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CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 2 - MCQExams.com
CBSE
Class 10 Maths
Some Applications Of Trigonometry
Quiz 2
At a point 15 metres away from the base of a 15 metres high house, the angle of elevation of the top is
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$$45^0$$
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$$30^0$$
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$$60^0$$
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$$90^0$$
Explanation
$$\tan\theta = \dfrac{15}{15} = 1$$
$$\theta$$ = $$45^0$$
The angle of elevation of a cloud from a point h m above the level of water in a lake is $$\alpha$$ and the angle of depression of its reflection in the lake is $$\beta$$.Then the height of the cloud above the water level is
$$\dfrac{h\,sin\, (\beta - \alpha)}{sin\,(\beta + \alpha)}$$
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True
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False
Explanation
From Figure
Point $$A$$ is the cloud
Point $$D$$ is view point
$$FD $$ is parallel to $$BE$$ and $$CD$$ transversal
So $$ \angle FDC=\angle DCE$$
From Reflection property
$$ \angle BCA =\angle DCE =\beta $$
From $$ \Delta ABC $$
$$ \sin \beta =\dfrac {AB}{AC} \cdots(1)$$
From $$ \Delta CDE $$
$$ \sin \beta = \dfrac {DE}{DC} \cdots(2)$$
From $$ (1 ) \& (2)$$
$$ \implies \dfrac {AB}{AC} =\dfrac {DE}{DC} $$
$$ \implies AB =\dfrac {DE}{DC}\times AC \cdots(3)$$
From $$ \Delta ACD $$
By Exterior angle is sum of two interior opposite angles
$$ 2\beta =\alpha +\beta +\angle DAC$$
$$ \angle DAC = \beta -\alpha $$
From $$\sin$$ rule
$$ \implies \dfrac {AC}{\sin (\alpha +\beta)}=\dfrac {DC}{\sin (\beta -\alpha )}$$
$$ \implies \dfrac {AC}{DC}=\dfrac {\sin (\beta -\alpha )}{\sin (\alpha +\beta)} \cdots(4)$$
Put it in $$(3)$$
$$ \implies AC =\dfrac{DE \sin (\beta -\alpha )}{\sin (\alpha +\beta)} $$
From Question $$ DE=h$$
$$ \implies AC =\dfrac{h \sin (\beta -\alpha )}{\sin (\alpha +\beta)} $$
The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line are complementary. The height of the tower is $$\sqrt{ab}$$ metres.
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True
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False
Explanation
Let AB be the tower and let C and D be the two points of the observer. Then,
AC = a metres and AD = b metres
Let $$\angle ACB =\theta$$. Then, $$\angle ADB =(90^{o}-\theta)$$.
Let AB = h metres
In right $$\triangle CAB$$, we have
$$\tan \theta =\dfrac{AB}{AD}$$
$$\tan \theta =\dfrac{h}{a}$$
$$h=a\tan \theta$$ ......(1)
In right $$\triangle DAB$$, we have,
$$\tan(90^{o}-\theta)=\dfrac{AB}{AD}$$
$$\cot \theta =\dfrac{h}{b}$$
$$h=b\cot \theta$$ .......(2)
From 1 and 2, we get,
$$h^2=ab$$
$$h=\sqrt{ab} \ metres$$
$$AB$$ is vertical pole with $$B$$ at the ground level and $$A$$ at the top. A man find that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $$60^{o}$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7\ m$$. From $$D$$ the angle of elevation of the point $$A$$ is $$45^{o}$$. Then the height of the pole is:
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$$\dfrac { 7\sqrt { 3 } }{ 2 } \dfrac { 1 }{ \sqrt { 3 } -1 } m$$
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$$\dfrac { 7\sqrt { 3 } }{ 2 } \left( \sqrt { 3 } +1 \right) m$$
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$$\dfrac { 7\sqrt { 3 } }{ 2 } \left( \sqrt { 3 } -1 \right) m$$
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$$\dfrac { 7\sqrt { 3 } }{ 2 } \dfrac { 1 }{ \sqrt { 3 } +1 } m$$
Explanation
Then in triangle $$ABC$$
$$tan60^{\circ}=\dfrac{AB}{BC}$$
$$x=\frac{h}{\sqrt3}$$ ----- (i)
in $$\Delta ABD$$
$$tan45^{\circ}=\dfrac {AB}{BD}$$
$$1=\dfrac{h}{x+7}$$
$$h=x+7$$
$$h=\dfrac{h}{\sqrt3}+7$$ from (i)
$$h=\dfrac{7\sqrt3}{\sqrt3-1} $$
$$h=\dfrac{7\sqrt3}{2}(\sqrt3+1)$$
Therefore the height of the pole is $$\dfrac{{7\sqrt 3 }}{2}(\sqrt 3 + 1)\;m$$
A bridge above the river makes an angle of $${45}^{o}$$ with the bank of river. If length of bridge above the river is $$150\ m$$ then breadth of river will be
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$$75\ m$$
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$$50\sqrt {2}\ m$$
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$$150\ m$$
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$$75\sqrt {2}\ m$$
Explanation
Let $$AC$$ be the bridge and $$AB$$ be the width of river.
