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CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 7 - MCQExams.com
CBSE
Class 10 Maths
Some Applications Of Trigonometry
Quiz 7
The angle of elevation of a ladder leaning against a wall is $$\displaystyle 60^{\circ}$$ and the foot of the ladder is $$4.6\ m$$ away from the wall The length of the ladder is
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$$2.3\ m$$
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$$4.6\ m$$
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$$7.8\ m$$
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$$9.2\ m$$
Explanation
Consider the given figure. Let $$AC$$ denote the length of the ladder and $$AB$$ denote the wall. Now
$$cos60^{\circ}=\dfrac{BC}{AC}$$
$$AC=\dfrac{BC}{cos\ 60^{\circ}}$$
$$AC=BC.sec\ 60^{\circ}$$
$$=4.6\times 2$$
$$=9.2$$
Hence the length of the ladder is $$9.2\ m$$.
Two objects are one side of a Tower. Their angles of depression from the top of the Tower are $$\displaystyle45^{\circ}$$ and $$\displaystyle60^{\circ}$$ respectively. If the distance between two objects will be-
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$$6.34$$ metre
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$$63.4$$ metre
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$$63$$ metre
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$$64$$ metre
Explanation
In $$\displaystyle \Delta ABC$$
$$\displaystyle \tan 45^{\circ}=\frac{AB}{BD}$$
$$\displaystyle 1=\frac{150}{BD}$$
$$\displaystyle BD=150m$$
In $$\displaystyle \Delta ABC$$ ,
$$\displaystyle \tan 60^{\circ}=\frac{AB}{BC}$$
$$\displaystyle \sqrt{3}=\frac{150}{BC}$$
$$\displaystyle BC=50\sqrt{3}=86.60m$$
Hence, the distance between two objects CD will be
$$\displaystyle =150-86.60$$
=63.4 metre
The angle of elevation of the top of a tower from two points distant a and b ( a > b )from its foot and in the same straight line from is are $$ \displaystyle 30^{\circ} $$ and $$ \displaystyle 60^{\circ} $$, The height of the tower is
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$$ \displaystyle \frac{a}{b} $$
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$$ \displaystyle \sqrt{}\frac{a}{b} $$
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ab
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$$ \displaystyle \sqrt{ab} $$
Explanation
From $$\displaystyle \bigtriangleup AOT $$
$$\displaystyle h=a\tan 30^{\circ} $$
From $$\displaystyle \bigtriangleup BOT $$
$$\displaystyle h=a\tan 60^{\circ} $$ $$\displaystyle \therefore h^{2}=ab\tan 30^{\circ}\tan 60^{\circ} $$
$$\displaystyle =ab\frac{1}{\sqrt{3}}.\sqrt{3} $$
or $$\displaystyle h=\sqrt{ab} $$
$$AB$$ is a straight road leading to $$C$$, the foot of a tower. $$A$$ is at a distance $$125\text{ m}$$ from $$B$$ and $$B$$ at $$75\text{ m}$$ meters from $$C$$. If the angle of elevation of the tower at $$B$$ be double the angle of elevation at $$A$$, then the find the value of $$\cos2\alpha,$$
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$$\dfrac{4}{5}$$
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$$\dfrac{1}{5}$$
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$$\dfrac{2}{5}$$
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$$\dfrac{3}{5}$$
Explanation
In $$\triangle APB,$$
$$ \angle CBP=\angle A+\angle APB $$
$$\Rightarrow2\alpha =\alpha +\angle APB $$
$$\Rightarrow\angle APB= \alpha $$
This implies that
$$AB=BP=125 m$$ as sides opposite to equal angles in a triangle are equal.
