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CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 9 - MCQExams.com
CBSE
Class 10 Maths
Some Applications Of Trigonometry
Quiz 9
The angle of elevation of a ladder leaning against a wall is $$60^{\circ}$$ 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
Let $$AB$$ be the wall and $$BC$$ be the ladder.
Then, $$\angle ACB = 60^{\circ}$$ and $$AC = 4.6\ m$$.
$$\dfrac {AC}{BC} = \cos 60^{\circ} = \dfrac {1}{2}$$
$$\Rightarrow BC = 2\times AC$$
$$= (2\times 4.6)m$$
$$= 9.2\ m$$
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of $$30^{\circ}$$ with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes $$60^{\circ}$$. What is the distance between the base of the tower and the point P?
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$$4\sqrt {3} units$$
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$$8\ units$$
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$$12\ units$$
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Data inadequate
Explanation
From the data, we conclude that one of $$AB, AD$$ and $$CD$$ length are not provided.
So, the data is inadequate.
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are $$30^{\circ}$$ and $$45^{\circ}$$ 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 lighthouse and C and D be the positions of the ships.
Then, $$AB = 100\ m, \angle ACB = 30^{\circ}$$ and $$\angle ADB = 45^{\circ}$$.
$$\dfrac {AB}{AC} = \tan 30^{\circ} = \dfrac {1}{\sqrt {3}} \Rightarrow AC = AB \times \sqrt {3} = 100\sqrt {3}m$$.
$$\dfrac {AB}{AD} = \tan 45^{\circ} = 1\Rightarrow AD = AB = 100\ m$$.
$$\therefore CD = (AC + AD) = (100\sqrt {3} + 100)m$$
$$= 100(\sqrt {3} + 1)$$
$$= (100\times 2.73)m$$
$$= 273\ m$$.
The length of shadow of a tree is $$16\ m$$ when the angle of elevation of the sun is $$60^{\circ}$$. What is the height of the tree?
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$$8$$ m
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$$16$$ m
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$$16\sqrt {3}$$ m
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$$\dfrac {16}{\sqrt {3}}$$ m
Explanation
Let $$ AB=h$$ be
height of the tree and $$BC$$ be it's shadow.
In $$\triangle ABC$$
$$\dfrac {h}{16}=\tan 60^o$$
$$h=\tan 60^0\times16$$
$$h=16\sqrt 3$$
$$\therefore $$ Height of the tree is $$16\sqrt 3\ m$$
A building that is 150 ft tall casts a shadow of 20 feet long. At the same time a tree casts a shadow of 2 ft. How tall is the tree?
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10
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15
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20
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25
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30
Explanation
Given: A building that is $$150 ft$$ tall casts a shadow of $$20$$ feet long. At same time a tree casts a shadow of $$2$$ ft.
Height of a building $$= 150 ft$$
Shadows height casted by building $$= 20 ft$$
Shadow height casted by tree $$= 2ft$$
According to the question
$$\Rightarrow \tan \theta = \dfrac{n}{2}= \dfrac{150}{20}$$
$$\Rightarrow h = \dfrac{150 \times 2}{20}$$
$$\Rightarrow h = 15 ft$$
Height of the tree $$= 15 ft$$
$$\therefore$$ Option B is correct.
The ramp for unloading a moving truck, has an angle of elevation of $$30^{\circ}$$. If the top of the ramp is $$0.9$$m above the ground level, then find the length of the ramp.
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$$1.8\ m$$
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$$1\ m$$
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$$0.8\ m$$
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None of these
To find the cloud ceiling, one night an observer directed a spotlight vertically at the clouds. Using a theodolite placed $$100\text{ m}$$ from the spotlight and $$1.5$$m above the ground, he found the angle of elevation to be $$60^{\circ}$$. How high was the cloud ceiling? ( Hint : See figure)
( Note : Cloud ceiling is the lowest altitude at which solid cloud is present. The cloud ceiling at airports must be sufficiently high for safe take offs and landings. At night the cloud ceiling can be determined by illuminating the base of the clouds by a spotlight pointing vertically upward.)
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$$174.7\text{ m}$$
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$$14.7\text{ m}$$
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$$147.7\text{ m}$$
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None of these
A person $$X$$ standing on a horizontal plane, observes a bird flying at a distance of $$100\ m$$ from him at an angle of elevation of $$30^{o} $$. Another person $$Y$$ standing on the roof of a $$20$$m high building, observes the bird at the same time at an angle of elevation of $$45^{o}$$. If $$X$$ and $$Y$$ are on the opposite sides of the bird, then find the distance of the bird from $$Y$$.
