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CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 12 - MCQExams.com
CBSE
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:
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$$5\sqrt{3}\ m$$
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$$5\ m$$
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$$\dfrac{10\sqrt{3}}{3}\ m$$
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$$\dfrac{20\sqrt{3}}{3}\ m$$
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 :
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$$\sqrt{3}m$$
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$$5\sqrt{3}m$$
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$$10\sqrt{3}m$$
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$$20\sqrt{3}m$$
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 $$10\sqrt3\ m.$$
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:
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$$24(\sqrt{3}+1)$$ metre
0%
24 metre
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$$14\sqrt{2}$$ metre
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96 metre
Explanation
Let $$QA=xm\\QB=AB+QA=(48+x)m$$
In $$\Delta PQA,\\ \angle PQA=90°,\angle PAQ=45°\\ \tan 45°=\cfrac{PQ}{QA}=\cfrac{PQ}{x}\\ \therefore PQ=xm\rightarrow (1)$$
In$$\Delta PQB, \\ \angle PQB=90°,\angle PBQ=30°\\ \tan 30°=\cfrac{PQ}{QB}=\cfrac{PQ}{(48+x)}\\ \Rightarrow PQ=\cfrac{(48+x)}{\sqrt3}m\rightarrow(2)$$
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
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 :
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225 meters
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200 meters
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230 metres
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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.
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?
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$$6$$
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$$10$$
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$$20$$
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$$5$$
Explanation
R.E.F image
let $$AB = x,QB = y,PQ = z$$
In $$\triangle PAQ $$
$$ \tan 30^{\circ} = \dfrac{z}{x+y} $$
$$ \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{z}{x+y} $$
$$ \Rightarrow z = \dfrac{x+y}{\sqrt{3}} $$
In $$ \triangle PBQ $$
$$ \tan 60^{\circ} = \dfrac{z}{y} $$
$$ \Rightarrow z = \sqrt{3y} $$
$$ \therefore \dfrac{x+y}{\sqrt{3}} = \sqrt{3y} $$
$$ \Rightarrow x+y = 3y $$
$$ \Rightarrow x = 3y-y $$
$$ \Rightarrow x = 2y \Rightarrow y = \dfrac{x}{2} $$
So to go with x, it takes around 10 minutes.
with y, it takes around 5 minutes.
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?
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$$152.5$$
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$$155$$
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$$165$$
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$$162.5$$
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?
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$$30\ m$$
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$$30\sqrt{3}\ m$$
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$$60\ m$$
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$$60\sqrt{3}\ m$$
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 $$30\sqrt{3}\: m$$.
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:
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$$24\sqrt{3}$$ m
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24 m
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$$24\sqrt{2}$$ m
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96 m
Explanation
$$AB$$ is the tower
In $$\triangle{ABC}, \tan{{15}^{\circ}}=\dfrac{AB}{AC}$$
$$\Rightarrow 2-\sqrt{3}=\dfrac{h}{d+48}$$ .....$$\left(1\right)$$
$$\Rightarrow \tan{{30}^{\circ}}=\dfrac{AB}{BD}=\dfrac{h}{d}$$
$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{h}{d}$$ 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 :
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$$30^{\circ}$$
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$$45^{\circ}$$
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$$60^{\circ}$$
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$$90^{\circ}$$
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 _________________.
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$$h\ \cos{60^{o}}$$
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$$(h/3)$$ $$\cot{30^{o}}$$
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$$(h/3)\ \cot{60^{o}}$$
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$$h$$ $$\cot{30^{o}}$$
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 _________________.
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$$10 m$$
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$$15 m$$
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$$20 m$$
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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?
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$$60^{o}$$
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$$30^{o}$$
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$$45^{o}$$
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$$90^{o}$$
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 $$\text{ m/sec}$$.
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$$100\text{ m/sec}$$
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$$200\text{ m/sec}$$
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$$300\text{ m/sec}$$
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$$400\text{ m/sec}$$
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 $$x \text{ m/sec}$$.
