MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 12

$A,\ B,\ C$ are three collinear points on the ground such that

$B$ lies between

$A$ and

$C$ and

$AB=10\ m.$ If the angles of elevation of the top of a vertical tower standing at

$C$ are respectively

${30}^{\circ}$ and

${60}^{\circ}$ as seen from

$A$ and

$B,$ then the height of the tower is:

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

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

$30^0$ . On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to

$60^0$ . The height of the tower is :

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

Explanation

Let the height of the tower be 'h' m.

In right

$\triangle{DCB}$

$\tan60=\dfrac{DC}{CB}$

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

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

In right

$\triangle{DCA}$

$\tan30=\dfrac{DC}{CA}$

$\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+20}$

$\Rightarrow \dfrac{h}{\sqrt3}+20=\sqrt{3}h$

$\therefore h=10\sqrt{3}$

Hence the height of the tower is

$10\sqrt3\ m.$

1113994_1208081_ans_288562a550734c59803876301ea4e9a4.png

The angles of elevation of the top of a tower from two points A and B lying on the horizontal plane through the foot of the tower are respectively

$45^0$ and

$30^0$ . If A and B are on the same side of the tower and

$AB=48 \,metre$ , then the height of the tower is:

0%
$24(\sqrt{3}+1)$ metre
0%
24 metre
0%
$14\sqrt{2}$ metre
0%
96 metre

Explanation

Let

$QA=xm\\QB=AB+QA=(48+x)m$

$\Delta PQA,\\ \angle PQA=90°,\angle PAQ=45°\\ \tan 45°=\cfrac{PQ}{QA}=\cfrac{PQ}{x}\\ \therefore PQ=xm\rightarrow (1)$

$\Delta PQB, \\ \angle PQB=90°,\angle PBQ=30°\\ \tan 30°=\cfrac{PQ}{QB}=\cfrac{PQ}{(48+x)}\\ \Rightarrow PQ=\cfrac{(48+x)}{\sqrt3}m\rightarrow(2)$

From

$(1)\&(2)$

$x=\cfrac{(48+x)}{\sqrt3}\Rightarrow x=\cfrac{48}{\sqrt3-1}m\\ \therefore\quad height\quad of\quad tower (PQ)=x=\cfrac{48}{\sqrt3-1}=24(\sqrt3+1)m$

Option A is correct

1131700_1205597_ans_72353e37275344fcbe963edb906e147f.png

At a point A, the angle of elevation of a tower of a lower is such that its tangent is

$\dfrac{5}{{12}};$ on walking 120 meters nearer the tower the tangent of the angle of elevation is

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

0%
225 meters
0%
200 meters
0%
230 metres
0%
None of these

Explanation

Here AB is the tower, it is given that

$\tan\phi =\dfrac{5}{12}$ ……….(i)

On walling

$240$ m near to tower, it reaches point C, then

$\tan\theta =\dfrac{3}{4}$ ………(ii)

Let

$BC=x$ m,

$AB=y$ m

Now in

$\Delta$ ABC,

$\tan\theta =\dfrac{y}{x}=\dfrac{3}{4}$

$\Rightarrow x=\dfrac{4}{3}y$

Now

$\tan\theta =\dfrac{y}{240+x}=\dfrac{5}{12}$

$=\dfrac{y}{240+\dfrac{4}{3}y}=\dfrac{5}{12}$

$\Rightarrow 12y=1200+\dfrac{20}{3}y$

$\Rightarrow 36y-20y=3600$

$\Rightarrow 16y=3600$

$\Rightarrow y=225$ m.

1379358_1216137_ans_a3d28e4017074fd3904a63bbaea76bac.JPG

A man is walking towards a vertical pillar in a straight path, at uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is

$30^o$ . After walking for

$10$ minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is

$60^o$ . Then the time taken(in minutes) by him, from B to reach the pillar, is?

0%
$6$
0%
$10$
0%
$20$
0%
$5$

Explanation

R.E.F image

let

$AB = x,QB = y,PQ = z$

$\triangle PAQ$

$\tan 30^{\circ} = \dfrac{z}{x+y}$

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

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

$\triangle PBQ$

$\tan 60^{\circ} = \dfrac{z}{y}$

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

$\therefore \dfrac{x+y}{\sqrt{3}} = \sqrt{3y}$

$\Rightarrow x+y = 3y$

$\Rightarrow x = 3y-y$

$\Rightarrow x = 2y \Rightarrow y = \dfrac{x}{2}$

So to go with x, it takes around 10 minutes.

with y, it takes around 5 minutes.