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$$30\sqrt{2}m$$
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$$20\sqrt{2}m$$
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$$30\sqrt{3}m$$
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$$None\ of\ these$$
Explanation
R.E.F image
For $$ \Delta XBA,$$
$$ sin\,30^{\circ} = \dfrac{BA}{XB} $$ as $$ sin\,\theta = \dfrac{OPP}{hyp} $$
$$ \dfrac{1}{2} = \dfrac{BA}{100} $$
$$ \dfrac{100}{2} = BA $$
$$ 50\,m = BA $$
For $$ \Delta BYD, $$
$$ sin\,45^{\circ} = \dfrac{BD}{BY} $$
$$ \dfrac{1}{\sqrt{2}} = \dfrac{BA-AD}{BY} $$
$$ \dfrac{1}{\sqrt{2}} = \dfrac{BA-YC}{BY} $$ [as $$ AD = YC $$]
$$ \dfrac{1}{\sqrt{2}} = \dfrac{50-20}{a} $$
$$ a = 30\sqrt{2}m $$
$$ \therefore $$ The distance of the bird from
Y is $$ 30\sqrt{2}m $$
A person in a helicopter flying at a height of $$700$$m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are $$30^{\circ}$$ and $$45^{\circ}$$ respectively. Find the width of the river. $$\left ( \sqrt{3}=1.732 \right )$$
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$$1.9\ km$$
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$$1.4\ km$$
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$$2.1\ km$$
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None of these
A lamp-post stands at the centre of a circular park. Let $$P$$ and $$Q$$ be two points on the boundary such that $$PQ$$ subtends and angle $$90^{\circ}$$ at the foot of the lamp-post and the angle of elevation of the top of the lamp post from $$P$$ is $$30^{\circ}$$. If $$PQ=30$$ m, then find the height of the lamp post.
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$$5\sqrt 6$$ m
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$$4\sqrt 6$$ m
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$$6\sqrt 5$$ m
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$$8\sqrt 3$$ m
Explanation
Let The center of Park be $$O$$ which is bottom of the temple. Let $$A$$ be the top point of the temple.
Here $$OP$$ and $$OQ$$ are radius of park
$$\implies OP=OQ$$
$$ PQ=30\ m$$
Since the $$\triangle POQ$$ is Right angled triange
$$\implies PQ^2=OP^2+OQ^2$$
$$\implies 900=2OP^2$$
$$\implies OP^2=450$$
$$\implies OP= \sqrt{450}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =15\sqrt 2\ m$$
Now $$\triangle POA$$ will also form a right triangle.
$$\tan \angle OPA=\dfrac{Height }{Distance }$$
$$\tan 30^o=\dfrac{Height}{15\sqrt 2}$$
$$\implies Height =\dfrac 1{\sqrt 3}\times 15\sqrt 2$$
$$\implies Height=\dfrac{15\sqrt 6}{3}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =5\sqrt 6\ m$$
The angles of elevation of an artificial earth satellite is measured from two earth stations, situated on the same side of the satellite, are found to be $$30^{\circ}$$ and $$60^{\circ}$$. The two earth stations and the satellite are in the same vertical plane. If the distance between the earth stations is $$4000$$ km, find the distance between the satellite and earth. $$\left ( \sqrt{3}=1.732 \right )$$
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$$3446$$ km
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$$3464$$ km
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$$3488$$ km
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$$3400$$ km
Explanation
Consider A and B be two earth stations
Let AB $$=4000$$ km
Let BC $$=y$$ km and CD $$=x$$ km, where CD is the height of the satellite from the earth
From $$\triangle ACD$$
$$\tan 30^{\circ}=\dfrac{x}{4000+y}$$
$$\implies \dfrac{x}{4000+y}=\dfrac{1}{\sqrt3}$$
$$\implies \sqrt{3}x=4000+y$$ ........(1)
From $$\triangle BCD$$
$$\tan 60^{\circ}=\dfrac{x}{y}$$
$$\implies \dfrac{x}{y}=\sqrt3$$
$$\implies y=\dfrac{x}{\sqrt{3}}$$ ........(2)
From (1) and (2)
$$\sqrt{3}x=4000+\dfrac{x}{\sqrt 3}$$
$$\implies 3x=4000\sqrt{3}+x$$
$$\implies 2x=4000\sqrt3$$
$$\implies x=2000\sqrt3\text {km}$$
$$\implies x=2000\times 1.732=3464\text{ km}$$ $$[\because \sqrt 3=1.732]$$
Hence, the distance of satellite from the earth is $$3464\text{ km}$$
An observer from the top of the light house, the angle of depression of two ships $$P$$ and $$Q$$ anchored in the sea to the same side are found to have measure $$35$$ and $$50$$ respectively. Then from the light house ...............
