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CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 6 - MCQExams.com
CBSE
Class 10 Maths
Some Applications Of Trigonometry
Quiz 6
The length of a string between a kite and a point on the ground is 90m. The string makes an angle of $$\displaystyle 60^{\circ}$$ with the level ground if there is no slack in the string, the height of the kite is
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$$\displaystyle 90\sqrt{3}m$$
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$$\displaystyle 45\sqrt{3}m$$
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$$180 m$$
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$$45 m$$
Explanation
Let $$A$$ be the kite and $$AC$$ is $$90$$m
In $$ΔABC$$
$$\sin 60^{0}=\dfrac{AB}{AC}$$
$$=\dfrac{h}{90}$$
$$\therefore \dfrac{\sqrt 3}{2}=\dfrac{h}{90}$$
$$\therefore h = 45\sqrt 3$$m
The angle of elevation of the top of a tower at a distance of $$\displaystyle \frac{50\sqrt{3}}{3}$$ metres from the foot is $$\displaystyle 60^{\circ}$$. Find the height of the tower
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$$\displaystyle 50\sqrt{3}$$ metres
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$$\displaystyle \frac{20}{\sqrt{3}}$$ metres
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$$-50$$ metres
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$$50$$ metres
Explanation
Let the height of the tower be $$h$$.
In $$ΔABC$$
$$\tan 60^{0}=\dfrac{AB}{BC}$$
$$=\dfrac { h }{ \dfrac { 50\sqrt { 3 } }{ 3 } } $$
$$=>\sqrt 3=\dfrac { 3\times h }{ { 50\sqrt { 3 } } } $$
$$=\dfrac { 3\times h }{ { 50\sqrt { 3 } \times \sqrt { 3 } } } $$
$$=\dfrac { 3\times h }{ { 50\times 3 } } $$
∴$$h=50$$m
If from the top of a tower 50 m high, the angles of depression of two objects due north of the tower are respectively $$\displaystyle 60^{\circ}$$ and $$\displaystyle 45^{\circ}$$, then the approximate distance between the objects is :
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$$\displaystyle 50\left ( \sqrt{2}-2 \right )m$$
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$$\displaystyle 50\left ( \sqrt{3}-3 \right )m$$
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$$31 \ m$$
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None of these
Explanation
Let the two objects be C, D and the height of the tower is $$50$$m
In $$ΔABC$$
$$tan60^{0}=\dfrac{AB}{BC}$$
$$=\dfrac{50}{x}$$
$$\therefore x =\dfrac{ 50}{\sqrt 3}$$ ... {i}
Now, In $$ΔABD$$
$$tan45^{0}=\dfrac{AB}{BD}$$
$$\Rightarrow 1 =\dfrac{50}{x+y}$$
$$\Rightarrow x + y = 50$$
$$\Rightarrow y=x-50$$
$$\Rightarrow y=50 - \dfrac{50}{\sqrt 3}$$ ...[putting the value from {i}]
$$\Rightarrow y = \dfrac{50}{\sqrt 3}(\sqrt 3−1)$$
∴ Distance between two objects is $$=$$ $$\dfrac{50}{\sqrt 3}(\sqrt 3−1)$$
Rationalising the denominator we get the Distance$$=\dfrac{50 \sqrt{3}(\sqrt{3}-1)}{3}$$ m
Option D is correct.
A man on a cliff observes a boat at an angle of depression of $$\displaystyle 30^{\circ} $$ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later the angle of depression of the boat is found to be $$\displaystyle 60^{\circ}. $$ Find the total time taken by the boat from the initial point to reach the shore.
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$$9$$ min
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$$7$$ min
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$$10$$ min
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$$6$$ min
The angle of elevation of the top of tower as observed from a pint 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 of 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 the height of the tower be h.