In $$\Delta ABC$$
$$\dfrac{AB}{AC}=\sin 45^{\circ}$$
$$\dfrac{AB}{150}=\dfrac{1}{\sqrt2}$$
$$AB=\dfrac{150\sqrt2}{2}=75\sqrt2$$
therefore width of the river is $$75\sqrt2\;m$$.
A flagstaff stands on the middle of a square tower. A man on the ground, opposite to the middle of one face and distant from it $$100$$ m, just see the flag ; on his receding another $$100$$ m, the tangents of the elevation of the top of the tower and the top of the flagstaff are found to be $$\dfrac {1}{2}$$ and $$\dfrac {5}{9}$$. Find the height of the flagstaff, the ground being horizontal
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$$20$$
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$$25$$
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$$30$$
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$$35$$
Explanation
$$\dfrac{h_2}{b}=\dfrac{h_1}{100}$$
$$\dfrac{h_1}{200}=\dfrac{1}{2}$$
$$\Rightarrow h_1=100$$
$$\dfrac{h_1+h_2}{200+b}=\dfrac{5}{9}$$
$$\dfrac{100+b}{200+b}=\dfrac{5}{9}$$
$$b=\dfrac{100}{4}$$
$$\Rightarrow b=h_2=25$$
As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. An equation to model the motion is y=20cos($$\frac {\pi}{4} (t-3))+23$$. Predict your height above the ground at a time of 1 seconds.
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$$20.86 ft$$
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$$23 ft$$
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$$8.14 ft$$
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$$18.96$$
Explanation
$$y=20\ cos (\cfrac{\pi}{4}(t-3))+23$$
At $$t=1$$ second
Height above the ground,
$$y=20\ cos (\cfrac{\pi}{4}(1-3))+23$$
$$y=20\ cos (\cfrac{-2\pi}{4})+23$$
$$\implies y=23 ft$$. [ Since ,
$$\ cos (\cfrac{-\pi}{4})=0$$ ]
The shadow of a tower, when the angle of elevation of the sun is $$45^{o}$$, is found to be 10 metres longer than when the angle of elevation is $$60^{o}$$. Find the height of the tower.
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$$15+5\sqrt{3} \ m$$
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$$12+5\sqrt{3} \ m$$
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$$15+\sqrt{3} \ m$$
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$$12+\sqrt{3} \ m$$
Explanation
Let AB be the tower and let AC and AD be its shadows when the angles of elevation of the sum are $$60^{o}$$ and $$45^{o}$$, respectively. Then,
$$\angle ACB=60^{o}, \angle ADB = 45^{o}, CD=10 \ m$$
Let $$AB=h \ metres$$ and $$AC=x \ metres$$
In right $$\triangle$$CAB, we have,
$$\tan 60^{o}=\dfrac{AB}{AC}$$
$$\sqrt{3}=\dfrac{h}{x}$$
$$x=\dfrac{h}{\sqrt{3}}$$ ......(1)
In right $$\triangle$$ DAB, we have,
$$\tan 45^{o}=\dfrac{AB}{AD}$$
$$1=\dfrac{h}{10+x}$$
$$x=h-10$$ .....(2)
From 1 and 2, we get,
$$\dfrac{h}{\sqrt{3}}=h-10$$
$$h=\sqrt{3}h-10\sqrt{3}$$
$$\sqrt{3}h-h=10\sqrt{3}$$
$$h(\sqrt{3}-1)=10\sqrt{3}$$
$$h=5\sqrt{3}(\sqrt{3}+1)$$
$$h=15+5\sqrt{3} \ m$$
The shadow of a $$6\ m$$ high tower is $$15\ m$$ and at the same point of time length of shadow of a tree is $$25\ m$$. What is the height of the tree?