1372492_1226086_ans_e15283d221cc491e881b57b504ee1033.JPG

To cross a river a person covers a straight forward distance of 325 m along a bridge over the river. If bridge subtends 30

$^{o}$ angle with edge of the river, find the width of the river?

0%
$152.5$
0%
$155$
0%
$165$
0%
$162.5$

In a storm, a tree got broken by the wind whose top meets the ground at an angle of 30

$^{o}$ , at a distance of 30 meters from the root. What was the height of the tree before breaking?

0%
$30\ m$
0%
$30\sqrt{3}\ m$
0%
$60\ m$
0%
$60\sqrt{3}\ m$

Explanation

Based on the given information, we can draw the figure shown above.

Here, line segment

$AB$ represents the height of the tree.

The tree breaks at point

$D$ .

$\therefore$ Segment

$AD$ is the broken part of tree which then takes the position of

$DC$ .

$\therefore AD = DC$

Given

$m\angle DCB = 30^\circ, BC=30\:m$

In right angled

$\Delta DBC$ ,

$tan\:30^\circ=\dfrac{BD}{BC}$

$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{BD}{30}$

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

$\therefore BD=10\sqrt{3}\:m$

Also

$cos\:30^\circ=\dfrac{BC}{DC}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{30}{DC}$

$\Rightarrow DC=\dfrac{60}{\sqrt {3}}$

$\therefore DC=20\sqrt{3}\:m$

Hence

$AD = DC = 20\sqrt{3}\: m$

From the figure,

$AB = AD + DB$

$\therefore \ AB = 20\sqrt{3} + 10\sqrt{3}$

$\Rightarrow AB = 30\sqrt{3}\: m$

Hence, the height of tree is

$30\sqrt{3}\: m$ .

2114642_1226767_ans_87cd253e846e46e0acf4b82ccbc64331.JPG

The angles of elevation of the top of a tower from two points A and B lying on the horizontal through the foot of the tower are respectively

$15^{\circ}$ and

$30^{\circ}$ . If A and B are on the same side of the tower and AB = 48 m, Then the height of the tower is:

0%
$24\sqrt{3}$ m
0%
24 m
0%
$24\sqrt{2}$ m
0%
96 m

Explanation

$AB$ is the tower

$\triangle{ABC}, \tan{{15}^{\circ}}=\dfrac{AB}{AC}$

$\Rightarrow 2-\sqrt{3}=\dfrac{h}{d+48}$ .....

$\left(1\right)$

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

$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{h}{d}$ or

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

Eqn

$\left(1\right)\Rightarrow 2-\sqrt{3}=\dfrac{\dfrac{d}{\sqrt{3}}}{d+48}$

$\Rightarrow 2-\sqrt{3}=\dfrac{d}{\sqrt{3}\left(d+48\right)}$

$\Rightarrow 2\sqrt{3}-3=\dfrac{d}{\left(d+48\right)}$

$\Rightarrow \left(2\sqrt{3}-3\right)\left(d+48\right)=d$

$\Rightarrow d\left(2\sqrt{3}-3\right)+48\left(2\sqrt{3}-3\right)=d$

$\Rightarrow d\left(2\sqrt{3}-3-1\right)=-48\left(2\sqrt{3}-3\right)$

$\Rightarrow d\left(2\sqrt{3}-4\right)=48\left(3-2\sqrt{3}\right)$

$\Rightarrow 2d\left(1\sqrt{3}-2\right)=48\left(3-2\sqrt{3}\right)$

$\Rightarrow d=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(\sqrt{3}-2\right)}$

We have

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

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

$\Rightarrow \sqrt{3}h=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(\sqrt{3}-2\right)}$ where

$d=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(1\sqrt{3}-2\right)}$ from above

$\Rightarrow h=\dfrac{24\left(3-2\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}-2\right)}$

$\Rightarrow h=\dfrac{24\left(3-2\sqrt{3}\right)}{\left(3-2\sqrt{3}\right)}$

$\therefore h=24$ m

The shadow of a tree is

$17\sqrt {3}\ \text{m}$ . If the height of the tree is

$17\ \text{m}$ , then the sun's altitude is :

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

Explanation

Length of the tree

$= 17\sqrt{3}\ \text{m}$

Height of the tree

$= 17 \ \text{m}$

then , the sun's altitude is

$\tan \theta = \dfrac{\text{Height of tree}}{\text{Length of shadow}}$

$\tan\theta = \dfrac{17}{17\sqrt{3}}$

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

$\theta = 30^{\circ}$

A tower substends an angle of

$30^{o}$ at a point on the same level as the foot of the tower. At a second point,

$h$ meter above first, point the depression of the foot of the tower is

$60^{o}$ , the horizontal distance of the tower from the point is _________________.

