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Some Applications of Trigonometry test - 46

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Some Applications of Trigonometry test - 46
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  • Question 1
    1 / -0
    The angle of elevation of a cloud from a point 100 meter above the surface of a lake is 30030^0 and the angle of depression of its image in the lake is 60060^0 then height of the cloud above the lake is 

    Solution
    tan300=h100d=13tan 30^0=\dfrac{h-100}{d}=\dfrac{1}{\sqrt{3}}......(1)
    tan600=h+100d=3tan 60^0=\dfrac{h+100}{d}=\sqrt{3}......................(2)
    Dividing equation (1) by (2) , we get 
    h100h+100=13\dfrac{h-100}{h+100}= \dfrac{1}{3}
    3h300=h+100\Rightarrow 3h-300=h+100
    2h=400\Rightarrow 2h=400
    h=200h=200 metre
    \therefore Height above the lake=200 m
  • Question 2
    1 / -0
    The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30o30^o with horizontal, then the length of the wire is 
    Solution
    Given that : Heights of two poles are 2020 and 1414m respectively.
                        Angle wire makes with horizontal =30= 30^{\circ}

    Let distance between two poles be xx and length of wire be yy
    tan30=2014x=6xtan 30^{\circ} = \dfrac{20-14}{x} = \dfrac{6}{x}

    13=6x\dfrac{1}{\sqrt3} = \dfrac{6}{x}

    x=63\Rightarrow x = 6\sqrt{3}

    As length of the wire will form the hypotenuse of right angled triangle thus formed,
    y2=62+(63)2=144y^{2} = 6^{2}+(6\sqrt3)^{2} = 144
    y=12\therefore y = 12 m
  • Question 3
    1 / -0
    The angle of elevation of stationary cloud from a point 25 ml above the lake is 150 15^0 and the angle of depression of reflection  in the lake is 45045^0 .Then the height of the cloud above the level 
    Solution
    tan15=x25tan15^{\circ}=\dfrac{x}{25} 
    x=25tan15\therefore x=25tan15^{\circ}
     =25(313+1)=25\left ( \dfrac{\sqrt{3}-1}{\sqrt{3}+1} \right )
     \therefore Height =25+25(313+1)=25+25\left ( \dfrac{\sqrt{3}-1}{\sqrt{3}+1} \right ) 
     =5033+1=\dfrac{50\sqrt{3}}{\sqrt{3}+1}
  • Question 4
    1 / -0
    The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle 0 with level ground such that tanθ=158tan\, \theta = \frac {15}{8}, how high is the kite ? 
    Solution
    Given that : AC=90AC = 90m and tanθ=158=BCAB=hABtan \theta = \dfrac{15}{8} = \dfrac{BC}{AB} = \dfrac{h}{AB}

    h=ABtanθ....(1)h = AB tan \theta....(1)

    Now, In \triangle ABC,
    (AC)2=(AB)2+(BC)2(AC)^{2} = (AB)^{2}+(BC)^{2}

    902=AB2+h2\Rightarrow 90^{2} = AB^{2} + h^{2}
    AB=902h2\Rightarrow AB = \sqrt{90^{2}-h^{2}}

    Substituting in (1)(1), we get
    h=902h2tanθh = \sqrt{90^{2}-h^{2}} tan \theta

    h2(1+tan2θ)=902tan2θ\Rightarrow h^{2}(1+tan^{2}\theta) = 90^{2}tan^{2}\theta

    Now substituting the value of tanθtan \theta, we get
    h2=8100×225289\Rightarrow h^{2} = \dfrac{8100\times225}{289}

    h=90×1517=79.475\Rightarrow h = \dfrac{90\times15}{17} = 79.4 \approx 75m

  • Question 5
    1 / -0
    A man observes two objects in a line in the west. On walking a distance xx towards the north, the objects subtends an angle α\alpha in front of him and on walking a further distance xx to north they subtend an angle, β\beta, then the distance between the objects is
    Solution
    tanp=2xc \tan p = \dfrac{2x}c                tanq=xc    \tan q = \dfrac{x}c
    tana=2xc+d\tan a = \dfrac{2x}{c+d}         tanb=xc+d  \tan b = \dfrac{x}{c+d}