In $$\triangle BCP,$$
$$ \displaystyle \cos 2\alpha =\frac{75}{125} $$
$$ \Rightarrow \cos 2\alpha =\dfrac{3}{5} $$
The angle of elevation of the top of a tower ad observed from a point on the horizontal ground is $$x$$. If we move a distance $$'d'$$ towards the foot of the tower, the angle of elevation increases to $$y$$, then the height if the tower is
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$$ \displaystyle \frac{d \tan x \tan y }{\tan y - \tan x} $$
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$$ \displaystyle(d\tan y +\tan x) $$
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$$ \displaystyle(d\tan y -\tan x) $$
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$$ \displaystyle \frac{d \tan x \tan y }{\tan y + \tan x} $$
Explanation
Let height of the tower , $$AB=h$$
From $$ \displaystyle \bigtriangleup ADB, \tan y =\frac{h}{BD} $$
or $$ \displaystyle BD= h \cot y $$ ....$$(i)$$
From $$ \displaystyle \bigtriangleup ADB$$ and $$\triangle ACB $$
$$ \displaystyle \tan x=\frac{h}{d+DB} $$ or $$d+DB =h \cot x$$
Or
$$ \displaystyle h(\cot x-\cot y)=d $$
$$ \displaystyle \therefore h=\frac{d}{\cot x-\cot y} $$
$$ \displaystyle =\dfrac{d}{\dfrac{1}{\tan x}- \dfrac{1}{\tan y}}\\=\dfrac{d\tan x.\tan y}{\tan y-\tan x} $$
The angles of elevation of a tower from a point on the ground is $$ \displaystyle 30^{\circ} $$ . At a point on the horizontal line passing through the foot of the tower and $$100$$ meters closer to it than the previous point, if the angle of elevation is found to be $$ \displaystyle 60^{\circ} $$ , then height of the tower is
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$$ \displaystyle 50\sqrt{3} $$ meters
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$$ \displaystyle \frac{50}{\sqrt{3}} $$ meters
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$$ \displaystyle 100\sqrt{3} $$
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$$ \displaystyle \frac{100}{\sqrt{3}} $$
Explanation
Let $$AB=h$$ be the height of the tower, and let $$AD=x$$ m
From right angled $$ \displaystyle \bigtriangleup ABD $$
$$ \displaystyle \tan 60^{\circ}=\frac{AB}{AD} $$
or $$ \displaystyle \sqrt{3}=\frac{h}{x} $$ or $$ x= \dfrac{h}{\sqrt{3}} $$
From right angled $$ \displaystyle \bigtriangleup BCA $$
$$ \displaystyle \tan 30^{\circ}=\frac{AB}{CA} $$
or $$ \displaystyle \frac{1}{\sqrt{3}}=\frac{h}{100+x} $$
or $$ \displaystyle h=\frac{100+x}{\sqrt{3}}=\frac{100+\dfrac{h}{\sqrt{3}}}{\sqrt{3}} $$
or $$ \displaystyle \frac{2}{\sqrt{3}}h=100 $$
or $$ \displaystyle \therefore h=50\sqrt{3}m $$
From a point $$p$$ on a level ground the angle of elevation of the top of a tower is $$\displaystyle 30^{0}$$ If the tower is $$100\ m$$ high the distance of point $$p$$ from the foot of the tower is
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$$149\ m$$
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$$156\ m$$
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$$173\ m$$
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$$200\ m$$
Explanation
Let $$AB$$ be the tower then $$\displaystyle \angle APB=30^{0}$$ and $$\displaystyle AB=100m$$
$$\displaystyle \frac{AB}{AP}=\tan 30^{0}\\\ \ \ \ \ \ \ \ =\dfrac{1}{\sqrt{3}}$$
$$\displaystyle \Rightarrow AP=\left ( AB\times \sqrt{3} \right )=100\sqrt{3}m$$
$$\displaystyle =\left ( 100\times 1.73 \right )m$$
$$\displaystyle = 173m$$
An observer 1.6m tall is $$\displaystyle 20\sqrt{3}$$m away from a tower The angle of elevation from his eye to the top of the tower is $$\displaystyle 30^{0}$$ The height of the tower is
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21.6m
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23.2m
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24.72m
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None of these
Explanation
Let $$AB$$ be the observer and $$CD$$ tower
Draw $$BE$$ perpendicular to $$CD$$
Then $$CE=AB=1.6 \ m$$
And $$BE=AC=$$ $$20\sqrt{3}$$ m
Then right angle triangle DEB
$$\therefore tan 30^{o}=\frac{DE}{BE}$$
$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{DE}{20\sqrt{3}}$$
$$\Rightarrow DE=20\sqrt{3}$$m
Then $$CD=CE+DE=1.6+20=21.6\ m$$
A rope of length 5 meters is tightly tied with one end at the top of a vertical pole and other end to the horizontal ground. If the rope makes an angle $$ \displaystyle 30^{\circ} $$ to the horizontal , then the height of the pole is
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$$ \displaystyle \frac{5}{2}m $$
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$$ \displaystyle \frac{5}{\sqrt{2}}m $$
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$$ \displaystyle 5\sqrt{2}m $$
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5m
The angle of elevation of the sun when the length of the shadow of a tree is $$\displaystyle \sqrt{3}$$ times the height of the tree is:
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$$\displaystyle 30^{0}$$
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$$\displaystyle 45^{0}$$
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$$\displaystyle 60^{0}$$
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$$\displaystyle 90^{0}$$
Explanation
Let $$AB$$ be the tree and $$AC$$ be its shadow.