0%
$h\ \cos{60^{o}}$
0%
$(h/3)$ $\cot{30^{o}}$
0%
$(h/3)\ \cot{60^{o}}$
0%
$h$ $\cot{30^{o}}$

On the level ground, the angle of elevation of the top of angle nearer to it the angle of elevation becomes

$60 ^ { \circ } .$ The height is _________________.

0%
$10 m$
0%
$15 m$
0%
$20 m$
0%
none

If a tower

$30\ m$ high, casts a shadow

$10\sqrt{3}\ m$ long on the ground, then what is the angle of elevation of the sun?

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

Explanation

Given, Length = 30 m (Tower)

Base =

$10 \sqrt 3$ m (Shadow)

We have to find the angle, lets take that A

So,

tan A =

$\frac{Perpendicular}{Base}$

tan A =

$\frac{30}{10 \sqrt 3}$

tan A =

$\frac{3}{ \sqrt 3}$

tan A =

$\sqrt 3$

So, tan A = tan

$60^o$

So, A =

$60^o$

An airplane flying horizontally at a height of

$2500\sqrt{3} \text{ m}$ above that ground, is observed to be at an angle of elevation

$60^\circ$ from the ground. After a flight of

$25 \text{ sec}$ the angle of elevation is

$30^\circ$ . Find the speed of the plane in

$\text{ m/sec}$ .

0%
$100\text{ m/sec}$
0%
$200\text{ m/sec}$
0%
$300\text{ m/sec}$
0%
$400\text{ m/sec}$

Explanation

Let

$P$ and

$Q$ be the two positions of the airplane.

$QB=2500\sqrt{3} \text{ m}$

Let the speed of the aeroplane be

$x \text{ m/sec}$ .

$\triangle ABQ$ ,

$\begin{aligned}{}\tan {60^\circ} &= \frac{{QB}}{{AB}}\\\sqrt 3 & = \frac{{2500\sqrt 3 }}{x}\\x &= 2500\text{ m}\end{aligned}$

$\triangle APC$ ,

$\begin{aligned}{}\tan {30^\circ} &= \frac{{2500\sqrt 3 }}{{x + y}}\\\frac{1}{{\sqrt 3 }} &= \frac{{2500\sqrt 3 }}{{x + y}}\\x + y &= 3 \times 2500\\2500 + y& = 7500\\y &= 5000\text{ m}\end{aligned}$

To travel

$5000 \text{ m}$ the aeroplane takes $$25 \text{ sec}$$.

So, the speed of the plane will be,

$\dfrac{5000}{25}=200\text{ m/sec}$

1460120_1254536_ans_b0188f79dfe247eaa4d86bc818bc7e36.png

A chinny of

$20\ mt$ height, standing on the top of a building subtends an angle whose tangent is

$\dfrac{1}{6}$ at a distance of

$70\ mt$ from the foot of the building. The height of the building is

0%
$50\ mt$
0%
$100\ mt$
0%
$13.749\ mt$
0%
$\dfrac{80}{\sqrt{3}}mt$

The angles of elevation of the top of a mountain from two points A and B on a horizontal line are

${15}^{0}$ and

${75}^{0}$ .If AB=650 mt,then the height of the mountain is

0%
$\cfrac {325 \sqrt {3}} {3} mt$
0%
$\cfrac {325 \sqrt {3}} {2} mt$
0%
$\cfrac {324 \sqrt {3}} {3} mt$
0%
$\cfrac {324 \sqrt {3}} {2} mt$

The angles of elevation of the top of a building from the top and bottom of a tree are

$x$ and

$y$ respectively. If the height of the tree is

$h$ meter then, in meter, the height of the building is:

0%
$\dfrac{h cot x}{cot x+ cot y}$
0%
$\dfrac{h cot y}{cot x+ cot y}$
0%
$\dfrac{h cot x}{cot x - cot y}$
0%
$\dfrac{h cot y}{cot x- cot y}$

