31. Height and Distance Solutions Part 4#

  1. The diagram is given below:

    151st problem

    We follow the question 7474 and have a similar question. Following the same method we find that h=ab.BC2=AC2+AB2=a2+abh = \sqrt{ab}. BC^2 = AC^2 + AB^2 = a^2 + ab. Let CDCD subtend an angle of α\alpha at BB.

    In BCD\triangle BCD, using sine rule

    CDsinα=BCsin(90θ)\frac{CD}{\sin\alpha} = \frac{BC}{\sin(90^\circ - \theta)}

    sinα=CDBCcosθ=CDBC.ACBC=(ba)aa2+ab=baa+b\Rightarrow \sin\alpha = \frac{CD}{BC}\cos\theta = \frac{CD}{BC}.\frac{AC}{BC} = \frac{(b - a)a}{a^2 + ab} = \frac{b - a}{a + b}.

  2. The diagram is given below:

    152nd problem

    Let BCBC be the pillar given a height of hh and CDCD be the statue having a height of xx. Both the statue and the pillar make the same angle at AA which we have let to be θ\theta.

    In ABC,tanθ=hd\triangle ABC, \tan\theta = \frac{h}{d}

    In ABD,tan2θ=2tanθ1tan2θ=h+xd\triangle ABD, \tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{h + x}{d}

    2hd1h2d2=h+xdh+x=2hd2d2h2\Rightarrow \frac{\frac{2h}{d}}{1 - \frac{h^2}{d^2}} = \frac{h + x}{d} \Rightarrow h + x = \frac{2hd^2}{d^2 - h^2}

    x=h(d2+h2)d2h2\Rightarrow x = \frac{h(d^2 + h^2)}{d^2 - h^2}.

  3. The diagram is given below:

    153rd problem

    Let BEBE be the tower and CDCD be the pole such that base of the tower is at half the height of the pole. Given height of the tower is 5050'. Aangles of depression of the top and the foot of the pole from top of the tower are given as 1515^\circ and 4545^\circ. Let the distance between the pole and the tower be dd'.

    In ABC,tan45=1=50+h2dd=50+h2\triangle ABC, \tan45^\circ = 1 = \frac{50 + \frac{h}{2}}{d}\Rightarrow d = 50 + \frac{h}{2}

    In BGD,tan15=313+1=50h2d\triangle BGD, \tan15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3 + 1}} = \frac{50 - \frac{h}{2}}{d}

    h=100/3\Rightarrow h = 100/\sqrt{3} ft.

  4. The diagram is given below:

    154th problem

    Given AA is the initial point of observation and DD is the second point of observation which is 44 km south of AA. Let PP be the point in air where the plane is flying and QQ be the point directly beneath it. Given that QQ is directly east of AA and angles of elevation from AA and DD are respectively 6060^\circ and 3030^\circ. Let PQ=hPQ = h km be the height of the airplane. Clearly, DAQ\angle DAQ is a right angle.

    In APQ,tan60=hAQAQ=hcot60\triangle APQ, \tan60^\circ = \frac{h}{AQ} \Rightarrow AQ = h\cot60^\circ

    In DPQ,tan30=hDQDQ=hcot30\triangle DPQ, \tan30^\circ = \frac{h}{DQ} \Rightarrow DQ = h\cot30^\circ

    In ADQ,DQ2=AB2+AQ2h2cot230=h2cot260+42\triangle ADQ, DQ^2 = AB^2 + AQ^2 \Rightarrow h^2\cot^230^\circ = h^2\cot^260^\circ + 4^2

    h=6\Rightarrow h = \sqrt{6} km.

  5. The diagram is given below:

    155th problem

    Let PNPN be the flag-staff having a height of hh. ABAB is perpendicular to ANAN. Let AN=xAN = x and BN=yBN = y. Given angles of elevation from AA and BB to PP are α\alpha and β\beta respectively.