$$\displaystyle \therefore $$ In $$\displaystyle \Delta ABC $$
$$\displaystyle \tan y=\dfrac{h}{b}$$
b = $$\displaystyle \dfrac{h}{\tan y}$$
In $$\displaystyle \Delta ABD $$
$$\displaystyle \tan x=\dfrac{h}{b+d}$$
$$\displaystyle \tan x=\dfrac{h}{\dfrac{h}{\tan y}+d}$$
$$\displaystyle \dfrac{h}{\tan x}=\dfrac{h}{\tan y}+d$$
$$d =$$ $$\displaystyle h\left ( \dfrac{1}{\tan x}-\dfrac{1}{\tan y} \right )$$
$$d =$$ $$\displaystyle h\left ( \dfrac{\tan y-\tan x}{\tan x\tan y} \right )$$
$$h =$$ $$\displaystyle \dfrac{d\tan x\tan y}{\tan y-\tan x}$$
The angels of elevation of the top of a vertical tower from two points $$30$$ meter apart and on the same straight line passing through the base of tower are $$\displaystyle 30^{\circ}$$ and $$\displaystyle 60^{\circ}$$ respectively The height of the tower is
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$$10 m$$
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$$15 m$$
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$$\displaystyle 15\sqrt{3}m$$
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$$30 m$$
Explanation
In $$ΔAPQ$$, $$\angle PAQ={ 30 }^{ ∘ }$$
Therefore,
$$\tan { { 30 }^{ ∘ } } =\dfrac { PQ }{ AQ } $$
$$\Rightarrow $$$$\dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ x } $$
$$\Rightarrow $$$$x=\dfrac { h }{ \sqrt { 3 } } $$............(eqn 1)
Similarly, $$ΔPQB$$, $$\angle PBQ={ 60 }^{ ∘ }$$
Therefore,
$$\tan { { 60 }^{ ∘ } } =\dfrac { h }{ 30-x } $$
$$\Rightarrow $$$$\sqrt { 3 } =\dfrac { h }{ 30-x } $$
............(eqn 2)
From equations 1 and 2,
$$\sqrt { 3 } =\dfrac { h }{ 30-\dfrac { h }{ \sqrt { 3 } } } $$
$$\Rightarrow $$$$\sqrt { 3 } \left( 30-\dfrac { h }{ \sqrt { 3 } } \right) =h$$
$$\Rightarrow $$$$2h=30\sqrt { 3 } $$
$$\Rightarrow $$$$h=15\sqrt { 3 } m$$
Thus height of the tower is $$15\sqrt { 3 } m$$.
Hence option C is correct.
The angle of elevation of the top of a tower from a point on the ground $$\displaystyle 30^{\circ}$$ after walking $$200$$ m towards the tower the angle of elevation becomes $$\displaystyle 60^{\circ}$$. The height of the tower is
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$$\displaystyle 100\sqrt{3}$$ m
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$$\displaystyle 200\sqrt{3}$$ m
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$$100$$ m
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$$200$$ m
Explanation
Let $$h$$ be the height of the tower
In $$ΔABC$$
$$tan 60^{0}=\dfrac{AB}{BC}$$
$$=>\sqrt 3=\dfrac{h}{x}$$
$$=>x = \dfrac{h}{\sqrt 3}$$ .......(i)
In $$ΔABD$$
$$tan 30^{0}=\dfrac{AB}{BD}$$
$$=>\dfrac{1}{\sqrt 3}=\dfrac{h}{x+200}$$
$$=>x + 200 = \sqrt 3h$$
$$=>x = \sqrt 3h - 200 $$ .......(ii)
From (i) and (ii) we get,
$$\dfrac{h}{\sqrt 3}=\sqrt 3h−200$$
$$=>h = 3h - 200 \sqrt 3$$
$$=>2h = 200\sqrt 3$$
$$h=> 100\sqrt 3$$m
$$=>h = 100 x 1.732$$
$$=>h = 173.2$$ metre
∴ Height of tower =$$ 173.2$$ m.