    β+a=p\beta + a = p

    α+b=q\alpha + b = q

    hence tanβ \tan\beta = tan(pa)\tan ({p-a})

    tanα \tan\alpha = tan(qb)\tan ({q-b})

    so

    tanβ=tanptana1tanptana\tan\beta = \dfrac{\tan p-\tan a}{1-\tan p\tan a}

    =2xc2xc+d1+2xc2xc+d = \dfrac{\dfrac{2x}{c}-\dfrac{2x}{c+d}}{1+\dfrac{2x}c\dfrac{2x}{c+d}}

    =2xdc2+dc+4x2=\dfrac{2xd}{c^2+dc+4x^2}

    c2+dc+4x2=2xdcotβc^2+dc+4x^2= 2xd\cot\beta                                      ...........(1)

    tanα=tanqtanb1tanqtanb\tan\alpha = \dfrac{\tan q-\tan b}{1-\tan q\tan b}

    =xcxc+d1+xcxc+d = \dfrac{\dfrac{x}{c}-\dfrac{x}{c+d}}{1+\dfrac{x}c\dfrac{x}{c+d}}

    =xdc2+dc+x2=\dfrac{xd}{c^2+dc+x^2}

    c2+dc+x2=xdcotαc^2+dc+x^2= xd\cot\alpha                                      ...........(2)

    From 1 and 2

    3x2=xd(2cotβcotα)3x^2= xd(2\cot\beta-\cot\alpha)

    d=3x2cotβcotα\Rightarrow d = \dfrac{3x}{2\cot\beta - \cot\alpha}

  • Question 6
    1 / -0
    A sphere with centre OO sits atop of pole as shown in the figure. An observer on the ground is at a distance 50m50m from the foot of the pole. She notes that the angles of elevation from the observer to points PP and QQ on the sphere are 3030^{\circ} and 6060^{\circ}, respectively. Then, the radius of the sphere in meters is

    Solution
    in PST\triangle PST
    tan30o=PS50PS=503\tan30^o=\dfrac{PS}{50}\Rightarrow PS=\dfrac{50}{\sqrt3}


    Now, in QUT\triangle QUT
    QU=PS+rQU=PS+r and UT=50rUT=50-r
    tan60o=QUUS3=PS+r50rr=503PS3+1r=5035033+1=50(113)\tan60^o=\dfrac{QU}{US}\Rightarrow \sqrt3=\dfrac{PS+r}{50-r}\Rightarrow r=\dfrac{50\sqrt3-PS}{\sqrt3+1}\Rightarrow r=\dfrac{50\sqrt3-\dfrac{50}{\sqrt3}}{\sqrt3+1}=50\Bigg(1-\dfrac{1}{\sqrt3}\Bigg)

    Therefore, Answer is 50(113)50\Bigg(1-\dfrac{1}{\sqrt3}\Bigg)

  • Question 7
    1 / -0
    One side of a rectangular piece of paper is 6 cm6\text{ cm}, the adjacent sides being longer than 6 cm6\text{ cm}. One corner of the paper is folded so that it sets on the opposite longer side. If the length of the crease is ll cm\text{cm} and it makes an angle θ\theta with the long side as shown, then ll is

    Solution


    Let EC =x=x

    In ΔBCE\Delta BCE,

    xl=sinθx=lsinθ\displaystyle\frac{x}{l}=\sin \theta \Rightarrow x=l\sin \theta

    EF=x=lsinθEF=x=l\sin \theta

    In ΔABF,\Delta ABF,

    ABF=902θ\angle ABF=90-2\theta

    AFB=2θ\angle AFB=2\theta

    sin2θ=6lsinθ\sin 2\theta =\displaystyle\frac{6}{l\sin \theta}

    l=6sinθ.sin2θ\Rightarrow l=\displaystyle\frac{6}{\sin\theta .\sin 2\theta}

    =6sinθ(2sinθcosθ)=\displaystyle\frac{6}{\sin \theta(2\sin \theta\cdot \cos\theta)}

    l=3sin2θcosθl=\displaystyle\frac{3}{\sin^2\theta \cdot \cos\theta}.