    In APN,x=hcotα\triangle APN, x = h\cot\alpha. In BPN,y=hcotβ\triangle BPN, y = h\cot\beta

    In ABN,AB2+h2cot2α=h2cot2β\triangle ABN, AB^2 + h^2\cot^2\alpha = h^2\cot^2\beta

    h=ABcot2βcot2α=ABsinαsinβsin(α+β)sin(αβ)\Rightarrow h = \frac{AB}{\sqrt{\cot^2\beta - \cot^2\alpha}} = \frac{AB\sin\alpha\sin\beta}{\sin(\alpha + \beta)\sin(\alpha - \beta)}

  6. The diagram is given below:

    156th problem

    Let ACAC be the tower having a height of hh such that AB:BC::1:9AB:BC::1:9. Given the point at a distance of 2020 m is where both ABAB and BCBC subtend equal angle which we have let to be θ\theta.

    In ABD,tanθ=h1020h=200tanθ\triangle ABD, \tan\theta = \frac{h}{10*20} \Rightarrow h = 200\tan\theta

    In ACD,tan2θ=h20=10tanθ\triangle ACD, \tan2\theta = \frac{h}{20} = 10\tan\theta

    2tanθ1tan2θ=10tanθ\Rightarrow \frac{2\tan\theta}{1 - \tan^2\theta} = 10\tan\theta

    10tan2θ8=0tanθ=25\Rightarrow 10\tan^2\theta -8 = 0 \Rightarrow \tan\theta = \frac{2}{\sqrt{5}}

    h=805\Rightarrow h = 80\sqrt{5} m.

  7. The diagram is given below:

    157th problem

    Let BCBC be the tower inclined at angle an angle θ\theta from horizontal having a vertical height of hh. Let AA and DD be two equidistant points from base BB of the tower from where the angles of elevation to the top of the tower is α\alpha and β\beta respectively. Let AB=BD=dAB = BD = d. Let BE=xBE = x.

    In BCE,tanθ=hxx=hcotθ\triangle BCE, \tan\theta = \frac{h}{x} \Rightarrow x = h\cot\theta

    Clearly, AE=dhcotθAE = d - h\cot\theta and BE=d+hcotθBE = d + h\cot\theta

    In ACE,tanα=hdhcotθ\triangle ACE, \tan\alpha = \frac{h}{d - h\cot\theta} and in BCE,tanβ=hd+hcotθ\triangle BCE, \tan\beta = \frac{h}{d + h\cot\theta}

    1cotα+cotθ=1cotβcotθ\Rightarrow \frac{1}{\cot\alpha + \cot\theta} = \frac{1}{\cot\beta - \cot\theta}

    θ=tan1sin(αβ)2sinαsinβ\Rightarrow \theta = \tan^{-1}\frac{\sin(\alpha - \beta)}{2\sin\alpha\sin\beta}.

  8. The diagram is given below:

    158th problem

    Let ABCABC be the triangle in horizontal plane and PQPQ be the 1010 m high flag staff at the center of the ABC\triangle ABC. Given that each side subtends an angle of 6060^\circ at the top of flag staff i.e. QQ.

    AQC=60AQ=QC\therefore \angle AQC = 60^\circ \Rightarrow AQ = QC making AQC\triangle AQC equilateral.

    Let AQ=QC=AC=2aAQ = QC = AC = 2a. We know that centroid is a point on median from where the top of the vertex is at a distance of 23\frac{2}{3} rd times length of a side. We also know that median of an equilateral triangle is perpendicular bisector of the opposite side.

    AP=23.2a.sin60=2a3\Rightarrow AP = \frac{2}{3}.2a.\sin60^\circ = \frac{2a}{\sqrt{3}}

    APQ\triangle APQ is also a right angle triangle with right angle at PP.

    AQ2=AP2+PQ2a=532\Rightarrow AQ^2 = AP^2 + PQ^2 \Rightarrow a = 5\sqrt{\frac{3}{2}}

    \therefore Length of a side =2a=56= 2a = 5\sqrt{6} m.

  9. The diagram is given below:

    159th problem

    Let ABAB be the pole having a height of hh then the height of the second pole CDCD would be 2h2h. OO is the point of observation situated at mid-point between the poles i.e. at a distance of 6060 m from each pole. Let AOB=θ\angle AOB = \theta and therefore COD=90θ\angle COD = 90^\circ - \theta.

    In AOB,tanθ=h60\triangle AOB, \tan\theta = \frac{h}{60}

    In COD,tan(90θ)=cotθ=2h60=2tanθtanθ=12\triangle COD, \tan(90^\circ - \theta) = \cot\theta = \frac{2h}{60} = 2\tan\theta \Rightarrow \tan\theta = \frac{1}{\sqrt{2}}

    h=302\Rightarrow h = 30\sqrt{2} m and 2h=6022h = 60\sqrt{2} m.