The angles of elevation of the top of a vertical tower from two points 30 metres apart and on the same straight line passing through the base of tower are $$\displaystyle 30^{\circ}$$ and $$\displaystyle 60^{\circ}$$ respectively. The height of the tower is
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$$10 m$$
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$$15 m$$
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$$\displaystyle 15\sqrt{3}m$$
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$$30 m$$
Explanation
Let the height of the tower is h
In $$ΔABC$$
$$\tan 60^{0}=\dfrac{h}{x}$$
$$=>\sqrt 3=\dfrac{h}{x}$$
$$=>x = \dfrac{h}{\sqrt 3}$$4 ......(i)
In $$ΔABD$$
$$\tan 30^{0}=\dfrac{h}{x+30}$$
$$=>\dfrac{1}{\sqrt 3} =\dfrac{h}{x+30}$$
$$=>x+30 = h\sqrt 3$$ ........(ii)
$$=>x=h\sqrt 3-30$$
From (i) and (ii)
$$\dfrac{h}{\sqrt 3}=h\sqrt 3−30$$
$$=>h = 3h - 30\sqrt 3$$
$$=>2h=30\sqrt 3$$
$$=>h = 15\sqrt 3$$
So, the height of tower = $$15\sqrt 3$$m
If the altitude of the sun is at $$\displaystyle 60^{\circ}$$ then the height of the vertical tower that will cast a shadow of length 20 m is
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$$\displaystyle 20\sqrt{3}m$$
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$$\displaystyle \frac{20}{\sqrt{3}}m$$
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$$\displaystyle \frac{15}{\sqrt{3}}m$$
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$$\displaystyle 40\sqrt{3}m$$
Explanation
Let the height of the tower be $$h$$
$$\tan 60^{0}=\dfrac{h}{20}$$
$$\Rightarrow \sqrt 3=\dfrac{h}{20}$$
$$\Rightarrow h = 20\sqrt 3$$m
$$\Rightarrow h=20\times 1.732$$m
$$\therefore h= 34.64$$m
The length of a ladder is exactly equal to the height of the wall it is leaning against. If the lower end of the ladder is kept on a bench of height 3 m. and the bench is kept 9 m. away from the wall, the upper end of the ladder coincides with the top of the wall. The height of the wall is
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$$11 m$$
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$$12 m$$
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$$15 m$$
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$$18 m$$
Explanation
$$AB^2+BC^2=AC^2$$
$$(h-2)^2+9^2=h^2$$
$$h=15 m$$
The shadow of a tower is 30 meters when the sun's altitude is $$\displaystyle 30^{\circ}$$. When the sun's altitude is $$\displaystyle 60^{\circ}$$ then the length of shadow will be
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$$60 m$$
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$$15 m$$
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$$10 m$$
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$$5 m$$
Explanation
In fig. AB is the tower , BC is the length of the shadow of tower when altitude is $$30^{\circ}$$ i.e the angle of elevation of the top of the tower from the tip of the shadow is $$30^{\circ}$$ and DB is the length of the shadow, when the angle of elevation is $$60^{\circ}$$
Given, BC = 30 m
In $$\triangle ABC $$
$$\tan 30^{\circ}= \dfrac {AB}{BC} $$
$$\Rightarrow \dfrac 1{\sqrt 3}= \dfrac {AB}{30} $$
$$\Rightarrow AB= \dfrac {30}{\sqrt 3} = 10\sqrt 3 m $$
Now,
In $$\triangle ABD $$
$$\tan 60^{\circ}= \dfrac {AB}{BD} $$
$$\Rightarrow \sqrt 3= \dfrac {10\sqrt 3}{BD} $$
$$\Rightarrow BD = 10 m $$
Option C is correct.
The angle of depression of a car moving with uniform speed towards the building as observed from the top of the building is found to be $$30^{\circ}$$. The same angle of depression changes to $$60^{\circ}$$ after $$12$$ seconds. How much more time would the car take to reach the base?
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$$6$$ sec
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$$8$$ sec
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$$4$$ sec
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$$12$$ sec
Explanation
Let $$'h'$$ be the height of the building.
BC $$=x$$ and CD $$=y$$
In $$\triangle ABC,$$
$$\Rightarrow \tan { 60° } = \dfrac { h }{ x } $$
$$\Rightarrow \sqrt { 3 } = \dfrac { h }{ x } $$
$$\Rightarrow h=\sqrt { 3 }x \longrightarrow \left( 1 \right) $$
In $$\triangle ADB, $$
$$ \Rightarrow \tan { 30° } =\dfrac { h }{ x+y } $$
$$ \Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ x+y } $$
$$ \Rightarrow x+y=\sqrt { 3 } h$$
From $$(1),$$
$$ \Rightarrow x+y=\sqrt { 3 } \times \sqrt { 3 } x$$
$$\Rightarrow x+y=3x$$
$$\Rightarrow y=2x$$
$$\Rightarrow x=\dfrac{y}{2}$$
Since $$y$$ is the distance traveled in $$12$$ sec from point D to point C.
so, the distance $$x$$ is travelled in $$\dfrac{12}{2}=6$$ sec. $$[\because$$ distance $$x$$ is half the distance $$y]$$
Hence, the answer is $$6$$ seconds
The angle of elevation of the top of a building from the foot of the tower is $$30$$ and the angle of the elevation of the top of the tower from the foot of the building is $$60$$. If the tower is $$50$$ m high, then the height of the building is :
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$$\displaystyle \frac { 50 }{ 3 } $$ m
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$$\displaystyle \frac { 35 }{ 3 }$$ m
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$$\displaystyle \frac { 47 }{ 3 }$$ m
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$$\displaystyle \frac { 52 }{ 3 }$$ m
Explanation
Let $$AB$$ be the tower and $$CD$$ be the building.