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$$21\ m$$
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$$10\ m$$
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$$35\ m$$
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$$none\ of\ these$$
Explanation
$$\dfrac{Hight \ of \ the \ tower}{Length \ of \ shadow \ of \ the \ tower} = \dfrac{Hight \ of \ the \ tree}{Length \ of \ shadow \ of \ the \ tree}$$
$$\implies \dfrac{6}{15} = \dfrac{H}{25}$$
$$\implies H = 25 \times \dfrac{6}{15}$$
$$\therefore H = 10 \ m$$
OPtion B is correct.
If a ladder $$13 m $$ is placed against a wait such that its roots at a distance from the wall, then the height of the top of the ladder from the ground :
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$$ 10 m $$
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$$ 11 m $$
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$$ 12 m $$
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None of these
Explanation
If a ladder is $$13$$ m long is placed at a distance of $$12$$ m from a wall, then the weight of the wall is:
$$\begin{array}{l} A.T.Q, \\ h=\sqrt { { { \left( { 13 } \right) }^{ 2 } }-{ { \left( { 12 } \right) }^{ 2 } } } \\ =\sqrt { 169-144 } \\ =\sqrt { 25 } \\ =5\, \, m \end{array}$$
A tower stands vertically on the ground. From a point on the ground which is $$30\ m$$ away from the foot of a tower, the angle of elevation of the top of the tower is found to be $$45^{o}$$. Find the height of tower.
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$$15$$
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$$40$$
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$$30$$
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$$20$$
Explanation
REF.Image.
$$30 m$$ away / $$45^{\circ}$$ inclination.
Let the height of tower be h m
so $$AB = h \, m$$
Distance of the point from
foot of tower = $$30 m$$
$$CB = 30 m$$
Angle of elevation $$ = 45^{\circ}$$
$$ \angle ACB = 45^{\circ}$$
Since tower is vertical to ground
$$ \angle ABC = 90^{\circ}$$
Now $$ tanC = \dfrac{side\, opp\, to\, angle\, C}{side\, adjacent \,to\, angle \,C}$$
$$ \tan C = \dfrac{AB}{CB} $$
$$ \tan45 = \dfrac{AB}{CB}$$
$$ 1 = \dfrac{H}{30}$$
$$ H = 30 $$
Hence height of tower = $$30 m$$
Answer = option $$C = 30 m $$
A man observes the angle of elevation of a balloon to be $$30^{0}$$ at a point $$A$$. He then walks towards the balloon and at a certain place $$B$$, he finds the angle of elevation to be $$60^{0}$$. He further walks in the direction of the balloon and finds it to be directly over him at a height of $$\dfrac12\ km$$, then the distance$$AB $$ is:
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$$\displaystyle \frac{1}{\sqrt{2}}\ km$$
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$$\displaystyle \frac{1}{\sqrt{3}}\ km$$
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$$\displaystyle \frac{1}{\sqrt{4}}\ km$$
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$$\displaystyle \frac{1}{\sqrt{5}}\ km$$
Explanation
$$h = 500\ m $$
$$\tan60^0 = \dfrac{h}{b}$$
$$\Rightarrow b = \dfrac{500}{\sqrt {3}}$$
$$ \tan30^0 = \dfrac{h}{d+b}$$
$$\Rightarrow d+ \dfrac{500}{\sqrt {3}} = 500 \sqrt {3}$$
$$\Rightarrow d = \dfrac{1000}{\sqrt {3}}\ m $$
$$\Rightarrow d = \dfrac{1}{\sqrt {3}}\ km$$
The horizontal distance between two towers is $$60\ m$$ and angular depression of the top of the first as seen from the second, which is $$150\ m$$ in height, is $$30^{0}$$. The height of the first tower is
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$$(150+20\sqrt{3})\ m$$
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$$(150+15\sqrt{3})\ m$$
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$$(150-20\sqrt{5})\ m$$
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$$(150-20\sqrt{3})\ m$$
Explanation
Let AB and CD be the two towers
Height of the second tower, AB =150m and
Distance between two towers BD =60m
Let's assume the height of the first tower is $$h\ m$$.