  10. This problem is similar to 158158, and has been left as an exercise.

  11. This problem is similar to 134134, and has been left as an exercise.

  12. The diagram is given below:

    162nd problem

    Since ABAB and CDCD are two banks of a straight river they would be parallel. We have shown alternate angles for β\beta and γ\gamma in the diagram other than given angles. In ABC,ACB=π(α+β+γ)sinACB=sin(α+β+γ)\triangle ABC, \angle ACB = \pi - (\alpha + \beta + \gamma) \Rightarrow \sin ACB = \sin(\alpha + \beta + \gamma).

    Using sine formula in ABC\triangle ABC,

    ABsinACB=ACsinABCAC=asinγsin(α+β+γ)\frac{AB}{\sin ACB} = \frac{AC}{\sin ABC} \Rightarrow AC = \frac{a\sin\gamma}{\sin(\alpha + \beta + \gamma)}

    Using sine formula in ACD\triangle ACD,

    CDsinα=ACsinβCD=asinαsinγsinβsin(α+β+γ)\frac{CD}{\sin\alpha} = \frac{AC}{\sin\beta} \Rightarrow CD = \frac{a\sin\alpha\sin\gamma}{\sin\beta\sin(\alpha + \beta + \gamma)}

  13. The diagram is given below:

    163rd problem

    Let PQPQ be the bank of river having a width of bb and RR be the point in line with PQPQ at a distance of aa from QQ. QSQS is the distance of 100100 m to which the person walks at right angle from initial line.

    In PRS,tan40=a+b100\triangle PRS, \tan40^\circ = \frac{a + b}{100}

    In QRS,tan25=b100\triangle QRS, \tan25^\circ = \frac{b}{100}

    b=100(tan40tan25)\Rightarrow b = 100(\tan40^\circ - \tan25^\circ).

  14. This problem is similar to 9696, and has been left as an exercise.

  15. This problem is similar to 9696, and has been left as an exercise.

  16. The diagram is given below:

    166th problem

    Let PP and QQ be the tops of two spires, PP' and QQ' be their reflections. From question OA=hOA = h. Let BP=BP=h1,CQ=CQ=hBP = BP' = h1, CQ = CQ' = h

    Let the distance between spires be x=MN=OMONx = MN = OM - ON.

    In OMP,tanβ=h+h1OMOMtanβ=h+h1\triangle OMP', \tan\beta = \frac{h + h1}{OM} \Rightarrow OM\tan\beta = h + h1

    In OMP,tanα=h1hOMOMtanα=h1h\triangle OMP, \tan\alpha = \frac{h1 - h}{OM} \Rightarrow OM\tan\alpha = h1 - h

    OM(tanβtanα)=2hOM=2htanβtanα\Rightarrow OM(\tan\beta - \tan\alpha) = 2h \Rightarrow OM = \frac{2h}{\tan\beta - \tan\alpha}

    Similarly, ON=2htanγtanαON = \frac{2h}{\tan\gamma - \tan\alpha}

    x=OMON=2h[1tanβtanα1tanγtanα]\Rightarrow x = OM - ON = 2h\left[\frac{1}{\tan\beta - \tan\alpha} - \frac{1}{\tan\gamma -\tan\alpha}\right]

    On simplification we arrive at the desired result.

  17. The diagram is given below:

    167th problem

    Let OO be the center of the square and OPOP be the pole having a height of hh. Let OQOQ be the shdow of the pole. Given CO=xCO = x and BQ=yBQ = y. Then BC=x+yBC = x + y. Let ORBCOR\perp BC.

    OR=BR=x+y2\therefore OR = BR = \frac{x + y}{2} and QR=xy2QR = \frac{x - y}{2}

    In POQ,tanα=hOQOQ=hcotα\triangle POQ, \tan\alpha = \frac{h}{OQ}\Rightarrow OQ = h\cot\alpha

    In ORQ,OQ2=OR2+QR2h2cot2α=(x+y2)2+(xy2)2\triangle ORQ, OQ^2 = OR^2 + QR^2 \Rightarrow h^2\cot^2\alpha = \left(\frac{x + y}{2}\right)^2 + \left(\frac{x - y}{2}\right)^2

    h=x2+y22tanα\Rightarrow h = \sqrt{\frac{x^2 + y^2}{2}}\tan\alpha

  18. The diagram is given below:

    168th problem

    Let OPOP be the vertical height cc of the candle. OO' is the point vertically below OO therefore OO=bOO' = b as given in the question. Let EFEF represent the line of intersection of the wall and the horizontal ground. Draw ODEFO'D\perp EF then OD=aO'D = a.