Explanation

$\triangle AED$

$\tan x =\dfrac {DE}{AE}\quad \therefore \ \tan x =\dfrac {H-h}{AE}$ ------- (1)

$AE=\dfrac {H-h}{\tan x}$ ------(A)

$\triangle DBC$

$\tan y=\dfrac {DC}{BC}\quad \therefore \ \tan y =\dfrac {H}{AE}$ ------- (2)

$AE=\dfrac {H}{\tan y}$ ------ (B)

Comparing

$A$ and

$B$

$\dfrac {H-h}{\tan x}=\dfrac {H}{\tan y}$

$(H-h)\tan y=H.\tan x$

$H\ \tan y-H\ \tan x=h\ \tan y$

$H=\dfrac { h\tan { y } }{ \tan { y } -\tan { x } }$

$=\dfrac { \dfrac { h }{ \cot { y } } }{ \dfrac { 1 }{ \cot { y } } -\dfrac { 1 }{ \cot { x } } }$

$=\dfrac { \dfrac { \dfrac { h }{ \cot { y } } }{ \cot { x } -\cot { y } } }{ \cot { x } .\cot { y } }$

$=\dfrac { h\cot { x } }{ \cot { x } -\cot { y } }$

1344633_1254780_ans_1eec2d9af7ef4137a58e773fe8c0cb69.png

The shadow of the tower becomes 60 meters longer when the altitude of the sun changes from

$45^{\circ} \ \text to\ 30^{\circ}$ . Then the height of the tower is:

0%
$20(\sqrt{3}+1)m$
0%
$24(\sqrt{3}+1)m$
0%
$30(\sqrt{3}+1)m$
0%
$30(\sqrt{3}-1)m$

Explanation

$\Delta TOW$ ,

$\tan 45^o = \dfrac{h}{x}$

$\therefore x = h$

$\Delta TXW$ ,

$\tan 30^o = \dfrac{h}{60 + x}$

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

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

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

$h = \dfrac{60}{\sqrt{3} - 1} = \dfrac{60 (\sqrt{3} + 1)}{3 - 1} = 30 (\sqrt{3} + 1) m$

1144446_1261875_ans_4ed2e9e3c6f641978861d064b6138cbe.png

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

$30^{o}$ 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 present height of the tree is

0%
$15\ m$
0%
$10\sqrt{3}\ m$
0%
$20\ m$
0%
$None\ of\ these$

Explanation

Height of the tree is =x

$\begin{array}{l} \tan \theta =\dfrac { P }{ B } \\ \tan { 30^{ 0 } } =\dfrac { x }{ { 15 } } \\ \therefore x=15\times \dfrac { 1 }{ { \sqrt { 3 } } } \\ =8.66 \end{array}$

1215166_1286844_ans_cd1ce41b65f649da8e37aa9ee0e6bcf0.png

A ladder is inclined to a well at an angle

$\alpha$ to the horizontal. Its foot is drawn up to a distance and as such, it slides a distance

$b$ along the well and makes angle

$\beta$ with the horizontal, then

$a$ is equal to:

0%
$\frac{b}{2}\tan (\alpha + \beta )$
0%
$b\tan (\alpha + \beta )$
0%
$b\tan (\frac{{\alpha + \beta }}{2})$
0%
$\frac{b}{2}\tan (\frac{{\alpha + \beta }}{2})$

Explanation

R.E.F image

$tan\alpha =\frac{a}{\sqrt{l^{2}-a^{2}}}$

$tan\beta =\frac{b+a}{\sqrt{l^{2}-(b+a)^{2}}}$

on solving,

a = b tan

$(\alpha +\beta )$

1212266_1294259_ans_bd26841f16f248f8b366975a458d7623.JPG

A flagstaff

$5$ m high is placed on a building

$25$ m high. If the flag and building be subtend equal angels on the observer at a height

$30$ m, then the distance between observer and the top of flag is

0%
$\dfrac {5\sqrt {3}}{2}$
0%
$5\sqrt {\dfrac {2}{3}}$
0%
$5\sqrt {\dfrac {3}{2}}$
0%
$\dfrac {5\sqrt {2}}{3}$

Explanation

Given,

$∠DAE = ∠ DAC = \theta$ (Let's assume)

Thus,

$∠EAC = 2\theta$

Let's assume the distance between the building and the observer is

$'x' m$ .