Since the angle of elevation of the top of the
building from the foot of the tower is $$30^{0}$$.
$$∴∠CBD=306{0}$$
Again, the angle of elevation of the top of the tower from the foot of the building is $$60^{0}$$.
$$∴∠ADB=60^{0}$$
In right-angled $$ΔABD$$, we have
$$\tan60^{0}=\dfrac{AB}{BD}$$
$$⇒\sqrt 3=\dfrac{50}{BD}$$
$$⇒BD=\dfrac{50}{\sqrt 3}$$ ........(i)
In right-angled $$ΔBDC$$, we have
$$\tan 30^{0}=\dfrac{CD}{BD}$$
$$\Rightarrow CD=\dfrac{1}{\sqrt 3}\times \dfrac{50}{\sqrt 3}$$ ....{from {i}}
$$\Rightarrow CD=\dfrac{50}{3}$$m
The angles of elevation of the top of a tower from two points a and b from the base and in the same straight line with it are complementary. The height of the tower is :
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$$\displaystyle ab$$
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$$\displaystyle \sqrt { ab } $$
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$$\displaystyle { a }^{ 2 }{ b }^{ 2 }$$
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None of these
Explanation
Let AB be the tower and C and D be the two points such that BD = a and BC = b.
If $$\displaystyle \angle BDA=\theta ,$$ then $$\displaystyle \angle BCA={ 90 }^{ o }-\theta $$
Now in right-angled $$\displaystyle \Delta ABD$$,
$$\displaystyle \frac { AB }{ BD } =\tan { \theta } \Rightarrow \frac { h }{ a } =\tan { \theta } $$
$$\displaystyle \therefore \quad h=a\tan { \theta } ......(i)$$
In right-angled $$\displaystyle \Delta ABC$$,
$$\displaystyle \frac { AB }{ BC } =\tan { \left( { 90 }^{ o }-\theta \right) } =\cot { \theta } $$
$$\displaystyle \Rightarrow \quad \frac { h }{ b } =\cot { \theta } \Rightarrow h=b\cot { \theta } ......(ii)$$
multiplying (i) and (ii), we get
$$\displaystyle { h }^{ 2 }=ab\tan { \theta } \cot { \theta } =ab\tan { \theta } \times \frac { 1 }{ \tan { \theta } } =ab$$
$$\displaystyle \therefore \quad h=\sqrt { ab } $$
Two poles of equal heights are standing opposite to each other on either side of a road, which is $$100$$ metres wide. From a point between them on the road, the angles of elevation of their tops are $$30$$ and $$60$$. The height of each pole is :
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$$44$$ m
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$$43.25$$ m
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$$50$$ m
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$$40.5$$ m
Explanation
Let $$AB$$ and $$CD$$ be the two poles and $$M$$ be the point where the angle of elevation is made of $$30^{0}$$ and $$60^{0}$$ respectively.
Let the road $$CM$$ $$=x$$ m and $$AM$$ $$=100-x$$ m
In right angled $$\triangle MCD$$, we have
$$\tan60^{0}=\dfrac{DC}{CM}$$
$$\Rightarrow \sqrt 3=\dfrac{h}{x}$$
$$\Rightarrow h=x\sqrt 3$$ (i)
In right angled $$\triangle MAB$$, we have
$$\Rightarrow \tan30^{0}=\dfrac{BA}{AM}$$
$$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{h}{100-x}$$
$$\Rightarrow h \sqrt 3=100-x$$
$$\Rightarrow (x\sqrt 3)\sqrt 3=100-x$$ ....(from {i})
$$\Rightarrow 3x=100-x$$
$$\Rightarrow 4x=100$$
$$\Rightarrow x=25$$ m
Therefore, height of pole $$CD$$,
$$\Rightarrow CD=25\times 1.732$$
$$\Rightarrow CD=43.25$$m
Thus, height of each pole is $$43.25$$ m.
The angle of elevation of the top of a tower from a point on the ground, which is $$30$$ m away from the foot of the tower is $$30$$. The height of the tower is :
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$$\displaystyle 8\sqrt { 3 } $$ m
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$$\displaystyle 9\sqrt { 3 } $$ m
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$$\displaystyle 10\sqrt { 3 } $$ m
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$$\displaystyle 12\sqrt { 3 } $$ m
The shadow of a vertical tower on level ground increases by $$10$$ metres when the altitude of the sun changes from the angle of elevation $$45^0$$ to $$30^0$$. Find the height of the tower correct to
one place of decimal.