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The distance of $$P$$ is more than $$Q$$.
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The distance of $$Q$$ is more than $$P$$.
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$$P$$ and $$Q$$ are at equal distance.
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The relation about the distance of $$P$$ and $$Q$$ cannot be determined.
Explanation
We know that the less the angle of depression the more it is far from the base
Hence distance of $$P$$ is more than $$Q$$.
From the figure we can see that,
The distance of $$P$$ from the lighthouse is more than the distance of $$Q$$ from the lighthouse.
From the top of a building $$h$$ metre high, the angle of depression of an object on the ground has a measure $$\theta$$. The distance of the object from the building is
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$$h\cos { \theta } $$ metre
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$$h\sin { \theta } $$ metre
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$$\tan { \theta } $$ metre
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$$h\cot { \theta } $$
Explanation
Height of the building is AB $$=h$$ metre and the object is located at point $$C$$
Now, $$\tan{\theta}=\cfrac {\text{opposite side}}{\text{adjacent side}}$$
$$\tan{\theta}=\cfrac {h}{\text{adjacent side}}$$
We need to find the adjacent side,
$$\implies \text{adjacent side}=\cfrac { h }{ \tan { \theta } } =h\cot { \theta } $$
Hence, the answer is $$h\cot \theta$$.
On walking ............... metres on a slope at an angle of measure 30 with the ground, one can reach the height '$$a$$' metres from the ground.
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$$\cfrac { 2a }{ \sqrt { 3 } } $$
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$$\cfrac { \sqrt { 3 } }{ 2 } a$$
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$$2a$$
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$$\cfrac{a}{2}$$
Explanation
Assume that on walking '$$y$$' metres, one can reach the height of '$$a$$' metres from the ground.
$$\sin{30^{o}} =\cfrac{a}{y} $$
$$\implies \cfrac { 1 }{ 2 } =\cfrac { a }{ y } $$
$$\implies y=2a$$
Hence, the answer is $$2a$$
On walking $$x$$ meters, making an angle of $${30}^{o}$$ with the ground, to find a ball fallen in a valley, one can reach a depth of '$$y$$' meters below the ground, then
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$$x=y$$
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$$x=2y$$
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$$2x=\sqrt { 3 } y$$
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$$2x=y$$
Explanation
According to the given fig.
$$\sin{30^{o}} =\cfrac{1}{2}$$
$$\cfrac{y}{x}=\cfrac{1}{2}$$
$$\therefore x=2y$$
When the length of the shadow of a pole is equal to the height of the pole, the angle of elevation of the Sun has measure of ................
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$${ 30 }^{ o }$$
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$${ 45 }^{ o }$$
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$${ 60 }^{ o }$$
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$${ 75 }^{ o }$$
Explanation
Let $$AB=h$$ be the height of the pole.
The shadow of a pole will form base $$BC$$
Let $$\theta$$ be the angle of elevation.
According to the fig.
$$\tan{\theta} =\cfrac{AB}{BC} $$
$$\implies \tan{\theta}=\cfrac{h}{h} =1$$
$$\implies \theta =\tan^{-1}(1)$$
$$\implies \theta=45^{o}$$
Hence, the angle elevation is $$45^{o}$$.
In given figure, the minimum distance to reach from point "C" to point "A" will be ....................
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$$a^2$$
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$$\sqrt 2$$
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2
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2a
Explanation
$$tan A = \dfrac{BC}{AB}$$
$$tan 60^o = \dfrac{BC}{a}$$
$$BC = a tan 60^o$$
$$BC = \sqrt 3 a$$
In $$\Delta ABC$$,
Using Pythagoras theorem we get,
$$AC^2 = AB^2 + BC^2$$
$$AC^2 = a^2 + (\sqrt 3a)^2$$
$$AC^2 = 4a^2$$
$$AC = 2a$$
An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from his/her eyes has measureWhat is the height of the tower?