    Clearly, EF=2DEEF = 2DE as shadow is symmetrical about line ODO'D,

    In similar triangles AOPAOP and POEPO'E,

    OAOE=OPOPaOE=cb+cOE=a(b+c)c\frac{OA}{O'E} = \frac{OP}{O'P} \Rightarrow \frac{a}{O'E} = \frac{c}{b + c} \Rightarrow O'E = \frac{a(b + c)}{c}

    In ODE\triangle O'DE,

    OE2=a2+DE2DE=acb2+2bcO'E^2 = a^2 + DE^2 \Rightarrow DE = \frac{a}{c}\sqrt{b^2 + 2bc}

    EF=2DE2acb2+2bc\Rightarrow EF = 2DE \frac{2a}{c}\sqrt{b^2 + 2bc}

  19. The diagram is given below:

    169th problem

    Let PABCDPABCD be the pyramid, PQPQ the flag-staff having a height of 66 m. Let OP=hOP = h and the shadow touches the side at LL.

    Proceeding like problem 167167, we have in OML\triangle OML,

    OL2=OM2+LM2(h+6)2cot2α=(x+y2)2+(xy2)2OL^2 = OM^2 + LM^2 \Rightarrow (h + 6)^2\cot^2\alpha = \left(\frac{x + y}{2}\right)^2 + \left(\frac{x - y}{2}\right)^2

    h=x2+y22tanα6\Rightarrow h = \sqrt{\frac{x^2 + y^2}{2}}\tan\alpha - 6

  20. The diagram is given below:

    170th problem

    Let PQPQ be the tower with given height h,Ch, C be the initial point of observation from where angle of elevation is θ\theta. When the man moves a distance dd let him reach point BB from where angle of elevation is 2θ2\theta and then final point be AA which is at a distance of 34d\frac{3}{4}d from BB, having an angle of elevation 3θ3\theta.

    QCB=CQB=θBC=BQ=d\angle QCB = \angle CQB = \theta \therefore BC = BQ = d

    In PQB,sin2θ=hdh=2dsinθcosθ\triangle PQB, \sin2\theta = \frac{h}{d} \Rightarrow h = 2d\sin\theta\cos\theta

    Using sine rulel in ABQ,3d4sinθ=dsin(180θ)\triangle ABQ, \frac{3d}{4\sin\theta} = \frac{d}{\sin(180^\circ - \theta)}

    34sinθ=13sinθ4sin3θ\Rightarrow \frac{3}{4\sin\theta} = \frac{1}{3\sin\theta - 4\sin^3\theta}

    sin2θ=512cos2θ=712\Rightarrow \sin^2\theta = \frac{5}{12} \therefore \cos^2\theta = \frac{7}{12}

    h2=4d2sin2θcos2θ36h2=35d2\Rightarrow h^2 = 4d^2\sin^2\theta\cos^2\theta \Rightarrow 36h^2 = 35d^2

  21. The diagram is given below:

    171st problem

    Let OO be the mid-point of ABAB having a measure of 88 m. Let OPOP be the 22 m long object, PQPQ be its position after 11 second and RSRS be the position after 22 seconds.

    PAQ=α,RAS=β\angle PAQ = \alpha, \angle RAS = \beta as given in the problem. Also given,

    dsdt=2t+1ds=(2t+1)dts=t2+t+k\frac{ds}{dt} = 2t + 1 \Rightarrow \int ds = \int(2t + 1)dt \Rightarrow s = t^2 + t + k where kk is the constant of acceleration. At t=0,s=0k=0t = 0, s = 0 \Rightarrow k = 0

    At t=1,s=2t = 1, s = 2 and t=2,s=6OP=PQ=QR=RS=2t = 2, s = 6 \therefore OP = PQ = QR = RS = 2 m.