(take $$\displaystyle \sqrt { 3 } =1.732$$)
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$$13.67\ m$$
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$$15\ m$$
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$$18.67\ m$$
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$$20\ m$$
Explanation
Let height of tower $$CD$$ be $$h$$ metres
Now in right-angled
$$\displaystyle \Delta BCD$$, we have
$$\displaystyle \frac { h }{ x } =\tan { { 45 }^{ o } } $$
$$\displaystyle \Rightarrow \frac { h }{ x } =1$$
$$\Rightarrow h=x$$ ......(i)
Again in right-angled $$\displaystyle \Delta ACD$$, we have
$$\displaystyle \frac { h }{ x+10 } =\tan { { 30 }^{ o } } $$
$$\displaystyle \Rightarrow \frac { h }{ x+10 } =\frac { 1 }{ \sqrt { 3 } } $$
$$\displaystyle \Rightarrow \sqrt { 3 } h=x+10$$
$$\displaystyle \Rightarrow \sqrt { 3 } h=h+10$$ ......using (i)
$$\Rightarrow \displaystyle h\left( \sqrt { 3 } -1 \right) =10$$
$$\displaystyle \Rightarrow h=\frac { 10 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } $$
$$\displaystyle =\frac { 10\left( \sqrt { 3 } +1 \right) }{ { \left( \sqrt { 3 } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 } } =\frac { 10\left( 1.732+1 \right) }{ 3-1 } $$
$$\displaystyle =5\times 2.732=13.67$$ m
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $$30$$ with it. The distance between the foot of the tree to the point where the top touches the ground is $$8$$ m. The height of the tree is:
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$$\displaystyle 5\sqrt { 3 } $$ m
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$$\displaystyle 8\sqrt { 3 } $$ m
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$$\displaystyle 10\sqrt { 3 } $$ m
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$$\displaystyle 6\sqrt { 3 } $$ m
Explanation
Let $$AB$$ be the tree.
Let it be broken by the wind at the point $$C$$ and $$CB$$ taking the position $$CD$$ strikes the ground at $$D$$
.
Now, $$\displaystyle \angle ADC={ 30 }^{ o }$$
Let $$AC = x $$ and $$CD = y $$
In right $$\displaystyle \Delta ADC$$, we have $$\displaystyle \frac { AC }{ AD } =\tan { { 30 }^{ o } } $$
$$\displaystyle \Rightarrow \frac { x }{ 8 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow x=\frac { 8 }{ \sqrt { 3 } } $$
Again $$\displaystyle \frac { AD }{ CD } =\cos { { 30 }^{ o } } \Rightarrow \frac { 8 }{ y } =\frac { \sqrt { 3 } }{ 2 } \Rightarrow y=\frac { 16 }{ \sqrt { 3 } } $$
$$\displaystyle \therefore $$ Height of the tree $$= (x + y)$$ metres.
$$\displaystyle =\frac { 8 }{ \sqrt { 3 } } +\frac { 16 }{ \sqrt { 3 } } =\frac { 24 }{ \sqrt { 3 } } =8\sqrt { 3 } $$ metres.
From the top of a building h metres high, the angle of elevation of a monument is $$45^o$$ and angle of depression of its foot is $$30^o$$. The height of the monument is
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$$\sqrt 3h$$
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$$\dfrac {(\sqrt 3+1)}{2}h$$
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$$\dfrac {\sqrt 3-1}{2}h$$
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$$(\sqrt 3+1)h$$
Explanation
In $$\Delta ADB$$
$$AB=\dfrac {h}{tan 30^\circ}$$
In $$\Delta DCF, FC=CD tan 45^\circ$$
But $$AB=CD$$
$$\rightarrow\dfrac {h}{tan 30^\circ}=\sqrt 3h$$
Thus, $$FC=\sqrt3h$$
Total height of monument $$=BC+FC$$
$$=h+\sqrt 3h$$
$$=(\sqrt 3+1)h$$
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a $$20$$ m high building are 45 and 60 respectively. The height of the tower is:
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$$\displaystyle 25\left( \sqrt { 3 } -1 \right)\ m$$
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$$\displaystyle 20\left( \sqrt { 3 } -1 \right)\ m$$
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$$\displaystyle 20\ m$$
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$$\displaystyle 10\ m$$
A kite is flying at a height of $$60$$ m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $$60$$. Find the length of the string, assuming that there is no slack in the string.