Thus, for the triangle

$ΔACE$ ;

$EC = ED + DC = 5+25 = 30m$

Opposite/adjacent

$= \dfrac{EC}{EA} = \dfrac{30}{x} = tan2\theta$

For the

$ΔEDA;$

Opposite/adjacent

$= \dfrac{ED}{EA} = \dfrac{5}{x} = tan\theta$

Now,

$tan2\theta = \dfrac{2tan\theta }{ (1-tan^2\theta)}$

or,

$\dfrac{30}{x} = \dfrac{2(\dfrac{5}{x}) }{ (1 - \dfrac{25}{x^2})}$ [As

$tan\theta = \dfrac{5}{x}$ and

$tan2\theta = \dfrac{30}{x}]$

or,

$3 =\dfrac{ 1 }{ (1-\dfrac{25}{x^2}}$

or,

$3x^2 - 75 = x^2$

or,

$2x^2 = 75$

or,

$x = 5\sqrt{(\dfrac{3}{2})} = 6.124m$

Thus the distance between the building and the observer is

$5\sqrt{(\dfrac{3}{2})} m$

1937235_1280757_ans_cf1a8d6038dd4b1dbad7b5b59b25bb2b.png

The length of the shadow of a vertical tower on level ground increase by 10 meters when the altitude of the sun changes from

$45^o$ to

$30^o$ . Then the height of the tower is:

0%
$5\sqrt{3}$
0%
$10(\sqrt{3} + 1) meter$
0%
$5(\sqrt{3} + 1)$ meter
0%
$10\sqrt{3} meter$

Explanation

$Let\text{ the distance when the sun elevation is 45 be x}$

$\text{whereas when sun elevation is 30 the distance is 10+x}$

$\text{let h be the height of the tower}$

$\text{when the sun is at elevation of 45}$

$\text{tan45=h/x}$

$\text{1=h/x}$

$\text{h=x}\,\,\,\,\,\,\,.........\left( i \right)$

$when\,the\,sun\,elevation\,is\,{{30}^{0}}$

$\tan {{30}^{0}}=h/\left( x+10 \right)$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+10}$

$\Rightarrow h=\frac{x+10}{\sqrt{3}}\,\,\,\,\,\,......\left( ii \right)$

$from\,\left( i \right)and\,\left( ii \right)$

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

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

$\Rightarrow x\left( \sqrt{3}-1 \right)=10$

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

$\,\,\,\,\,\,\,\,\,\,\,=\frac{10}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\,\,\,\,\,\,\,\,\,\,=5\left( \sqrt{3}+1 \right)$

$hence\,height=5\left( \sqrt{3}+1 \right)m$

A tower is

$50\sqrt{3}$ m high. The angle of elevation of its top from a point 50 m away from its foot has measure............degree

0%
65
0%
60
0%
30
0%
15

A ladder leans against a wall making an angle of measure

$60^\circ$ with the ground. If the foot of the ladder is

$3$ metre away from the wall, then the length of the ladder is ..... metre.

0%
$\dfrac{3}{2}$
0%
$3\sqrt{3}$
0%
$6$
0%
$\dfrac{3\sqrt{3}}{2}$

Explanation

Given distance between the foot ofthe ladder and wall is

$3\ m$

Let the length of ladder be

$l$

$\implies \cos 60^{\circ}=\dfrac{3}{l}$

$\implies l=3\text{sec} 60^{\circ}$

$\implies l=6$

So the length of the ladder is

$6\ m$

The tops of two towers of height x and y standing on a level round subtends angle of 30 and 60 respectively at the line joining their feet then x:y is.

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

Explanation

$\tan 30^o=\dfrac{x}{a}$

$\tan 60^o=\dfrac{y}{a}$

$x=a\tan 30^o$

$y=a\tan 60^o$

$\dfrac{x}{y}=\dfrac{\tan 30^o}{\tan 60^o}=\dfrac{1}{3}$

$x : y =1 : 3$ .

1210236_1316528_ans_f4ce83082f274017845c34024fe987fa.jpg

The measure of the angle of elevation of sun increases from

$30\ to\ 45$ , then the length of the shadow of a

$100\ meter$ high building reduces by

$x\ meter$ . Then

$x$ is..... meter.

0%
$100$
0%
$100\sqrt {3}$
0%
$100(\sqrt {3}-1)$
0%
$\dfrac {100}{\sqrt {3}}$

A balloon is connected to a meteorological ground station by a cable of length 215 m inclined at

$60^o$ to the horizontal. Determine the height of the balloon from the ground. Assume that there is no stack in the cable.