    Let OAP=θ1,OAQ=θ2,OAR=θ3\angle OAP = \theta_1, \angle OAQ = \theta_2, \angle OAR = \theta_3 and OAS=θ4\angle OAS = \theta_4

    tanα=tan(θ2θ1)=tanθ2θ11+tanθ1tanθ2=1121+12=13\Rightarrow \tan\alpha = \tan(\theta_2 - \theta_1) = \frac{\tan\theta_2 - \theta_1}{1 + \tan\theta_1\tan\theta_2} = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3}

    Similarly, tanβ=18\tan\beta = \frac{1}{8}

    cos(αβ)=526\Rightarrow \cos(\alpha - \beta) = \frac{5}{\sqrt{26}}

  22. The diagram is given below:

    172nd problem

    Let ODOD be the pole having a height of hh. Given that ABC\triangle ABC is isosceles and BB and CC subtend same angle at PP which is feet of the observer, therefore AB=ACAB = AC. Let BD=DC=xBD = DC = x. Given APO=β,CPQ=α\angle APO = \beta, \angle CPQ = \alpha and OP=dOP = d.

    ΔABC=12BC.AD=x.AD\Delta ABC = \frac{1}{2}BC.AD = x.AD

    In AOP,tanβ=h+ADdAD=dtanβh\triangle AOP, \tan\beta = \frac{h + AD}{d} \Rightarrow AD = d\tan\beta - h

    In CQP,tanα=hPQPQ=hcotα\triangle CQP, \tan\alpha = \frac{h}{PQ} \Rightarrow PQ = h\cot\alpha

    In OPQ,OQ2=PQ2OP2OQ=h2cot2αd2\triangle OPQ, OQ^2 = PQ^2 - OP^2 \Rightarrow OQ = \sqrt{h^2\cot^2\alpha - d^2}

    ΔABC=(dtanβh)h2cot2αd2\Rightarrow \Delta ABC = (d\tan\beta - h)\sqrt{h^2\cot^2\alpha - d^2}

  23. The diagram is given below:

    173rd problem

    In the diagram A,Q,BA, Q, B are in the plane of paper and PQPQ is perpedicular to the plane of paper.

    In APQ,tan(90θ)=hAQ\triangle APQ, \tan(90^\circ - \theta) = \frac{h}{AQ}

    In BPQ,tanθ=hBQ\triangle BPQ, \tan\theta = \frac{h}{BQ}

    h=AQ.BQ\Rightarrow h = \sqrt{AQ.BQ}

    Since BQBQ is north-west AQB=45=QBAAQ=AB=100\therefore \angle AQB = 45^\circ = \angle QBA\Rightarrow AQ = AB = 100 m.

    In ABQ,OB=AQ2+AB2=1002\triangle ABQ, OB = \sqrt{AQ^2 + AB^2} = 100\sqrt{2} m.

    h=10024\Rightarrow h = 100\sqrt[4]{2} m.

  24. The diagram is given below:

    174th problem

    Let ABAB and CDCD be the vertical poles having heights of aa and bb respectively and angle of elevation α\alpha from OO which is same for both of them. Also, the angles of elevation from PP are β\beta and γ\gamma along with APC=90\angle APC = 90^\circ.

    In ABQ,tanα=aAQAQ=acotα\triangle ABQ, \tan\alpha = \frac{a}{AQ} \Rightarrow AQ = a\cot\alpha

    In CDQ,tanα=bCQCQ=bcotα\triangle CDQ, \tan\alpha = \frac{b}{CQ} \Rightarrow CQ = b\cot\alpha

    AC=AQ+CQ=(a+b)cotα\Rightarrow AC = AQ + CQ = (a + b)\cot\alpha

    In ABP,tanβ=aAPAP=acotβ\triangle ABP, \tan\beta = \frac{a}{AP} \Rightarrow AP = a\cot\beta

    In CDP,tanγ=aCPCP=bcotγ\triangle CDP, \tan\gamma = \frac{a}{CP}\Rightarrow CP = b\cot\gamma

    In APC,AC2=AP2+CP2\triangle APC, AC^2 = AP^2 + CP^2

    (a+b)2cot2α=a2cot2β+b2cot2γ\Rightarrow (a + b)^2\cot^2\alpha = a^2\cot^2\beta + b^2\cot^2\gamma