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$$\displaystyle 40\sqrt { 3 } $$ m
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$$\displaystyle 30\sqrt { 3 } $$ m
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$$\displaystyle 20\sqrt { 3 } $$ m
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$$\displaystyle 10\sqrt { 3 } $$ m
Explanation
Given, the height of the kite $$AB$$ is $$60$$ m and it makes an angle of elevation at point $$C$$ of $$60^{0}$$.
In right angled $$\triangle ABC$$, we have
$$cosec 60^{0}=\cfrac{AC}{AB}$$
$$\Rightarrow \cfrac{2}{\sqrt 3}=\cfrac{AC}{60}$$
$$\Rightarrow AC=60\times \cfrac{2}{\sqrt 3}$$
$$\Rightarrow AC=60\times \cfrac { 2\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } $$
$$\Rightarrow AC=40\sqrt 3$$ m
The angle of elevation of the top of a hill at the foot of a tower is $$60$$ and the angle of elevation of top of the tower from the foot of the hill is $$30$$. If the tower is $$50$$ m high, then the height of the hill is :
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$$148$$ m
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$$150$$ m
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$$152$$ m
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$$160$$ m
Explanation
Let $$AB$$ be the hill and $$CD$$ be the tower.
Since the angle of elevation of the top of the
tower from the foot of the hill is $$30^{0}$$.
$$∴∠CBD=30^{0}$$
Again, the angle of elevation of the top of the hill from the foot of the tower is $$60^{0}$$.
$$∴∠ADB=60^{0}$$
In right-angled $$ΔBDC$$
$$\tan30^{0}=\dfrac{CD}{BD}$$
$$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{50}{BD}$$
$$⇒BD=50\sqrt3$$ m .....(i)
In right-angled $$ΔABD$$, we have
$$\tan60^{0}=\dfrac{AB}{BD}$$
$$\Rightarrow \sqrt 3=\dfrac{AB}{50\sqrt 3}$$
....
{from (i)}
$$\Rightarrow AB={\sqrt 3}\times 50\sqrt 3$$ ........(ii)
$$\Rightarrow AB=150$$ m
The shadow of a pole standing on a horizontal plane is $$a$$ meters longer when the sun's elevation is $$\theta$$ than when it is $$\phi$$. The height of the pole will be: {Use $$\sin(A-B) = \sin A \cos B - \sin B \cos A$$}
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$$a\displaystyle\frac{\cos\theta\cdot\cos\phi}{\cos(\theta-\phi)}$$ meter
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$$a\displaystyle\frac{\sin\theta\sin\phi}{\sin(\phi-\theta)}$$ meter
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$$a\displaystyle\frac{\sin\theta\cos\phi}{\sin(\theta-\phi)}$$ meter
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$$a\displaystyle\frac{\sin\phi\cos\theta}{\cos(\theta-\phi)}$$ meter
Explanation
Let the height of the pole standing on a horizontal plane be $$h$$
Let the length of the shadow be $$x$$ and $$(x+a)$$, when the sun's elevation is $$\phi $$ and $$\theta$$ respectively.
In $$\triangle APQ$$,
$$\dfrac{x}{h}=\cot \phi $$
$$\Rightarrow x=h \cot \phi $$ ....(1)
In $$\triangle BPQ$$,
$$\dfrac{x+a}{h}=\cot \theta$$
$$\Rightarrow (x+a)=h \cot \theta$$ ....(2)
$$\Rightarrow (x+a)-x=h \cot \theta-h \cot \phi$$ {using (1)}
$$\Rightarrow a=h(cot \theta-cot \phi)$$
$$\Rightarrow h=\cfrac { a }{ \left( \dfrac { \cos \theta }{ \sin \theta } -\dfrac { \cos \phi }{ \sin \phi } \right) } $$
$$\Rightarrow h=a\cfrac { \sin \theta \sin \phi }{ \left( \sin \phi \cos \theta -\cos \phi \sin \theta \right) } $$
$$\Rightarrow h=a\cfrac { \sin \theta \sin \phi }{ \sin\left( \phi - \theta \right) } $$
From the top of house $$32$$ meter high, if the angle of elevation of the top of a tower is $$ \displaystyle 45^{\circ} $$ and the angle of depression of the foot of the tower is$$ \displaystyle 30^{\circ} $$ , then the height of the tower is
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$$ \displaystyle \frac{32}{\sqrt{3}}(\sqrt{3+1}) $$ meter
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$$ \displaystyle 32(\sqrt{3+1}) $$ meter
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$$ \displaystyle 32\sqrt3 $$ meter
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$$ \displaystyle \frac{32}{3}(\sqrt{3+1}) $$ meter
Explanation
From $$\displaystyle \bigtriangleup OBH $$
$$\cot 30^{\circ}=\dfrac{BH}{OB}$$
$$\Rightarrow \sqrt 3 =\dfrac{BH}{OB} $$
$$\Rightarrow \sqrt 3 =\dfrac{BH}{32} $$
or $$\displaystyle BH =32\sqrt{3} $$
Height of Tower
$$\displaystyle =OT=OB+BT=32(\sqrt{3}+1)\ meters $$
An aeroplane flying horizontally at a height of $$1.5$$ km above the ground is observed at a certain point on earth to subtend an angle of $$60$$. After $$15$$ seconds, its angle of elevation is observed to be $$30$$. Calculate the speed of the, aeroplane in km/h.