Also, $$AC=\sqrt3AB$$.
Let the angle of elevation of the sun $$=\displaystyle \angle ACB=\theta $$
Then in $$\triangle ABC,$$
$$\displaystyle \tan \theta = \frac{AB}{AC}=\dfrac{1}{\sqrt{3}} $$ {
$$\because AC=\sqrt3AB$$}
$$ \tan 30^o =\dfrac{1}{\sqrt{3}}$$
$$\displaystyle \Rightarrow \theta =30^{0}$$
Two ships are sailing in the sea on the two sides of a lighthouse The angles of elevation of the top of the lighthouse as observed from the two
ships are $$\displaystyle 30^{0}$$ and $$\displaystyle 45^{0}$$ respectively If the lighthouse is $$100$$ m high the distance between the two ships is
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$$173$$m
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$$200$$m
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$$273$$m
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$$300$$m
Explanation
Let $$AB$$ be the height of lighthouse and $$C$$ and $$D$$ be the position of the ships.
Given $$AB=100$$ m and $$\angle ACB=30^{0}$$$$\angle ADB=45^{0}$$
In $$\Delta BCA$$
$$\dfrac{AB}{AC}=tan 30^{0}\\\ \ \ \ \ \ \ =\dfrac{1}{\sqrt{3}}$$
$$\Rightarrow AC=\sqrt{3}AB\\\ \ \ \ \ \ \ \ \ \ =100\sqrt{3}$$
In $$\Delta BAD$$
$$\dfrac{AB}{AD}=tan 45^{0}\\\ \ \ \ \ \ \ \ =1$$
$$\Rightarrow AD=AB=100 m$$
$$\therefore CD=AC+AD\\=100\sqrt{3}+100\\=100(\sqrt{3}+1)\\=100(1.73+1)\\=273 m$$
The top of a $$15$$ metre high tower makes an angle of elevation of $$\displaystyle 60^{0}$$ with the bottom of an electric pole and angle of elevation of $$\displaystyle 30^{0}$$ with the top of the pole. What is the height of the electric pole?