We know, In right angled $$\triangle$$,
$$\tan(\Theta ) = \dfrac {Opposite} {Adjacent}$$
$$\tan 30^0 = \dfrac{AE}{EC}$$
$$\tan 30^0 = \dfrac{150-h}{60}$$
.....[AE=AB-ED] and [EC=BD=60]
$$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{150-h}{60}$$
$$\Rightarrow h \sqrt{3} = 150 \sqrt{3} -60$$
$$\Rightarrow h = 150-20 \sqrt{3}$$
The height of the first tower is
$$150-20 \sqrt{3}$$ m.
A kite is flying with the string inclined at $$30^{o}$$ to the horizon. The height of the kite above the ground, when the string is $$15\ m$$ long is
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$$15\ m$$
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$$30\ m$$
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$$\dfrac{15}2\ m$$
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$$\dfrac{15}3\ m$$
Explanation
$$\sin{30^0} = \dfrac{AB}{AC} = \dfrac h{15}$$
$$\Rightarrow \dfrac12 = \dfrac h{15}$$
$$\Rightarrow h = \dfrac {15}2\ m$$
From the top of the tree, a man observes the angle of depression of a point which is at a distance of $$40\ m$$ from the foot is $$75^{0}$$. The height of the tree is
:
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$$40\sqrt{3}\ m$$
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$$40(2+\sqrt{3})\ m$$
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$$21\sqrt{3}\ m$$
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$$3\sqrt{21}\ m$$
Explanation
Let say AC be the tree and the Observer is at A who is observing point B.
Given: BC = 40m
$$\angle B = 75^o $$ = angle of depression {alternate angles}
In $$\triangle ABC$$,
$$\tan 75^0 = \dfrac{h}{40}$$
$$\Rightarrow \tan (45^o+30^o) = \dfrac{h}{40}$$
$$\Rightarrow \dfrac{\tan45^o+ \tan 30^o}{1-\tan45^o\tan 30^o} = \dfrac{h}{40}$$
$$\because \dfrac{\tan A+ \tan B}{1-\tan A\tan B} = tan(A+B)$$
$$\Rightarrow \dfrac{1+\dfrac{1}{\sqrt{3}}}{1- 1\times \dfrac{1}{\sqrt{3}}}= \dfrac{h}{40}$$
$$\Rightarrow \dfrac{(\sqrt{3}+1)}{(\sqrt{3}- 1)} = \dfrac{h}{40}$$
$$\Rightarrow\left(\dfrac{3+1+2\sqrt{3}}{2}\right) = \dfrac{h}{40}$$ {rationalizing by multiplying and dividing by $$\sqrt 3+1$$}
$$\Rightarrow (2+\sqrt{3}) = \dfrac{h}{40}$$
$$\Rightarrow h = 40 (2+\sqrt{3})m$$
From the top of a hill $$h$$ meters high, the angle of depression of the top and the bottom of a pillar are $$\alpha,\ \beta$$ respectively. Then the height(in meters) of the pillar is
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$$\displaystyle \dfrac{h(\tan\beta-\tan\alpha)}{\tan\beta}$$
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$$\displaystyle \dfrac{h(\tan\alpha-\tan\beta)}{\tan\alpha}$$
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$$\displaystyle \dfrac{h(\tan\beta+\tan\alpha)}{\tan\beta}$$
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$$\displaystyle \dfrac{h(\tan\beta+\tan\alpha)}{\tan\alpha}$$
Explanation
$$\tan \alpha = \dfrac{h-d}{a}$$
$$\tan \beta = \dfrac{h}{a}$$
$$\dfrac{\tan \alpha}{\tan \beta} = \dfrac{h-d}{h}$$
$$\Rightarrow \dfrac{\tan \alpha}{\tan \beta} = 1- \dfrac{d}{h}$$
$$\Rightarrow \dfrac{d}{h}= \dfrac{\tan \beta-\tan \alpha}{\tan \beta}$$
$$\Rightarrow d= h\left(\dfrac{\tan \beta-\tan \alpha}{\tan \beta}\right)$$
From the top of a building $$h$$ metres, the angle of depression of an object on the ground is $$\alpha$$, the distance of the object from the foot of the building is
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$$h\cot\alpha$$
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$$h\tan\alpha$$
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$$h\cos\alpha$$
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$$h\sin\alpha$$
Explanation
$$\tan \alpha = \dfrac{h}{d}$$
$$\Rightarrow d = \dfrac{h}{\tan \alpha}$$
$$\Rightarrow d = h \cot \alpha $$
The angle of elevation of an object from a point $$P$$ on the level ground is $$\alpha$$. Moving $$d$$ meters on the ground towards the object, the angle of elevation is found to be $$\beta$$, then the height (in meters) of the object is
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$$ d\tan\alpha$$
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$$ d\cot\beta$$
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$$\dfrac d{\cot\alpha+\cot\beta}$$
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$$\dfrac d{\cot\alpha-\cot\beta}$$
Explanation
Let the height of the object $$AB$$ be $$h$$ metres.