In $$\triangle ABQ$$,
$$\begin{aligned}{}\tan {60^\circ} &= \frac{{QB}}{{AB}}\\\sqrt 3 & = \frac{{2500\sqrt 3 }}{x}\\x &= 2500\text{ m}\end{aligned}$$
In $$\triangle APC$$,
$$\begin{aligned}{}\tan {30^\circ} &= \frac{{2500\sqrt 3 }}{{x + y}}\\\frac{1}{{\sqrt 3 }} &= \frac{{2500\sqrt 3 }}{{x + y}}\\x + y &= 3 \times 2500\\2500 + y& = 7500\\y &= 5000\text{ m}\end{aligned}$$
To travel $$5000 \text{ m}$$ the aeroplane takes $$25
\text{ sec}
$$.
So, the speed of the plane will be,
$$\dfrac{5000}{25}=200\text{ m/sec}$$
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
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$$50\ mt$$
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$$100\ mt$$
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$$13.749\ mt$$
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$$\dfrac{80}{\sqrt{3}}mt$$
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
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$$ \cfrac {325 \sqrt {3}} {3} mt $$
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$$ \cfrac {325 \sqrt {3}} {2} mt $$
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$$ \cfrac {324 \sqrt {3}} {3} mt $$
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$$ \cfrac {324 \sqrt {3}} {2} mt $$
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:
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$$\dfrac{h cot x}{cot x+ cot y}$$
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$$\dfrac{h cot y}{cot x+ cot y}$$
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$$\dfrac{h cot x}{cot x - cot y}$$
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$$\dfrac{h cot y}{cot x- cot y}$$
Explanation
In $$\triangle AED$$
$$\tan x =\dfrac {DE}{AE}\quad \therefore \ \tan x =\dfrac {H-h}{AE}$$------- (1)
$$AE=\dfrac {H-h}{\tan x}$$------(A)
In $$\triangle DBC$$
$$\tan y=\dfrac {DC}{BC}\quad \therefore \ \tan y =\dfrac {H}{AE}$$------- (2)
$$AE=\dfrac {H}{\tan y}$$------ (B)
Comparing $$A$$ and $$B$$
$$\dfrac {H-h}{\tan x}=\dfrac {H}{\tan y}$$
$$(H-h)\tan y=H.\tan x$$
$$H\ \tan y-H\ \tan x=h\ \tan y$$
$$H=\dfrac { h\tan { y } }{ \tan { y } -\tan { x } }$$
$$ =\dfrac { \dfrac { h }{ \cot { y } } }{ \dfrac { 1 }{ \cot { y } } -\dfrac { 1 }{ \cot { x } } }$$
$$ =\dfrac { \dfrac { \dfrac { h }{ \cot { y } } }{ \cot { x } -\cot { y } } }{ \cot { x } .\cot { y } }$$
$$ =\dfrac { h\cot { x } }{ \cot { x } -\cot { y } } $$
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:
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$$20(\sqrt{3}+1)m$$
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$$24(\sqrt{3}+1)m$$
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$$30(\sqrt{3}+1)m$$
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$$30(\sqrt{3}-1)m$$
Explanation
In $$\Delta TOW$$,
$$\tan 45^o = \dfrac{h}{x}$$ $$\therefore x = h$$
In $$ \Delta TXW$$,
$$\tan 30^o = \dfrac{h}{60 + x}$$
$$\dfrac{1}{\sqrt{3}} = \dfrac{h}{60 + h}$$
$$60 + h = \sqrt{3} h$$
$$60 = h (\sqrt{3} - 1)$$
$$h = \dfrac{60}{\sqrt{3} - 1} = \dfrac{60 (\sqrt{3} + 1)}{3 - 1} = 30 (\sqrt{3} + 1) m$$
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
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$$15\ m$$
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$$10\sqrt{3}\ m$$
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$$20\ m$$
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$$None\ of\ these$$
Explanation
Height of the tree is =x
$$\begin{array}{l} \tan \theta =\dfrac { P }{ B } \\ \tan { 30^{ 0 } } =\dfrac { x }{ { 15 } } \\ \therefore x=15\times \dfrac { 1 }{ { \sqrt { 3 } } } \\ =8.66 \end{array}$$
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:
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$$\frac{b}{2}\tan (\alpha + \beta )$$
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$$b\tan (\alpha + \beta )$$
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$$b\tan (\frac{{\alpha + \beta }}{2})$$
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$$\frac{b}{2}\tan (\frac{{\alpha + \beta }}{2})$$
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 $$(\alpha +\beta )$$
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
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$$\dfrac {5\sqrt {3}}{2}$$
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$$5\sqrt {\dfrac {2}{3}}$$
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$$5\sqrt {\dfrac {3}{2}}$$
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$$\dfrac {5\sqrt {2}}{3}$$
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 $$5\sqrt{(\dfrac{3}{2})} m$$
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:
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$$5\sqrt{3}$$
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$$10(\sqrt{3} + 1) meter$$
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$$5(\sqrt{3} + 1)$$ meter
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$$10\sqrt{3} meter$$
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
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65
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60
0%
30
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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.