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$$\displaystyle 240\sqrt { 3 }$$ km/h
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$$\displaystyle 230\sqrt { 3 }$$ km/h
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$$\displaystyle 210$$ km/h
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$$\displaystyle 220$$ km/h
Explanation
Let $$P$$ and $$Q$$ be the two position of the plane.
Given angles of elevation of the plane in two position $$P$$ and $$Q$$ as $$\angle PAB=60^{0}$$ and $$\angle QAC=30^{0}$$
$$PB=QC=$$ $$15$$ km
Now in right angled $$\triangle ABP$$
$$\tan 60^{0}=\dfrac{BP}{AB}$$
$$\Rightarrow \sqrt 3=\frac{1.5}{AB}$$
$$\Rightarrow AB=\dfrac{1.5}{\sqrt 3}$$
$$\Rightarrow (0.5)\sqrt 3$$ km
Again in right angled triangle $$\triangle ACQ$$,
$$\tan30^{0}=\dfrac{QC}{AC}$$
$$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{1.5}{AB}$$
$$\Rightarrow AC=(1.5)\sqrt 3$$ km
$$PQ=BC=AC-AB$$
$$=(1.5)\sqrt 3-(0.5)\sqrt 3$$
$$=(1.5-0.5)\sqrt 3$$
$$=\sqrt 3$$ km
The plane travels $$PQ=$$ $$\sqrt 3$$ km in $$15$$ secs
$$\therefore$$ Speed of the aeroplane $$=$$ $$\dfrac{distance}{time}$$
$$\Rightarrow \dfrac { \sqrt { 3 } }{ \frac { 15 }{ 3600 } } $$
$$\Rightarrow 240 \sqrt 3$$ km/h
If the angle of elevation of an object from a point 200 meter above the 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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$$ \displaystyle \frac{200\left ( \sqrt{3-1} \right )}{\left ( \sqrt{3+1} \right )} $$ meter
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$$ \displaystyle \frac{200\left ( \sqrt{3-1} \right )}{\sqrt{3}} $$
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$$ \displaystyle \frac{200\left ( \sqrt{3+1} \right )}{\sqrt{3}} $$
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$$ \displaystyle \frac{200\left ( \sqrt{3+1} \right )}{\left ( \sqrt{3+1} \right )} $$ meter
Explanation
Let OL=h, therefore LO'=H
From $$\displaystyle \bigtriangleup PQO',PQ=QO'=QL+LO' $$
=200+h [OL=LO'=h]
From $$\displaystyle \bigtriangleup PQO,\frac{OQ}{PQ}=\tan 30^{\circ} $$
or $$\displaystyle OQ=PQ \tan 30^{\circ} $$
or $$\displaystyle h-200=(200+h)\frac{1}{\sqrt{3}} \therefore h=\frac{200(\sqrt{3}+1)}{\sqrt{3}-1} $$
A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height $$h$$. At a point on the plane, the angle of elevation of the bottom of the flagstaff is $$\displaystyle \alpha $$ and that of the top of the flagstaff is $$\displaystyle \beta $$ Then the
height of the tower is :
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$$\displaystyle \frac { h }{ \tan { \alpha } } $$
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$$\displaystyle \frac { h\tan { \alpha } }{ \tan { \beta } -\tan { \alpha } } $$
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$$\displaystyle \frac { \tan { \beta } }{ h } $$
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$$\displaystyle \frac { \tan { \alpha } }{ \tan { \beta } -\tan { \alpha } } $$
Explanation
Let $$BC$$ be the tower and $$CD$$ be the flagstaff.