Given, $$\angle ACB = \alpha, \angle ADB = \beta,\ CD = d$$ metres
Let $$DB = x$$ metres.
Then, in right-angled $$\triangle ABD,$$
$$ \cot β=\dfrac{x}{h}$$
$$⇒ x = h \cot \beta$$ $$...(i)$$
In right-angled $$\triangle ACB,$$
$$\cot \alpha =\dfrac{x+d}{h}$$
$$⇒ x + d = h \cot \alpha$$ $$...(ii)$$
Let us subtract the equations: $$(ii)\ – (i)$$
$$\Rightarrow d = h \cot \alpha \ – h \cot \beta$$
$$\Rightarrow d=h(\cot \alpha \ -\cot \beta)$$
$$\Rightarrow h = \dfrac{d}{ \cot \alpha - \cot \beta }$$
The flag staff of height $$10$$ metres is placed on the top of a tower of height $$30$$ metres. At the top of a tower of height $$40$$ metres, the flag staff and the tower subtend equal angles then the distance between the two towers (in metres) is
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$$40\sqrt{2}$$
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$$10\sqrt{2}$$
0%
$$20\sqrt{2}$$
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$$30\sqrt{2}$$
Explanation
$$\tan 2 \theta = \dfrac{40}{d}$$ ...(i)
$$\Rightarrow \dfrac{2\tan \theta}{1-\tan^{2} \theta} = \dfrac{40}{d}$$
$$\tan \theta = \dfrac{10}{d}$$ ...(ii)
Substituting (ii) in (i), we get
$$\dfrac{2\dfrac{10}{d}}{1-\dfrac{(10)^{2}}{d^{2}}}= \dfrac{40}{d}$$
$$\Rightarrow \displaystyle \frac{d}{d^{2}-10^{2}}= \frac{2}{d}$$
$$\Rightarrow d^{2} = 2d^{2} - 2 \times 10^{2}$$
$$\Rightarrow d^{2} = 2 \times 10^{2}$$
$$\Rightarrow d = 10 \sqrt{2}\ m$$
Straight pole(AB) subtends a right angle at a point $$D$$ of another pole at a distance of $$30$$ meters from $$A$$, the top of $$A$$ being $$60^{0}$$ above the horizontal line joining the point $$B$$ to the pole $$A$$. The length of the pole $$A$$ is, in meters
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$$20 \sqrt{3}$$
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$$40 \sqrt{3}$$
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$$60 \sqrt{3}$$
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$$\displaystyle \frac{40}{\sqrt{3}}$$
Explanation
$$\tan60=\dfrac{h_{2}}{30}$$
$$h_{2}= 30\sqrt{3}$$
$$\tan30^{0}=\dfrac{h_{1}}{30}$$
$$h_{1}=10\sqrt{3}$$
$$h_{1}+h_{2} $$= length of the pole =$$40\sqrt{3}$$
The upper part of a tree broken over by the wind makes an angle of $$60^{0}$$ with the ground and touches the ground at a distance of 50 metres from the foot. The height of the tree in metres is
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$$124.2$$
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$$186.6$$
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$$243.2$$
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$$164.2$$
Explanation
We will consider that the tree is broken at A.
So the total heigh of the tree will be $$AC + AB$$
Let $$AC=L_1$$
and $$AB =L_2$$
$$\Rightarrow$$Height of the tree $$= L_{1} + L_{2}$$
In the $$\triangle ABC, \angle C = 90^\circ$$,
$$\sin 60^\circ = \dfrac{AC}{AB}$$
$$L_{2} \sin 60^{0} = L_{1}$$ .....(1)
$$\tan 60^{0} = \displaystyle\frac{L_{1}}{50}$$
$$\Rightarrow L_{1} = 50 \sqrt{3} m$$
Substituting the value of $$L_1$$ in (1)
$$L_2 \times \dfrac{\sqrt 3}{2} = 50\sqrt3$$
$$\Rightarrow L_{2} = 100 m$$
$$\Rightarrow h =L_{1}+L_{2}$$
$$\Rightarrow h = 100+50 \sqrt{3}$$
$$=186.6 m$$
$$\therefore$$ Height of the tree =
$$=186.6 m$$
The angle of elevation of the top of a flagstaff when observed from a point at a distance $$60$$ meters from its foot is $$30^{0}$$. The height of the flagstaff (in meters) is:
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$$20\sqrt{3}$$
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$$10\sqrt{3}$$
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$$60\sqrt{3}$$
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$$30\sqrt{3}$$
Explanation
Let $$AB=h$$ be the height of the flagstaff and $$C$$ be the point at a distance of $$60m$$ from the foot.