  25. The diagram is given below:

    175th problem

    Given the pole is PQPQ, let hh be its height. PQPQ is perpedicular to the plane of paper i.e ABC.QPA=QPB=QPC=90ABC. \therefore \angle QPA = \angle QPB = \angle QPC = 90^\circ

    In APQ,tanθ=hPAPA=hcotθ\triangle APQ, \tan\theta = \frac{h}{PA} \Rightarrow PA = h\cot\theta

    Similarly, PB=PC=hcotθ=PAPB = PC = h\cot\theta = PA

    Hence, PP is the circumcenter of the ABC\triangle ABC and PAPA is circum-radius of the circumcircle.

    h=PAtanθ=abc4Δtanθ\therefore h = PA\tan\theta = \frac{abc}{4\Delta}\tan\theta

  26. The diagram is given below:

    176th problem

    Let PQPQ be the tower having a height of hh and AOP=θ\angle AOP = \theta. Given that tanθ=12\tan\theta = \frac{1}{\sqrt{2}}

    In AOP,tanθ=APAOAP=1502\triangle AOP, \tan\theta = \frac{AP}{AO} \Rightarrow AP = 150\sqrt{2} m.

    In POQ,tan30=hOPOP=h3\triangle POQ, \tan30^\circ = \frac{h}{OP} \Rightarrow OP = h\sqrt{3}

    In AOP,OP2=OA2+AP23h2=3002+(1502)2\triangle AOP, OP^2 = OA^2 + AP^2 \Rightarrow 3h^2 = 300^2 + (150\sqrt{2})^2

    h=1502\Rightarrow h = 150\sqrt{2} m,

    tanϕ=hAP=1ϕ=45\therefore \tan\phi = \frac{h}{AP} = 1 \Rightarrow \phi = 45^\circ.

  27. The diagram is given below:

    177th problem

    Let OB=h,OA=xOB = h, OA = x. In AOB,tanα=xh\triangle AOB, \tan\alpha =\frac{x}{h}

    x=htanα\Rightarrow x = h\tan\alpha

    In BOC,tanβ=hdxd=htanα+hcotβ\triangle BOC, \tan\beta = \frac{h}{d - x} \Rightarrow d = h\tan\alpha + h\cot\beta

    In BOD,tanγ=hd+xd=hcotγhtanα\triangle BOD, \tan\gamma = \frac{h}{d + x}\Rightarrow d = h\cot\gamma - h\tan\alpha

    htanα+hcotβ=hcotγhtanα\therefore h\tan\alpha + h\cot\beta = h\cot\gamma - h\tan\alpha

    2tanα=cotγcotβ\Rightarrow 2\tan\alpha = \cot\gamma - \cot\beta.

  28. The diagram is given below:

    178th problem

    Let MM be the mid-point of ESES such that SM=ME=xSM = ME = x and OPOP be the tower having a height of hh.

    In EOP,tanα=hOEOE=hcotα\triangle EOP, \tan\alpha = \frac{h}{OE} \Rightarrow OE = h\cot\alpha

    Similarly in OPS,OS=hcotβ\triangle OPS, OS = h\cot\beta and in MOP,OM=hcotθ\triangle MOP, OM = h\cot\theta

    Since OEOE is eastward and OSOS is southward EOS=90\Rightarrow EOS = 90^\circ

    ES2=OS2+OE24x2=h2(cot2β+cot2α)\Rightarrow ES^2 = OS^2 + OE^2 \Rightarrow 4x^2 = h^2(\cot^2\beta + cot^2\alpha)

    Since MM is mid-point of ES,OMES, OM would be the median.

    OS2+OE2=2MS2+2OM2\Rightarrow OS^2 + OE^2 = 2MS^2 + 2OM^2

    h2cot2β+h2cot2α=h2(cot2β+cot2α)2+2h2cot2θ\Rightarrow h^2\cot^2\beta + h^2\cot^2\alpha = \frac{h^2(\cot^2\beta + \cot^2\alpha)}{2} + 2h^2\cot^2\theta

    cot2β+cot2α=4cot2θ\Rightarrow \cot^2\beta + \cot^2\alpha = 4\cot^2\theta

  29. The diagram is given below:

    179th problem

    Let APAP be the tree having a height of hh and ABAB be the width of canal equal to xx. Given, BC=20BC = 20 m and BAC=120\angle BAC = 120^\circ.