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$$5$$ metres
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$$8$$ metres
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$$10$$ metres
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$$12$$ metres
Explanation
Let $$AB$$ be the tower and $$CD$$ be the electric pole
Then $$\displaystyle \angle ACB=60^{0},\angle EDB=30^{0}$$ and
$$\displaystyle AB=15$$ m
Let $$\displaystyle CD=h.$$ Then $$\displaystyle BE=\left ( AB-AE \right )$$
$$\displaystyle =\left ( AB-CD \right )=\left ( 15-h \right )$$
We have $$\displaystyle \frac{AB}{AC}=\tan 60^{0}=\sqrt{3}$$
$$\displaystyle \Rightarrow AC=\frac{AB}{\sqrt{3}}=\frac{15}{\sqrt{3}}$$
And $$\displaystyle \frac{BE}{DE}=\tan 30^{0}=\frac{1}{\sqrt{3}}$$
$$\displaystyle \Rightarrow DE=\left ( BE\times \sqrt{3} \right )$$
$$\displaystyle =\sqrt{3}\left ( 15-h \right )$$
$$\displaystyle \Rightarrow 3h=\left ( 45-15 \right )$$
$$\Rightarrow h=10$$ m
If the height of a pole is $$\displaystyle 2\sqrt{3}$$ metres and the length of its shadow is 2 metres then the angle of elevation of the sun is equal to
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$$\displaystyle 30^{0}$$
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$$\displaystyle 45^{0}$$
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$$\displaystyle 60^{0}$$
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$$\displaystyle 90^{0}$$
Explanation
Height of a pole is $$2\sqrt { 3 } $$
length of its shadow is $$2$$ metre
Let ABC be a triangle where AB be
$$2\sqrt { 3 } $$ height of a pole
AC be $$2$$ metre length of shadow
Angle of elevation of the sun will be
$$\tan { \theta = } \dfrac { AB }{ AC } =\dfrac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } $$
$$\tan { { 60^ o } } =\sqrt { 3 } $$
Angle of elevation of the sun is equal to
$$ { 60^o } $$
The upper part of a tree broken by the wind makes an angle of $$\displaystyle 60^{0}$$ with the ground and the distance from the roots to the point where the top of the tree meets the ground is $$20$$ m. The length of the broken part of the tree is
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$$20$$ m
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$$40$$ m
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$$\displaystyle 20\sqrt{3}$$ m
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$$\displaystyle 40\sqrt{3}$$ m
Explanation
As shown in the fig. let length of the tree be $$AB$$ and $$DC$$ is the length of broken part.
$$\displaystyle BC=DC=\dfrac{20}{\cos 60^{0}}=20\times 2m=40m$$
If a flag-staff 6 metres high placed on the top pf a tower throws a shadow of $$\displaystyle 2\sqrt{3}$$ metres along the ground then the angle that the sun makes with the ground is
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$$\displaystyle 60^{0}$$
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$$\displaystyle 30^{0}$$
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$$\displaystyle 45^{0}$$
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None of these
Explanation
OA and AB be the shadows of the tower OP and the flag-staff PQ respectively on the grounds Let the sum makes an angle $$\displaystyle \theta $$ with the ground
$$\displaystyle AB=2\sqrt{3};OA=x;OP=h;PQ=6$$
In triangles OAP and OBQ We have
$$\displaystyle \tan \theta =\frac{h}{x}$$ and $$\displaystyle \tan \theta =\frac{h+6}{x+2\sqrt{3}}$$
$$\displaystyle \Rightarrow \frac{h}{x}=\frac{h+6}{x+2\sqrt{3}}$$
$$\displaystyle \Rightarrow x=\frac{h}{\sqrt{3}}$$
or $$\displaystyle \frac{h}{x}=\tan \theta =\sqrt{3}$$ so $$\displaystyle \theta=60^{0} $$
A man standing on the bank of the river observes that the angle subtended by a tree on the opposite bank is $$\displaystyle 60^{0}$$. When he retires $$36$$ metres from the bank he finds the angle to be $$\displaystyle 30^{0}$$. The breadth of the river is
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$$\displaystyle 12\sqrt{3}$$m
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$$18\ m$$
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$$12\ m$$
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$$27\ m$$
Explanation
Let the breadth of the river be AC = 'x' metres
Clearly $$\displaystyle \tan 30^{0}=\frac{AB}{AD}=\frac{h}{x+36}$$ and
$$\displaystyle \tan 60^{0}=\frac{AB}{AD}=\frac{h}{x}$$
$$\displaystyle \Rightarrow \frac{\tan 60^{0}}{\tan 30^{0}}=\frac{h}{x}\div \frac{h}{x+36}$$
$$\displaystyle \Rightarrow 3=\frac{x+36}{x}$$
$$\displaystyle \Rightarrow x=18m$$
In case of angle of depression the observer has to look _____ to view the object.
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Straight
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Anywhere
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Down
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Up
Explanation
In case of angle of depression, the observer has to look down to view the object.