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28.5 m
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30 m
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27 m
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1.5 m
Explanation
We can form the above figure by given data
$$\implies BD = CE = 1.5$$
In $$\Delta ABC, \angle CBA=90^{o}$$
$$\implies \tan 45^{o} = \dfrac{AB}{BC}$$
$$\implies 1 = \dfrac{AB}{28.5}$$
$$\implies AB = 28.5$$
Now, $$h = AB + BD = 28.5 + 1.5 = 30$$
Hence, height of the tower is $$30\ m$$
The angle of elevation of a tower at a level ground is $$30^o$$ . The angle of elevation becomes $$\theta$$ when moved 10 m towards the tower. If the height of tower is $$ 5 \sqrt{3} m $$, then what is $$\theta$$ equal to ?
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$$45^o$$
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$$60^o$$
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$$75^o$$
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None of the above
Explanation
In $$\triangle ABC,$$
$$\tan(30^{o})=\dfrac { 5\sqrt { 3 } }{ 10+x } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { 5\sqrt { 3 } }{ 10+x } $$
$$\Rightarrow x+10=15$$
$$\Rightarrow x=5$$
In $$\triangle ADC,$$
$$\tan(\theta )=\dfrac{5\sqrt{3}}{x}=\dfrac { 5\sqrt { 3 } }{ 5 } =\sqrt { 3 }$$
$$\Rightarrow \theta =60^{o}$$
Hence, option B is correct.
What is the angle subtended by 1 m pole at a distance 1 km on the ground in sexagesimal measure ?
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$$\frac{9} {50 \pi}$$ degree
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$$\frac{9} {5 \pi}$$ degree
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3.4 minute
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3.5 minute
From the top of a lighthouse $$70m$$ high with its base at sea level, the angle of depression of a boat is $${ 15 }^{ o }$$. The distance of the boat from the foot of the lighthouse is:
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$$70(2-\sqrt { 3 } )m$$
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$$70(2+\sqrt { 3 } )m$$
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$$70(3-\sqrt { 3 } )m$$
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$$70(3+\sqrt { 3 } )m$$
Explanation
Let AB be the light house and the boat is at point C
Angle BCA is also equal to angle of depression $$=15^{o}$$
$$\Rightarrow \angle BCA=15^{o}$$
Now, $$\tan (15^{o})=\dfrac { AB }{ BC }$$
$$\Rightarrow BC=AB(\tan (15^{o}))\\ \tan (15^{o})=\tan (60^{o}-45^{o})\\ =\dfrac { \tan 60^{o}-\tan 45^{o} }{ 1+\tan 60^{o}\tan 45^{o} } =\dfrac { \sqrt { 3 } -1 }{ 1+\sqrt { 3 } } =\dfrac { { (\sqrt { 3 } -1) }^{ 2 } }{ 3-1 } =2-\sqrt { 3 }$$
So, $$BC=70(2-\sqrt { 3 } )m$$
Option A is correct
The angles of depression of two boats as observed from the mast head of a ship $$50$$ metres high are $${45}^{o}$$ and $${30}^{o}$$. The distance between the boats, if they are on the same side of mast head in line with it, is
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$$50\sqrt { 3 } $$ metres
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$$0(\sqrt { 3 } +1)$$ metres
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$$50(\sqrt { 3 } -1)$$ metres
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$$50(1-\sqrt { 3 } )$$ metres
Explanation
From $$\triangle ADC,\quad \dfrac { 50 }{ x } =Tan{ 45 }^{ 0 }\Rightarrow x=50m$$
From $$\triangle ABD,\quad \dfrac { 50 }{ h+50 } =Tan{ 30 }^{ 0 }$$
$$\Rightarrow \quad h=50\left( \sqrt { 3 } -1 \right) m$$
Two poles are 10 m and 20 m high. The line joining their tips makes an angle of $$15^o$$ with the horizontal. What is the distance between the poles ?