In $$\triangle ABC$$,
$$\tan{30^0} = \dfrac{AB}{BC} = \dfrac{h}{60}$$
$$\Rightarrow \dfrac1{\sqrt3} = \dfrac h{60}$$
$$\Rightarrow h = \dfrac{60}{\sqrt3}$$
$$\Rightarrow h = 20\sqrt3m$$
A tower subtends an angle $$\alpha$$ at a point$$A$$ on the same level as the foot of the tower $$B$$ is a point vertically above $$A$$ and $$AB=h$$ metres. The angle of depression of the foot of the tower from $$B$$ is $$\beta$$. The height of the tower is
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0%
$$h\tan\alpha\cot\beta$$
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$$h\tan\alpha\tan\beta$$
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$$h\cot\alpha\cot\beta$$
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$$h\cot\alpha\tan\beta$$
Explanation
$$\tan\alpha =\dfrac{h_{1}}{d}$$
$$\tan\beta =\dfrac{h}{d}$$
$$\dfrac{\tan\alpha }{\tan\beta }=\dfrac{h_{1}}{h}$$
$$h_1=h \tan \alpha \cot \beta$$
In a prison wall there is a window of $$1$$ metre height, $$24$$ metres from the ground. An observer at a height of $$10 m$$ from ground, standing at a distance from the wall finds the angle of elevation of the top of the window and the top of the wall to be $$45^{ 0}$$ and $$60^{0}$$ respectively. The height of the wall above the window is
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0%
$$15\sqrt{3}$$
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$$15\left(1-\displaystyle \frac{1}{\sqrt{3}}\right)$$
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$$15(\sqrt{3}-1)$$
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$$14(\sqrt{3}-1)$$
Explanation
We can see that , $$CE=DF=10$$
Therefore, $$BC=BE-CE=24-10=14$$
in $$\triangle BCD$$
$$\tan45^o=\dfrac{14}{b}\Rightarrow b=14$$
Now, In $$\triangle ACD$$
$$\tan60^o=\dfrac{a+14}{14}\Rightarrow \sqrt3=\dfrac{a+14}{14}\Rightarrow a=14(\sqrt3-1)$$
Therefore, Answer is $$14(\sqrt3-1)$$
The horizontal distance between two towers is $$30$$ meters. From the foot of the first tower the angle of elevation of the top of the second tower is $$60^{o}$$. From the top of the second tower the angle of depression of the top of the first is $$30^{o}$$. The height of the small tower is:
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$$20(\sqrt{3}+1)$$ mts
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$$20(\sqrt{3}-1)$$ mts
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$$20\sqrt{3}$$ mts
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$$20$$ mts
Explanation
$$\tan 60^o= \dfrac{h_{2}}{30}$$ ...{1}
$$\tan 30^o=\dfrac{h_{2}-h_{1}}{30}$$ ...{2}
$$\tan 30^{o}=\dfrac{h_{2}}{30}-\dfrac{h_{1}}{30}$$
From {1} and {2}
$$\Rightarrow \dfrac{h_{1}}{30}=\tan 60^o - \tan 30^o$$
$$\Rightarrow \dfrac{h_{1}}{30}=\dfrac{3-1}{\sqrt{3}}$$
$$\Rightarrow h_1=\dfrac{30\times2}{\sqrt{3}}$$
$$\Rightarrow h_{1}=20\sqrt{3} \,m$$
If from the top of a tower of $$60$$ metre high, the angles of depression of the top and floor of a house are $$\alpha$$ and $$\beta$$ respetivley and if the height of the house is $$\displaystyle \frac{60\sin(\beta-\alpha)}{x}$$, then $$x=$$
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0%
$$\sin\alpha\sin\beta$$
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$$\cos\alpha\cos\beta$$
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$$\sin\alpha\cos\beta$$
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$$\cos\alpha\sin\beta$$
Explanation
$$\tan \alpha=\dfrac{60-h}{d}$$
$$\tan \beta=\dfrac{60}{d}$$
$$\dfrac{\tan \alpha}{\tan \beta}=\dfrac{60-h}{60}$$
$$\dfrac{\tan \alpha}{\tan \beta}=1-\dfrac{h}{60}$$