A man on the top of a rock lying on a seashore observes a boat coming towards it. If it takes $$10$$ minutes for the angles of depression to change from $$\displaystyle 30^{\circ}$$ to $$\displaystyle 60^{\circ}$$ how soon will the boat reach the shore?
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$$20$$ minutes
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$$15$$ minutes
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$$10$$ minutes
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$$5$$ minutes
Explanation
AB is the rock C and D be the two positions of the boat
Given:
$$\angle EAC=30^{\circ}$$ and $$\angle EAD=60^{\circ}$$
Now,
$$\angle ADB=EAD$$ $$[\because AE \parallel BC, AD~ \text{is transversal}, ~\angle ADB$$ and $$\angle EAD $$ are alternate interior angles $$]$$
$$\implies \displaystyle \angle ADB=60^{\circ}$$
$$\angle EAC=ACB$$ $$[\because AE \parallel BC, AC~ \text{is transversal}, ~\angle EAC$$ and $$\angle ACB $$ are alternate interior angles $$]$$
$$\implies \displaystyle \angle ACB=30^{\circ}$$
Let $$CD=x, ~DB=y$$
From $$\triangle ABC$$
$$\displaystyle \tan 30^{0}=\frac{AB}{x+y}=\frac{1}{\sqrt{3}}$$
$$\displaystyle \Rightarrow AB=\frac{x+y}{\sqrt{3}}$$
From $$\triangle ABD$$
$$\displaystyle \tan 60^{0}=\frac{AB}{y}\Rightarrow AB=\sqrt{3}y$$
$$\displaystyle \therefore \frac{x+y}{\sqrt{3}}=\sqrt{3}y$$ or $$\displaystyle \frac{x}{2}=y$$
The boat takes $$10$$ minutes to cover distance $$x,$$
so it will take $$5$$ minutes to cover $$\displaystyle \frac{x}{2}$$
$$\displaystyle \therefore $$ Time to reach $$\displaystyle B=\left ( 10+5 \right )$$ min $$= 15$$ Minutes from point $$C$$
A flagstaff $$5$$ m high stands on a building $$25$$ m high. As observed from a point at a height of $$30$$ m, the flagstaff and the building subtend equal angles. The distance of the observer from the top of the flagstaff is
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$$\displaystyle \frac{5\sqrt{3}}{2}$$
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$$\displaystyle 5\sqrt{\frac{3}{2}}$$
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$$\displaystyle 5\sqrt{\frac{2}{3}}$$
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None of thses
Explanation
OB is the bisector of DOAC of DAOC
$$\displaystyle \therefore \frac{OC}{OA}=\frac{BC}{AB}=\frac{x}{\sqrt{x^{2}+30^{2}}}=\frac{5}{25}$$
$$\displaystyle \Rightarrow x=\frac{30}{\sqrt{24}}=\frac{30}{2\sqrt{6}}$$
$$\displaystyle \therefore x=5\sqrt{\frac{3}{2}}$$
The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of $$100\ m$$ from its base is $$\displaystyle 45^{o}$$ If the angle of elevation of the top of the complete pillar at the same point is to be $$\displaystyle 60^{o}$$ then the height of the incomplete pillar is to be increased by
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$$\displaystyle 50\sqrt{2}\ m$$
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$$100\ m$$
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$$\displaystyle 100\left ( \sqrt{3-1} \right )\ m$$
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$$\displaystyle 100\left ( \sqrt{3+1} \right )\ m$$
Explanation
Let BC be the incomplete and BD be the complete pillar In Triangles ABC and ABD we have
Given $$BC=100 $$m,
And $$\angle CAB=45^{0}$$$$\angle DAB=60^{0}$$
Let $$CD= x$$ m
In $$\Delta ABC$$
$$tan 45^{o}=\frac{BC}{AB}$$
$$\Rightarrow 1=\frac{BC}{100}$$
$$\Rightarrow BC=100 m$$
In $$ABD$$
$$tan60^{0}=\frac{AD}{AB}$$
$$\Rightarrow \sqrt{3}=\frac{BD}{100}$$
$$\Rightarrow BD=100\sqrt{3}$$m
Then $$BD=BC+CD=100+x$$
$$\therefore 100+x=100\sqrt{3}$$
$$\Rightarrow x=100(\sqrt{3}-1)$$m
The length of the shadow of a person is x when the angle of elevation of the sun is $$\displaystyle 45^{0}$$. If the length of the shadow increased by $$\displaystyle \left ( \sqrt{3}-1 \right )x$$ then the angle of elevation becomes
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$$\displaystyle 15^{0}$$
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$$\displaystyle 18^{0}$$
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$$\displaystyle 25^{0}$$
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$$\displaystyle 30^{0}$$
Explanation
Let $$AB$$ be the height of the person,
$$AC$$ and $$AD$$ be his shadows at two instants.