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$$\dfrac{3}{2}$$
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$$3\sqrt{3}$$
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$$6$$
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$$\dfrac{3\sqrt{3}}{2}$$
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.
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1 : 2
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2 : 1
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1 : 3
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3 : 1
Explanation
$$\tan 30^o=\dfrac{x}{a}$$ $$\tan 60^o=\dfrac{y}{a}$$
$$x=a\tan 30^o$$ $$y=a\tan 60^o$$
$$\dfrac{x}{y}=\dfrac{\tan 30^o}{\tan 60^o}=\dfrac{1}{3}$$
$$x : y =1 : 3$$.
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.
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$$100$$
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$$100\sqrt {3}$$
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$$100(\sqrt {3}-1)$$
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$$\dfrac {100}{\sqrt {3}}$$
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.
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175 m
0%
186 m
0%
190 m
0%
204 m
Explanation
Let $$A$$ be the position of the balloon.
In right triangle $$ABC,$$
$$\sin 60^o=\dfrac{AB}{AC}$$
$$AB=215 \sin 60^o$$
$$AB=215 \times \dfrac{\sqrt 3}{2}$$
$$AB =186 \ m$$
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:
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40 m
0%
60 m
0%
80 m
0%
20 m
Explanation
Here, $$\alpha =A- \beta $$
$$\therefore \beta = A - \alpha $$
Hence $$\tan \beta = \dfrac{{\tan A + \tan \alpha }}{{1 - \tan A\tan \alpha }}$$
$$ \Rightarrow \dfrac{3}{5} = \dfrac{{\dfrac{h}{{40}} + \left(- {\dfrac{{ h}}{{160}}} \right)}}{{1 - \left( {\dfrac{h}{{40}}} \right)\left( -{\dfrac{{ h}}{{160}}} \right)}}$$
$$ \Rightarrow {h^2} - 200h + 6400=0$$
$$ \Rightarrow \left( {h - 40} \right)\left( {h - 160} \right) = 0$$
$$ \Rightarrow h = 40\,\,or\,\,h = 160$$
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.
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$$40\sqrt2 \ m$$
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$$20 \sqrt 3 \ m$$
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$$50 \sqrt 3 \ m$$
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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 $$=40\sqrt 2 \text{ m}$$
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
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$$9(1+\sqrt{3})$$
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$$18(1+\sqrt{3})$$
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$$27(1+\sqrt{3})$$
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$$18(1+\sqrt{9})$$
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
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$$h\sqrt{2+\sqrt{3}}$$
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$$2h\sqrt{2+\sqrt{3}}$$
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$$4h\sqrt{2+\sqrt{3}}$$
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$$3h\sqrt{2+\sqrt{3}}$$
Explanation
$$\begin{array}{l} Q\tan { 15^{ \circ } } =\dfrac { h }{ { OQ } } \\ h\cot { 15^{ \circ } } =OQ \\ \dfrac { h }{ { 2-\sqrt { 3 } } } \, \, \, \left[ { \tan { { 15 }^{ \circ } } =2-\sqrt { 3 } } \right] \\ OP=\dfrac { h }{ { \tan { { 45 }^{ \circ } } } } =h \\ PQ=\sqrt { O{ P^{ 2 } }+O{ Q^{ 2 } } } \\ =h\sqrt { { { \left( { \dfrac { 1 }{ { 2-\sqrt { 3 } } } } \right) }^{ 2 } }+1 } \\ l\sqrt { { { \left( { \sqrt { 3 } +2 } \right) }^{ 2 } }+1 } \\ l\sqrt { 3+4+4\sqrt { 3 } +1 } \\ h\sqrt { 8+4\sqrt { 3 } } \\ 2h\sqrt { \sqrt { 3 } +2 } \\ 2h\sqrt { 2+\sqrt { 3 } } \\ Hence,\, option\, \, A\, is\, the\, \, correct\, answer. \end{array}$$
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Practice Class 10 Maths Quiz Questions and Answers
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