=10\sqrt{3} \times \sqrt{3}=30 \text{ m}$

Hence, distance of point P is

$30\text{ m}$

1430348_1190158_ans_c240139bf95543fdbf7259d06b51d54d.png

The length of a string between a kite and a point on the ground is 50 m. The string makes an angle of

$60^{\circ}$ with the level ground. If there is no slack in the string, the height of the kite is :

0%
$50\sqrt{3}$ m
0%
$25\sqrt{3}$ m
0%
$25 m$
0%
$\dfrac {25}{\sqrt{3}}$ m

Explanation

$\tan{\theta}=\tan{{60}^{\circ}}=\dfrac{h}{AB}$

$\Rightarrow h=AB\sqrt{3}$ .......

$\left(1\right)$

Now, in

$\triangle ABC,$

${AC}^{2}={BC}^{2}+{AB}^{2}$ by pythagoras theorem

$\Rightarrow {50}^{2}={h}^{2}+{AB}^{2}$

$\Rightarrow {AB}^{2}={50}^{2}-{h}^{2}$

$\Rightarrow AB=\sqrt{{50}^{2}-{h}^{2}}$

Substituting in eqn

$\left(1\right)$ we get

$\Rightarrow h=AB\sqrt{3}=\left(\sqrt{{50}^{2}-{h}^{2}}\right)\sqrt{3}$

squaring both sides, we get

$\Rightarrow {h}^{2}=3\left({50}^{2}-{h}^{2}\right)$

$\Rightarrow {h}^{2}=3\times 2500 -3{h}^{2}$

$\Rightarrow 4{h}^{2}=3\times 2500$

$\Rightarrow {h}^{2}=625\times 3$

$\Rightarrow h=25\sqrt{3}$ m

$\therefore h=25\sqrt{3}$ m

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

$30^{\circ}$ with the ground. The distance from the foot of the tree to the point where the top touches the ground is

$10 \text{ m}$ . The height of the tree is

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

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

$=x+y$

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

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

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

1103603_1200525_ans_d25f482590e846a8916f855cd135946c.png

A man observes that when he moves up a distance

$c$ metres on a slope, the angle of depression of a point on the horizontal plane from the base of the slope is

${30}^{o}$ and when he moves up further a distance

$c$ metres, the angle of depression of that point is

${45}^{o}$ . The angle of inclination of the slope with the horizontal is

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

Explanation

Applying m-n theorem of trigonometry, we get

$(c+c)\cot(\theta-30)=c\cot15-c\cot30$

$\Rightarrow \cot(\theta-30)=\dfrac{1}{2}\dfrac{\sin(30-15)}{\sin15\sin30}$

$\Rightarrow \cot(\theta-30)=\dfrac{1}{2}\dfrac{1}{\sin 30}$

$\Rightarrow \cot(\theta-30)=1$

$\Rightarrow \cot(\theta-30)=\cot 45$

$\Rightarrow \theta-30= 45$

$\therefore \theta=75^{o}$

So, angle of inclination if the slope with horizontal will be

$75^{o}$

1088128_1190692_ans_09c4c888c0174263972c31bc5d3e091b.png

From the top of a pillar of height

$20\ m$ , the angles of elevation and depression of the top and bottom of another pillar are

$30^0$ and

$45^0$ respectively. The height of the second pillar (in meters) is:

0%
$\dfrac{20(\sqrt{3}+1)}{\sqrt{3}}$
0%
$10$
0%
$10\sqrt{3}$
0%
$\dfrac{10}{\sqrt{3}}(\sqrt{3}+1)$

Explanation

REF Image:

$CB$ is the height of the second tower.

$PA$ is the height of the first tower and

$ABQP$ is the rectangle.

$BC=CQ+QB\\PQ=AB(\because\Box PABQ),\\ \angle CPQ=30°\\ \tan45°=\cfrac{PA}{AB}\Rightarrow AB=\cfrac{PA}{\tan 45°}=20m = PQ\\ \therefore\quad In \Delta PQC,\angle CPQ=30°\\ \therefore \tan 30°=\cfrac{CQ}{PQ}\Rightarrow CQ=20\tan30°\Rightarrow CQ=\cfrac{20}{\sqrt3}m\\ In\Delta PQB, \angle PBQ=45°\\ \tan45°=\cfrac{PQ}{QB}\Rightarrow QB=PQ=20m\\ \therefore CB=CQ+QB=20+\cfrac{20}{\sqrt3}=\cfrac{20(\sqrt3+1)}{\sqrt3}m$
1129982_1200419_ans_dbac52042fc247489af948f0db40acd1.png

1129982_1200419_ans_dbac52042fc247489af948f0db40acd1.png

The length of a shadow of a vertical tower is

$\dfrac{1}{\sqrt{3}}$ times its height. The angle of elevation of the sun is:

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

Explanation

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

The angle of elevation of the sun is

$30^{\circ}$

Option A is correct.

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

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

$10$ meters. 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%
$\dfrac { 10 }{ \sqrt { 3 } } m$

Explanation

Let

$AC$ be the broken part of tree and

$CB$ be the remaining part of tree.

$\Delta ABC$

${ \tan30 }^{ 0 }=\dfrac { BC }{ 10 }$

$\Rightarrow BC=\dfrac{10}{\sqrt{3}} m$

Again

${ \cos30 }^{ 0 }=\dfrac { AB }{ AC }$

$\Rightarrow \dfrac { \sqrt { 3 } }{ 2 } =\dfrac { 10 }{ AC }$

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

height of the tree will be

$=AC+BC=\dfrac { 20 }{ \sqrt { 3 } } +\dfrac { 10 }{ \sqrt { 3 } } =\dfrac { 30 }{ \sqrt { 3 } } =10\sqrt { 3 } m$

Correct answer : Option B.

1207354_1204505_ans_dddd904b7a15424893049e036a0929fb.png

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 11 - 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 11 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

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