$$\displaystyle AC=x$$
$$\displaystyle AD= AC+CD =x+\left ( \sqrt{3}-1 \right )x$$
$$\displaystyle =\sqrt{3}x$$
and $$\displaystyle \angle ACB=45^{0}$$
$$\displaystyle \frac{AB}{AC}=\tan 45^{0}=1\\ \Rightarrow AB=AC=x$$
$$\displaystyle \tan \theta =\frac{AB}{AD}\\ =\dfrac{x}{\sqrt{3x}}\\ =\dfrac{1}{\sqrt{3}}$$
$$\displaystyle \Rightarrow \theta =30^{0}$$
A tower subtends an angle $$\displaystyle \alpha $$ at point 'A' in the plane of its base and the angle of depression of the foot of the tower at height 'b' just above A is $$\displaystyle \beta $$. The height of the tower is
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$$\displaystyle b\tan \alpha \cot \beta $$
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$$\displaystyle b\cot \alpha \cot \beta $$
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$$\displaystyle b\tan \alpha \tan \beta $$
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$$\displaystyle b\cot \alpha \tan \beta $$
Explanation
Let $$\displaystyle AP=BR=x$$ then from $$\displaystyle \Delta APQ$$ we find that
the height of the tower $$\displaystyle =PQ=x\tan \alpha =\left ( b\cot \beta \right )\tan \alpha $$
($$\displaystyle \because $$ In triangle APB, $$\displaystyle \frac{x}{b}=\cot \beta $$)
From the top of the cliff $$150\ m$$ high the angles of depression of the top and bottom of a tower are observed to be $$\displaystyle 30^{\circ}$$ and $$\displaystyle 60^{\circ}$$ respectively. The height of the tower is
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$$100\ m$$
0%
$$\displaystyle 50\sqrt3\ {m}$$
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$$\displaystyle 133\frac{1}{3}$$
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$$\displaystyle 100\left ( \sqrt{3}-1 \right )$$
Explanation
$$\displaystyle BF=\left ( 150-x \right )m.$$ Let $$\displaystyle AC=y$$
$$\displaystyle \tan 30^{0}=\frac{150-x}{y}$$
$$\displaystyle \tan 60^{0}=\frac{150}{y}$$
$$\displaystyle \Rightarrow y=\left ( 150-x \right )\sqrt{3}=\frac{150}{\sqrt{3}}$$
$$\displaystyle \therefore 150-x=50$$ or $$\displaystyle x=100\ m$$
If the length of the shadow of a pole on ground is twice the length of the pole, then the angle of elevation of the sun is _____
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30$$^o$$
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45$$^o$$
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60$$^o$$
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None of these
Explanation
Referring figure, if $$AB$$ is pole and $$AC$$ is length
of the shadow of pole,
then $$AB = x$$, $$AC = 2x$$
and let $$\angle\; ACB=\theta$$
Then, $$\tan$$$$\theta$$ = $$\displaystyle{\frac{AB}{AC} \Rightarrow \tan \theta = \frac{x}{2x} = \frac{1}{2}}$$
$$\therefore$$ Value of $$\theta$$ is neither 30$$^o$$, 60$$^o$$ nor 45$$^o$$.
So, option D is correct.