The angle of elevation of the bottom of the flagstaff is $$\alpha $$ and that of the top of flagstaff is $$\beta$$.
Let $$h$$ be the height of the tower
In $$\triangle ABC$$, we have
$$\Rightarrow \tan \alpha=\dfrac{BC}{AB}$$ ....{i}
In $$\triangle ABD$$
$$\Rightarrow \tan \beta=\dfrac{BD}{AB}$$
$$\Rightarrow \dfrac{BC+h}{AB}=\dfrac{BD}{AB}$$ ....{ii}
Now dividing {ii} by {i}, we get
$$\dfrac{BC+h}{BC}=\dfrac{\tan \beta}{\tan \alpha}$$
$$\Rightarrow (BC+h)\tan \alpha=BC \tan \beta$$
$$\Rightarrow BC(\tan \beta-\tan \alpha)=h \tan \alpha$$
$$\Rightarrow BC=\dfrac{h \tan \alpha}{\tan \beta- \tan \alpha}$$
The angles of elevations of the top of the tower from two points in the same straight line and at a distance of $$9 \text{ m}$$ and $$16\text{ m}$$ from the base of the tower are complementary. The height of the tower is :
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$$18\text{ m}$$
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$$16\text{ m}$$
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$$10\text{ m}$$
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$$12\text{ m}$$
Explanation
From the given figure,
$$\displaystyle \tan { \theta } =\frac { h }{ 9 } $$ -------$$(1)$$
$$\displaystyle \tan { \left( 90-\theta \right) } =\frac { h }{ 16 }$$
$$\implies\cot \theta=\dfrac{h}{16} $$ .......$$(2)$$
Multiplying $$(1)$$ and $$(2)$$
$$\implies \displaystyle \tan { \theta } \times \cot { \theta } =\frac { { h }^{ 2 } }{ 9\times 16 } $$ $$[\because \tan \theta \times \cot \theta=1]$$
$$\implies \displaystyle h^2=9\times 16 \\\implies h=\displaystyle \sqrt{9\times 16}\\\implies h=12\text{ m}$$
$$\therefore h=12\text{ m}$$
The angle of elevation of the top of a tower at a distance of $$\displaystyle\frac{50\sqrt{3}}{3}$$ metres from the foot is $$60^\circ$$
. Find the height of the tower.
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$$50\sqrt{3}\:m$$
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$$\displaystyle\frac{20}{\sqrt{3}}\:m$$
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$$-50 m$$
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$$50 m$$
Explanation
Let the height of the tower be h metres
$$\displaystyle\therefore\dfrac{h}{\displaystyle\left(\frac{50\sqrt{3}}{3}\right)}=\tan{60^\circ}=\sqrt{3}$$
$$\therefore h=50\:m$$
The angles of elevation of the top of a tower from two points at the distances of $$4\ m$$ and $$9\ m$$ from the base of the tower and in the same straight line with it are complementary. The height of the tower is :
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$$8\ m$$
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$$5\ m$$
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$$6\ m$$
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$$4\ m$$
Explanation
Let AB be the tower and C and D be the two points such that BD =
4 m and BC = 9 m.
If $$\displaystyle \angle BDA=\theta ,then\quad \angle BCA={ 90 }^{ o }-\theta $$
Now in right-angled $$\displaystyle \Delta ABD$$
$$\displaystyle \frac { AB }{ BD } =\tan { \theta } \Rightarrow \frac { h }{ 4 } =\tan { \theta } $$
$$\displaystyle \therefore \quad h=4\tan { \theta } .....(i)$$
In right-angled $$\displaystyle \Delta ABC$$
$$\displaystyle \frac { AB }{ BC } =\tan { \theta } \left( { 90 }^{ o }-\theta \right) =\cot { \theta } $$
$$\displaystyle \Rightarrow \quad \frac { h }{ 9 } =\cot { \theta } \Rightarrow h=9\cot { \theta } .....(ii)$$
multiplying (i) and (ii), we get
$$\displaystyle { h }^{ 2 }=36\tan { \theta } \cot { \theta } =36\tan { \theta } \times \frac { 1 }{ \tan { \theta } } =36\\ $$
$$\displaystyle \therefore \quad h=\pm \sqrt { 36 } =6m$$
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Practice Class 10 Maths Quiz Questions and Answers
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