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$$10 ( \sqrt{3} - 1 ) m$$
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$$ 5 ( 4 + 2 \sqrt{3} -1 ) m$$
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$$20 ( \sqrt{3} + 1 ) m$$
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$$10 ( \sqrt{3} +2 ) m$$
Explanation
Now AC will be the distance between the poles
Here,
$$\tan(\frac { \theta }{ 2 } )=\pm\sqrt { \dfrac { 1-cos\theta }{ 1+cos\theta } } $$
Now,
$$\tan(15^{o})=\dfrac { 10 }{ AC } $$
$$\Rightarrow AC=\dfrac { 10 }{\tan(15^{o})}$$
But, $$\tan(15^{o})=\sqrt { \dfrac { 1-\cos(30^{o}) }{ 1+\cos(30^{o})}}$$
$$=\sqrt { \dfrac { 1-\dfrac { \sqrt { 3 } }{ 2 } }{ 1+\dfrac { \sqrt { 3 } }{ 2 } } } =\sqrt { \dfrac { 2-\sqrt { 3 } }{ 2+\sqrt { 3 } } } =\sqrt { \dfrac { 2-\sqrt { 3 } }{ 2+\sqrt { 3 } } \times \dfrac { 2-\sqrt { 3 } }{ 2-\sqrt { 3 } } } =2-\sqrt { 3 } $$
So, $$AC=\dfrac { 10 }{ 2-\sqrt { 3 } } =\dfrac { 10 }{ 2-\sqrt { 3 } } \times \dfrac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } } =20+10\sqrt { 3 } $$
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite of bank is $${60}^{o}$$. When he retires $$40 m$$ from the bank, he finds the angle to be $${30}^{o}$$. What is the breadth of the river?
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$$60 m$$
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$$40 m$$
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$$30 m$$
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$$20 m$$
Explanation
Let $$R$$ br the length of River
and $$T$$ be the length of tree
In $$\triangle ABC$$
$$\tan60^o=\dfrac{T}{R}\Rightarrow T=\sqrt3R.......(1)$$
in $$\triangle ABD$$
$$\tan30^o=\dfrac{T}{R+40}\Rightarrow T=\dfrac{R+40}{\sqrt3}.....(2)$$
From (1) and (2)
$$\sqrt3R=\dfrac{R+40}{\sqrt3}\Rightarrow 3R=R+40\Rightarrow R=20$$
Therefore, Answer is $$20m$$
The top of a hill observed from the top and bottom of a building of height $$h$$ is at angles of elevation $$\alpha$$ and $$\beta$$ respectively. The height of the hill is
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$$\cfrac { h\cot { \beta } }{ \cot { \beta } -\cot { \alpha } } $$
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$$\cfrac { h\cot { \alpha } }{ \cot { \alpha } -\cot { \beta } } $$
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$$\cfrac { h\tan { \alpha } }{ \tan { \alpha } -\tan { \beta } } $$
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None of the above
Explanation
Refer the image
AB is the hill and CD is the building
$$\tan (\alpha )=\dfrac { AB }{ BE } \\ \implies BE=AB\cot (\alpha )\\ \tan (\beta )=\dfrac { AB }{ BD } \\ \implies BD=AB\cot (\beta )$$
Also, $$\tan (\alpha )=\dfrac { CD }{ DE }$$
$$\implies DE=CD\cot (\alpha )=h\cot (\alpha )$$
Now, $$BE=BD+DE$$
$$\Rightarrow AB\cot (\alpha )=AB\cot (\beta )+h\cot (\alpha )\\ \Rightarrow AB=\dfrac { h\cot (\alpha ) }{ \cot (\alpha )-\cot (\beta ) } $$
Option B is correct
A man observes two objects in a line in the west. On walking a distance $$x $$ towards the north, the objects subtends an angle $$\alpha$$ in front of him and on walking a further distance $$x$$ to north they subtend an angle, $$\beta$$, then the distance between the objects is
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$$\cfrac { 2x }{ 2\cot { \beta } -\cot { \alpha } } $$
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$$\cfrac { 3x }{ 2\cot { \beta } -\cot { \alpha } } $$
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$$\cfrac { 3x }{ 2\cot { \beta } +\cot { \alpha } } $$
0%
none of these
Explanation
$$ \tan p = \dfrac{2x}c$$ $$ \tan q = \dfrac{x}c$$
$$\tan a = \dfrac{2x}{c+d}$$ $$ \tan b = \dfrac{x}{c+d}$$
$$\beta + a = p$$
$$\alpha + b = q$$
hence $$ \tan\beta$$ = $$\tan ({p-a})$$
$$ \tan\alpha$$ = $$\tan ({q-b})$$
so
$$\tan\beta = \dfrac{\tan p-\tan a}{1-\tan p\tan a}$$
$$ = \dfrac{\dfrac{2x}{c}-\dfrac{2x}{c+d}}{1+\dfrac{2x}c\dfrac{2x}{c+d}}$$
$$=\dfrac{2xd}{c^2+dc+4x^2}$$
$$c^2+dc+4x^2= 2xd\cot\beta$$ ...........(1)
$$\tan\alpha = \dfrac{\tan q-\tan b}{1-\tan q\tan b}$$
$$ = \dfrac{\dfrac{x}{c}-\dfrac{x}{c+d}}{1+\dfrac{x}c\dfrac{x}{c+d}}$$
$$=\dfrac{xd}{c^2+dc+x^2}$$
$$c^2+dc+x^2= xd\cot\alpha$$ ...........(2)
From 1 and 2
$$3x^2= xd(2\cot\beta-\cot\alpha)$$
$$\Rightarrow d = \dfrac{3x}{2\cot\beta - \cot\alpha}$$
An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\sqrt {3}\ km$$ above it, is observed at an elevation of $$60^{o}$$ from a point on the ground. If, after five seconds, its elevation from the same point, is $$30^{o}$$, then the speed (in $$km/ hr$$) of the aeroplane, is
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$$1500$$
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$$750$$
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$$720$$
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$$1440$$
Explanation
Let the observing point be $$O$$
The height at which the plane was flying above ground is constant at $$AP= QB= \sqrt{3}$$ km.