A tree is broken by the wind and now its upper part touches the ground at a point $$10$$ m from the foot of the tree and makes an angle of $$\displaystyle 60^{0}$$ with the ground. The entire length of the tree is:
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$$15$$ m
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$$20$$ m
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$$\displaystyle \left ( 10+\frac{20}{\sqrt3} \right )\sqrt{3}m$$
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$$\displaystyle \left ( 10+\frac{\sqrt{3}}{2} \right )m$$
Explanation
Let say the Length of the tree $$ = BT = BC + CT$$ and $$CT = CA$$ when the tree is broken.
Given: $$AB = 10m$$
In $$\triangle ABC$$,
$$\tan60^o = \dfrac{BC}{AB} $$ ............(1)
Also, $$sec\ 60^o = \dfrac{AC}{AB}$$ ...................(2)
$$ BT= $$ $$BC + AC$$
$$=$$ $$AB \displaystyle \tan 60^{0}+AB\ \text{sec} 60^{0}$$ {Using (1) and(2)}
$$= AB\sqrt3 + AB\sec60^o$$
$$\displaystyle =AB\left ( \sqrt{3}+ 2 \right ) =10\left ( 1+\frac{2}{\sqrt3} \right )\sqrt{3}m$$
$$\displaystyle =\left ( 10+\frac{20}{\sqrt3} \right )\sqrt{3}m$$
A flagstaff $$10\ m$$ high stands at the center of a equilateral triangle which is horizontal at the top of the flagstaff. Each side subtends at an angle of $$\displaystyle 60^{o}$$. The length of each side of the triangle is:
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$$\displaystyle \dfrac{6\sqrt{3}}{3}$$
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$$\displaystyle \dfrac{4\sqrt{6}}{3}$$
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$$\displaystyle \dfrac{20}{\sqrt{3}}$$
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$$\displaystyle \dfrac{6\sqrt{5}}{3}$$
Explanation
$$\Rightarrow$$ Considering Right angle $$\triangle ADC$$ we have,
$$\Rightarrow$$ $$\tan 60^o=\dfrac{AD}{DC}$$
$$\Rightarrow$$ $$\sqrt{3}=\dfrac{10}{DC}$$
$$\Rightarrow$$ $$DC=\dfrac{10}{\sqrt{3}}$$
$$\Rightarrow$$ $$DC=\dfrac{10}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}$$
$$\therefore$$ $$DC=\dfrac{10\sqrt{3}}{3}$$
$$\Rightarrow$$ $$BC=2\times DC$$
$$\therefore$$ $$BC=2\times \dfrac{10\sqrt{3}}{3}$$
$$\therefore$$ $$BC=\dfrac{20\sqrt{3}}{3}$$
$$\therefore$$ The length of each side of the triangle is $$\dfrac{20}{\sqrt{3}}\,m$$
If the angle of elevation of an object from a point $$100 \text{ m}$$ above a lake is found to be $$\displaystyle 30^{\circ}$$ and the angle of depression of its image in the lake is $$\displaystyle 45^{\circ}$$, then the height of the object above the lake is
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$$100\left ( 2-\sqrt{3} \right )\text{ m}$$
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$$100\left ( 2+\sqrt{3} \right )\text{ m}$$
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$$100\left ( \sqrt{3}-1 \right )\text{ m}$$
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$$1000\left ( \sqrt{3}+1 \right )\text{ m}$$
Explanation
Let $$AB$$ the surface of the lake $$O$$ be the point of observation $$P$$ be the object and $$P'$$ be its image.
Let the distance of object $$P$$ above lake be $$x.$$ So, it will be below the lake by same distance as they are mirror images.