    In ABP,tan60=hABAB=h3\triangle ABP, \tan60^\circ = \frac{h}{AB} \Rightarrow AB = \frac{h}{\sqrt{3}}

    In ACP,tan30=hACAC=3h\triangle ACP, \tan30^\circ = \frac{h}{AC}\Rightarrow AC = \sqrt{3}h

    Using cosine rule in ABC,\triangle ABC,

    cos120=AB2+BC2AC22.AB.BC12=h23+2023h22.20.h3\cos120^\circ = \frac{AB^2 + BC^2 - AC^2}{2.AB.BC}\Rightarrow -\frac{1}{2} = \frac{\frac{h^2}{3} + 20^2 - 3h^2}{2.20.\frac{h}{\sqrt{3}}}

    2h253h300=0h=53+15114\Rightarrow 2h^2 - 5\sqrt{3}h - 300 = 0\Rightarrow h = \frac{5\sqrt{3} + 15\sqrt{11}}{4} m.

    AB=5+5334\Rightarrow AB = \frac{5 + 5\sqrt{33}}{4} m.

  30. The diagram is given below:

    180th problem

    Let OPOP be the tower with PP being the top having a height of hh. According to question S1S2=S2S3,PS2S1=γ1,PS3S2=γ2,S1PS2=δ1,S2PS3=δ2,PS1O=β1S_1S_2 = S_2S_3, \angle PS_2S_1 = \gamma_1, \angle PS_3S_2 = \gamma_2, \angle S_1PS_2 = \delta_1, \angle S_2PS_3 = \delta_2, \angle PS_1O = \beta_1 and PS2O=β2\angle PS_2O = \beta_2.

    In OPS1,sinβ1=hPS1PS1=hsinβ1\triangle OPS1, \sin\beta_1 = \frac{h}{PS_1} \Rightarrow PS_1 = \frac{h}{\sin\beta_1}

    In OPS2,PS2=hsinβ2\triangle OPS_2, PS_2 = \frac{h}{\sin\beta_2}

    Using sine rule in PS1S2,S1S2sinδ1=PS1sinγ1PS_1S_2, \frac{S_1S_2}{\sin\delta_1} = \frac{PS_1}{\sin\gamma_1}

    hS1S2=sinβ1sinγ1sinδ1\Rightarrow \frac{h}{S_1S_2} = \frac{\sin\beta_1\sin\gamma_1}{\sin\delta_1}

    Similarly in PS2S3,hS2S3=sinβ2sinγ2sinδ2PS_2S_3, \frac{h}{S_2S_3} = \frac{\sin\beta_2\sin\gamma_2}{\sin\delta_2}

    Equalting last two results we have desired equality.

  31. The diagram is given below:

    181st problem

    Let PQPQ be the vertical pillar having a height of hh. According to question, tanα=2,AN=20\tan\alpha = 2, AN = 20 m and that PAM\triangle PAM is equilateral. Let QAP=β,QBP=γ\angle QAP =\beta, \angle QBP = \gamma

    In NPQ,tanα=hPN=2PN=h2\triangle NPQ, \tan\alpha = \frac{h}{PN} = 2 \Rightarrow PN = \frac{h}{2}

    In ANP,tan60=3=PNANPN=203h=403\triangle ANP, \tan60^\circ = \sqrt{3} = \frac{PN}{AN} \Rightarrow PN = 20\sqrt{3} \Rightarrow h = 40\sqrt{3} m.

    cos60=12=ANPAPA=40\cos60^\circ = \frac{1}{2} = \frac{AN}{PA} \Rightarrow PA = 40 m.

    PAM\triangle PAM is equilateral and PNAMAN=MN=20PN\perp AM \therefore AN = MN = 20 m AM=40\Rightarrow AM = 40 m, AB=80\Rightarrow AB = 80 m.

    PB=AB2PA2=403\therefore PB = \sqrt{AB^2 - PA^2} = 40\sqrt{3} m.

    β=60\Rightarrow \beta = 60^\circ and γ=45\gamma = 45^\circ.