0%
175 m
0%
186 m
0%
190 m
0%
204 m

Explanation

Let

$A$ be the position of the balloon.

In right triangle

$ABC,$

$\sin 60^o=\dfrac{AB}{AC}$

$AB=215 \sin 60^o$

$AB=215 \times \dfrac{\sqrt 3}{2}$

$AB =186 \ m$

1467740_1322564_ans_730b16fd161b4fbc892457e0e6fe8877.PNG

The upper

$\dfrac 34 ^{th}$ portion of a vertical pole subtends an angle

$\tan^{-1} \bigg(\dfrac 35\bigg)$ at a point in the horizontal plane through its foot and at a distance

$40\ m$ from the foot. A possible height of the vertical pole is:

0%
40 m
0%
60 m
0%
80 m
0%
20 m

Explanation

Here,

$\alpha =A- \beta$

$\therefore \beta = A - \alpha$

Hence

$\tan \beta = \dfrac{{\tan A + \tan \alpha }}{{1 - \tan A\tan \alpha }}$

$\Rightarrow \dfrac{3}{5} = \dfrac{{\dfrac{h}{{40}} + \left(- {\dfrac{{ h}}{{160}}} \right)}}{{1 - \left( {\dfrac{h}{{40}}} \right)\left( -{\dfrac{{ h}}{{160}}} \right)}}$

$\Rightarrow {h^2} - 200h + 6400=0$

$\Rightarrow \left( {h - 40} \right)\left( {h - 160} \right) = 0$

$\Rightarrow h = 40\,\,or\,\,h = 160$

1225730_1311397_ans_b994a2552ba54c4a90a243e3bd7f15f1.jpg

A 80 m long ladder is leaning on a wall. If the ladder makes an angle of

$45^o$ with the ground, find the distance of the ladder from the wall.

0%
$40\sqrt2 \ m$
0%
$20 \sqrt 3 \ m$
0%
$50 \sqrt 3 \ m$
0%
None of these

Explanation

Here,

$\cos \theta = \dfrac{Base}{Hypotenuse}$

$\cos 45^o=\dfrac{Base}{80}$

$Base = 80 \cos 45^o=\dfrac{80}{\sqrt 2}=40\sqrt 2$

Therefore, Distance of the ladder from the wall

$=40\sqrt 2 \text{ m}$

1478509_1339283_ans_ff23821c6ca045edbf50e1b7acbb0c10.PNG

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min for the angle of depression to change from

$30^{\circ}$ to

$45^{\circ}$ : then after this , the time taken (in min) by the car to reach the foot of the tower, is

0%
$9(1+\sqrt{3})$
0%
$18(1+\sqrt{3})$
0%
$27(1+\sqrt{3})$
0%
$18(1+\sqrt{9})$

A flag staff of height

$'h'$ stands in the centre of a rectangular field and subtends angle of

$15^{o}$ and

$45^{o}$ at the mid point of its sides. Find the distance between them

0%
$h\sqrt{2+\sqrt{3}}$
0%
$2h\sqrt{2+\sqrt{3}}$
0%
$4h\sqrt{2+\sqrt{3}}$
0%
$3h\sqrt{2+\sqrt{3}}$

Explanation

$\begin{array}{l} Q\tan { 15^{ \circ } } =\dfrac { h }{ { OQ } } \\ h\cot { 15^{ \circ } } =OQ \\ \dfrac { h }{ { 2-\sqrt { 3 } } } \, \, \, \left[ { \tan { { 15 }^{ \circ } } =2-\sqrt { 3 } } \right] \\ OP=\dfrac { h }{ { \tan { { 45 }^{ \circ } } } } =h \\ PQ=\sqrt { O{ P^{ 2 } }+O{ Q^{ 2 } } } \\ =h\sqrt { { { \left( { \dfrac { 1 }{ { 2-\sqrt { 3 } } } } \right) }^{ 2 } }+1 } \\ l\sqrt { { { \left( { \sqrt { 3 } +2 } \right) }^{ 2 } }+1 } \\ l\sqrt { 3+4+4\sqrt { 3 } +1 } \\ h\sqrt { 8+4\sqrt { 3 } } \\ 2h\sqrt { \sqrt { 3 } +2 } \\ 2h\sqrt { 2+\sqrt { 3 } } \\ Hence,\, option\, \, A\, is\, the\, \, correct\, answer. \end{array}$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 12 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 12 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.