$$PE=P'E=x$$
$$PD=x-100$$
$$P'D=x+100$$
$$OC=DE=100 \text{ m}$$
In $$\triangle OPD,$$
$$\tan 30^{\circ}=\dfrac{PD}{OD}$$
$$\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{x-100}{OD}$$
$$\Rightarrow OD=\sqrt3 (x-100)\text{ m}$$
In $$\triangle OP'D,$$
$$\begin{aligned}{}\tan {45^\circ } &= \frac{{P'D}}{{OD}}\\OD& = P'D\\\sqrt 3 (x - 100) &= x + 100\\x& = \frac{{100(\sqrt 3 + 1)}}{{\sqrt 3 - 1}}\\ &= 100(2 + \sqrt 3 ){\text{ m}}\end{aligned}$$
Hence, the distance of the object above the lake is $$100(2+\sqrt3)\text{ m}.$$
The angle of elevation $$\displaystyle \theta $$ of a vertical tower from a point on the ground is such that its tangent is $$\displaystyle \frac{5}{12}$$. On walking $$192$$ meters towards the tower in the same straight line the tangent of angle of elevation $$\displaystyle \phi $$ is found to be $$\displaystyle \frac{3}{4}$$. Find the height of the tower
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$$170$$ m
0%
$$175$$ m
0%
$$180$$ m
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$$185$$ m
Explanation
In $$\displaystyle \Delta PQR,\tan \theta =\frac{h}{192+x}=\frac{5}{12}$$
$$\displaystyle \Rightarrow x=\frac{1}{5}\left ( 12h-960 \right )$$ ...(i)
In $$\displaystyle \Delta PQS,\tan \phi =\frac{h}{x}=\frac{3}{4}$$
$$\displaystyle \Rightarrow x=\frac{4h}{3}$$ ...(ii)
From (i) and (ii), we get
$$\displaystyle \frac{4h}{3}=\frac{1}{5}\left ( 12h-960 \right )$$
$$\displaystyle \Rightarrow h=180$$
$$\displaystyle \therefore $$ The height of the tower is 180 m
From the top of a lighthouse, the angles of depression of two stations on opposite sides and $$a$$ distance apart are $$\displaystyle \alpha $$ and $$\displaystyle \beta $$. The height of the lighthouse is:
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$$\displaystyle \frac{a}{\cot \alpha \cot \beta }$$
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$$\displaystyle \frac{a}{\cot \alpha +\cot \beta }$$
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$$\displaystyle \frac{a\cot \alpha \cot\beta }{\cot \alpha +\cot \beta }$$
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$$\displaystyle \frac{a\tan \alpha \tan \beta }{\cot \alpha +\cot \beta }$$
Explanation
Let Height of Light House be $$h=AD$$, as shown in the figure above.
and $$\angle ABD=\alpha\ (Given)$$
$$\angle ACD =\beta\ (Given)$$
Now, in $$\triangle ABD,\ BD=h\cot \alpha\quad \quad \left[\cot \theta=\dfrac {\text{Adjacent Side}}{\text{Opposite Side}} \right]$$
Also, in $$\triangle ADC, \ CD=h\cot\beta\quad \quad \left[\cot \theta=\dfrac {\text{Adjacent Side}}{\text{Opposite Side}} \right]$$
Given that, $$BC=a$$
$$\therefore \ a=BC=BD+CD\\$$
$$\Rightarrow a=h\cot\alpha+h\cot\beta\\$$
$$\Rightarrow a=h(\cot\alpha+\cot\beta)\\$$
$$\therefore \ h=\dfrac{a}{\cot\alpha+\cot\beta}\\$$
Hence, option B is correct.
An aeroplane flying at a height of $$300\ metre$$ above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on ground are $$\displaystyle 60^{\circ}$$ and $$\displaystyle 45^{\circ}$$ respectively. The height of the lower plane above the above the ground in meters is
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0%
$$\displaystyle 100\sqrt{3}$$
0%
$$\displaystyle \frac{100}{\sqrt{3}}$$
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$$50$$
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$$\displaystyle 150\left ( \sqrt{3+1} \right ).$$
Explanation
$$P$$ and $$Q$$ are positions of planes.
From $$\displaystyle \Delta OMP;MP=300\ m$$
$$\displaystyle OM=MP\cot 60^{0}=300\times \dfrac{1}{\sqrt{3}}\ m$$
$$\displaystyle =100\sqrt{3}$$
In $$\displaystyle \Delta OMQ,MQ=OM$$
Height of the lower plane,$$ MQ = OM = 100\sqrt{3}\ m$$
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