  • Question 8
    1 / -0
    An observer standing at a point PP on the top of a hill near the sea-shore notices that the angle of depression of a ship moving towards the hill in a straight line at constant speed is 30o{30}^{o}. After 45 minutes, this angle becomes 45o{45}^{o}. If T (in minutes) is the total time taken by the ship to move to a point in the sea where the angle of depression from PP of the ship is 60o{60}^{o}, then T is equal to:
    Solution
    We know that, time taken to travel between any points is proportional to distance covered.
    Hence, TACTAB=ACAB\dfrac {T_{AC}}{T_{AB}}=\dfrac {AC}{AB}    [uniform speed]

    In PQA\triangle PQA, PQAQ=tan30\dfrac {PQ}{AQ}=\tan 30
    AQ=3h\Rightarrow AQ=\sqrt 3h    ....(1)

    In PBQ\triangle PBQ, tan45=PQBQ\tan 45=\dfrac {PQ}{BQ}
    BQ=h\Rightarrow BQ=h    ....(2)

    Similarly in PQC\triangle PQC, tan60=PQCQ\tan 60=\dfrac {PQ}{CQ}
    CQ=h3\Rightarrow CQ=\dfrac {h}{\sqrt 3}

    Now ACAB=AQCQAQBQ\dfrac {AC}{AB}=\dfrac {AQ-CQ}{AQ-BQ}

    =3hh33hh=\dfrac {\sqrt 3h-\frac {h}{\sqrt 3}}{\sqrt 3h-h}

    =31331=\dfrac {\sqrt 3-\frac {1}{\sqrt 3}}{\sqrt 3-1}

    =1+13=1+\dfrac {1}{\sqrt 3}

    Thus TAC=TAB×(1+13)T_{AC}=T_{AB}\times \left(1+\dfrac {1}{\sqrt 3}\right)

    T=45(1+13)\Rightarrow T=45\left(1+\dfrac {1}{\sqrt 3}\right)

    Hence, the time taken is 45(1+13)45\left(1+\dfrac {1}{\sqrt 3}\right).

  • Question 9
    1 / -0
    The height of a house subtends a right angle at opposite window from the base of the house is 60060^0. If the width of the road be 6 metres, then the height of the house is
    Solution
    Given DBC=600\angle DBC=60^0 and ADB=900\angle ADB=90^0 from figure we know that EDB=DBC=600\angle EDB=\angle DBC=60^0
    Since ADB=ADE+EDB=900\angle ADB=\angle ADE+\angle EDB=90^0
    ADE=900600=300\angle ADE =90^0-60^0=30^0
    From  AED ,
    tan30=AE6\tan30=\dfrac{AE}{6}
    AE23mAE-2\sqrt3 m
    From EDC,
    tan60=EB6\tan60=\dfrac{EB}{6}
    EB=6 3mEB=6\sqrt  3m
    Height of the house =AE+EB=23+63=83mAE+EB=2\sqrt3+6\sqrt3=8\sqrt3 m

  • Question 10
    1 / -0
    From the top of aspire the angle of depression of the top and bottom of a tower of height h are θ\theta and ϕ\phi respectively. Then height of the spire and its horizontal distance from the tower are respectively.
    Solution
    Given that ECD=θ,ECA=ϕ,AD=h\angle ECD=\theta,\angle ECA=\phi,AD=h
    From Geometry ECD=θ=FDC,ECA=ϕ=BAC\angle ECD=\theta =\angle FDC ,\angle ECA=\phi =\angle BAC
    tanθ=BChAB\tan \theta =\dfrac{BC-h}{AB} .............. (1)(1)
    tanϕ=BCABBC=ABtanϕ\tan \phi=\dfrac{BC}{AB}\Rightarrow BC=AB\tan\phi .......... (2)(2)
    Substitute equation (2)(2) in (1)(1)
    ABtanθ=ABtanϕhAB\tan\theta=AB\tan\phi -h
    Distance betweeen tower and sphire AB=htanϕtanθ=hsinϕcosϕsinθcosθ=hcosθcosϕsin(ϕθ)AB=\dfrac{h}{\tan\phi-\tan \theta}=\dfrac{h}{\dfrac{\sin\phi}{\cos\phi}-\dfrac{\sin\theta}{\cos\theta}}=\dfrac{h\cos\theta\cos\phi}{\sin(\phi-\theta)}
    Height of the spire BC=ABtanϕ=hcosθcosϕsin(ϕθ)×tanϕ=hcosθsinϕsin(ϕθ)BC=AB\tan\phi=\dfrac{h\cos\theta \cos\phi}{\sin(\phi-\theta)}\times \tan\phi=\dfrac{h\cos\theta\sin\phi}{\sin(\phi-\theta)}

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