In the first case, distance of projection of plane from point of observation is $$\tan{60^0}=\frac{AP}{OP}\implies OP= 1$$ km.
In the first case, distance of projection of plane from point of observation is $$\tan{30^0}=\frac{BP}{OQ}\implies OP= 3$$ km.
So, a distance of $$3-1=2$$ km. is covered in $$5$$ seconds. So the speed of the plane is $$\dfrac{2\times3600}{5} = 1440$$ km/hr.
So option D is the correct answer.
The height of a house subtends a right angle at opposite window from the base of the house is $$60^0$$. If the width of the road be 6 metres, then the height of the house is
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$$6 \sqrt 3 m$$
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$$8 \sqrt 3 m$$
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$$10 \sqrt 3 m$$
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$$3 \sqrt 6$$
Explanation
Given $$\angle DBC=60^0$$ and $$\angle ADB=90^0$$ from figure we know that $$\angle EDB=\angle DBC=60^0$$
Since $$\angle ADB=\angle ADE+\angle EDB=90^0$$
$$\angle ADE =90^0-60^0=30^0$$
From AED ,
$$\tan30=\dfrac{AE}{6}$$
$$AE-2\sqrt3 m$$
From EDC,
$$\tan60=\dfrac{EB}{6}$$
$$EB=6\sqrt 3m$$
Height of the house =$$AE+EB=2\sqrt3+6\sqrt3=8\sqrt3 m$$
Over a tower $$AB$$ of height $$h\ mt$$, there is a flag staff $$BC, AB$$ and $$BC$$ makes equal angles at a point distant $$'d' mt$$ from the foot $$A$$ of the tower. The height of the flag staff is
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$$\dfrac {h(d^{2} + h^{2})}{(d^{2} - h^{2})}mt$$
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$$\dfrac {h(d^{2} - h^{2})}{(d^{2} - h^{2})}mt$$
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$$\dfrac {h(d^{2} - h^{2})}{(d^{2} + h^{2})}mt$$
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$$\dfrac {h(d^{2} + h^{2})}{(d^{2} + h^{2})}mt$$
The angle of elevation of the top of a tower as observed from a point on the horizontal ground is $$x$$. If we move a distance $$d$$ towards the flux of the tower, the angle of elevation increases to $$y$$, then the height of the tower is
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0%
$$\cfrac { d\tan { x } \tan { y } }{ \tan { y- } \tan { x } } $$
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$$d\left( \tan { y } +\tan { x } \right) $$
0%
$$d\left( \tan { y } -\tan { x } \right) $$
0%
$$\cfrac { d\tan { x } \tan { y } }{ \tan { y } +\tan { x } } $$
Explanation
In $$\triangle ABC,$$
$$\Rightarrow \tan { y }=\dfrac hz$$
$$\Rightarrow h=z\tan { y } \quad ...(1)$$
In $$\triangle ADC,$$
$$\tan { x }=\dfrac{h}{d+z}$$
$$\Rightarrow \tan x=\dfrac { z\tan { y } }{ d+z } $$ [from (1)]
$$\Rightarrow z\left( \tan { y- } \tan { x } \right) =d\tan { x } $$
$$\therefore\ h=\cfrac { d\tan { x } \tan { y } }{ \tan { y- } \tan { x } } $$
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