$$\dfrac{h}{60}=\dfrac{\tan \beta-\tan \alpha}{\tan \beta}$$
$$\dfrac{h}{60}=\dfrac{\sin \beta \ \cos\alpha-\sin\alpha \ \cos\beta}{\cos\alpha \ \sin\beta}$$
$$\dfrac{h}{60}=\dfrac{\sin(\beta-\alpha)}{\sin\beta \ \cos\alpha}$$
$$h=\dfrac{60\ \sin(\beta-\alpha)}{\cos\alpha \sin\beta }$$
$$\therefore x=\cos\alpha \sin \beta $$
On the level ground the angle of elevation of the top of a tower is $$30^{0 }$$ On moving 20 metres nearer tower, the angle of elevation is found to be $$60^{0}$$ The height of the towerin metres is
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0%
10 $$\sqrt{3}$$
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$$8\sqrt{3}$$
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$$6\sqrt{3}$$
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$$5\sqrt{3}$$
Explanation
$$\tan\ 60=\dfrac{h}{d}$$
$$d=\dfrac{h}{\sqrt{3}}$$
$$\tan\ 30=\dfrac{h}{20+d}$$
$$\dfrac{1}{\sqrt{3}}=\dfrac{h}{20+\dfrac{h}{\sqrt{3}}}$$
$$20\sqrt{3}+h=3h$$
$$h=10\sqrt{3}$$
lf the shadow of a tower is $$\sqrt{3}$$ times of its height, the altitude of the sun is
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$$15^{0}$$
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$$30^{0}$$
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$$45^{0}$$
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$$60^{0}$$
Explanation
Let the height of tower be $$h$$.
$$\tan \theta = \dfrac{h}{\sqrt{3}h}$$
$$\Rightarrow \tan \theta = \dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \theta = 30^{o}$$
Two poles of heights $$6$$ m and $$11$$ m stand vertically on a plane ground. If the distance between their feet is $$12$$ m, find the distance between their tips.
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$$13$$ m
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$$15$$ m
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$$14$$ m
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none of the above
Explanation
From the given figure, the distance between the tips of the poles forms a right angled triangle with $$ 5 $$ m and $$ 12 m $$ as its hypotenuse.
Pythagoras theorem for right triangle,
$${Hyp}^{2} = ({Side1})^{2} + ({Side2})^{2}$$
So, required distance $$ = \sqrt {{5}^{2} + {12}^{2} } = \sqrt {25 + 144} = \sqrt {169} = 13 m $$
The angle of elevation of the top of a hill when observed from a certain point on the horizontal plane through its base is $$30^{0}$$. After walking 120 meters towards it on level ground the elevation is found to be $$60^{0}$$. Find the height of the hill(in meters).
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120
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$$60\sqrt{3}$$
0%
$$120\sqrt{3}$$
0%
60
Explanation
From the triangle $$ABC$$, let
Height of the hill $$(BC) = h$$
$$AD = 120\,\mathrm{m}$$ and $$DB = d$$
In the $$\triangle DBC, \angle B = 90^\circ$$
$$\tan 60^\circ = \dfrac{BC}{DB}$$
$$\,\Rightarrow\sqrt{3} = \dfrac{h}{d}$$ ........(1)
In the $$\triangle ABC, \angle B = 90^\circ$$
$$\tan 30^{\circ}=\dfrac{BC}{AB}$$
$$\,\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{h}{d+120}$$......(2)
Dividing (1) by (2)
$$\dfrac{\dfrac{h}{d}}{\dfrac{h}{d+120}} = \dfrac{\sqrt 3}{\dfrac{1}{\sqrt3}}$$
$$\Rightarrow \dfrac{d+120}{d} =3$$
$$\Rightarrow 3d = d + 120$$
$$\Rightarrow d = 60$$
Substituting the value of $$d$$ in (1), we get
$$\sqrt{3} = \dfrac{h}{60}$$
$$\Rightarrow h = 60\sqrt{3}$$
$$\therefore$$ Height of the hill $$=60\sqrt3 \mathrm{m}$$
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