  32. The diagram is given below:

    182nd problem

    Let ABCABC be the triangular park, OO be the mid-point of BCBC and OPOP be the television tower(out of the plane of paper). Given that, PAO=45,PBO=60,PCO=60,AB=AC=100\angle PAO = 45^\circ, \angle PBO = 60^\circ, \angle PCO = 60^\circ, AB = AC = 100 m. Also, let OP=hOP = h m.

    Clearly, POA=POB=POC=90\angle POA = \angle POB = \angle POC = 90^\circ.

    In POA,tan45=hOAOA=h\triangle POA, \tan45^\circ = \frac{h}{OA}\Rightarrow OA = h

    In POB,tan60=hOBOB=h3\triangle POB, \tan60^\circ = \frac{h}{OB}\Rightarrow OB = \frac{h}{\sqrt{3}}

    Similarly OCOC would be h3\frac{h}{\sqrt{3}}.

    ABC\because \triangle ABC is an isosceles triangle and OO is the mid-point of BC.AOBCBC. \therefore AO\perp BC.

    In AOB,AB2=OA2+OB2h=503\triangle AOB, AB^2 = OA^2 + OB^2 \Rightarrow h = 50\sqrt{3} m.

  33. The diagram is given below:

    183rd problem

    Let ABCDABCD be the base of the square tower whose upper corners are A,B,C,DA', B', C', D' respectively. From a point OO on the diagonal ACAC the three upper corners A,BA', B' and DD' are visible.

    According to question AOA=60,BOB=DOD=45\angle AOA' = 60^\circ, \angle BOB' = \angle DOD' = 45^\circ

    Also, AA=BB=hAA' = BB' = h and AB=aAB = a

    In AAO,tan60=hAOAO=h3\triangle AA'O, \tan60^\circ = \frac{h}{AO}\Rightarrow AO = \frac{h}{\sqrt{3}}

    In BBO,tan45=1=hBOBO=h\triangle BB'O, \tan45^\circ = 1= \frac{h}{BO} \Rightarrow BO = h

    Using cosine rule in AOB,\triangle AOB,

    cos135=AO2+AB2BO22AO.AB\cos135^\circ = \frac{AO^2 + AB^2 - BO^2}{2AO.AB}

    12=h23+a2h22.h3.a\Rightarrow -\frac{1}{\sqrt{2}} = \frac{\frac{h^2}{3} + a^2 - h^2}{2.\frac{h}{\sqrt{3}}.a}

    Considering h>0h > 0, on simplification we arrive at ha=6(1+5)4\frac{h}{a} = \frac{\sqrt{6}(1 + \sqrt{5})}{4}.

  34. The diagram is given below:

    184th problem

    In the diagram PPRRPP'R'R is a plane perpendicular to the plane of the paper. Let CC be the center of top of the cylindrical tower. Since AA is the point on the horizontal plane nearest to QQ, hence AA will be on the line QAQ'A where QAQQQ'A\perp QQ'. According to question QQ=h,CQ=r,QAQ=60QQ' = h, C'Q' = r, \angle QAQ' = 60^\circ and PAP=45\angle PAP' = 45^\circ.

    In AQQ,tan60=3=hAQAQ=h3\triangle AQQ', \tan60^\circ = \sqrt{3} = \frac{h}{AQ'} \Rightarrow AQ' = \frac{h}{\sqrt{3}}

    In APP,tan45=1=hAPAP=h\triangle APP', \tan45^\circ = 1 = \frac{h}{AP'} \Rightarrow AP' = h

    AC=AQ+CQ=h3+rAC' = AQ' + C'Q' = \frac{h}{\sqrt{3}} + r

    In ACP,AP2=AC2+CP2h2+(h3+r)2+r2\triangle AC'P', AP'^2 = AC'^2 + C'P'^2 \Rightarrow h^2 + \left(\frac{h}{\sqrt{3}} + r\right)^2 + r^2

    Taking into account that h>0h > 0, on simplification we arrive at

    hr=3(1+5)2\frac{h}{r} = \frac{\sqrt{3}(1 + \sqrt{5})}{2}.

  35. The diagram is given below:

    185th problem

    Let APAP be the pole having a height of hh m. Let PCA=θ,ADB=α\angle PCA = \theta, \angle ADB = \alpha and BDC=β\angle BDC = \beta. Then PBA=2θ\angle PBA = 2\theta and BPC=θ\angle BPC = \theta.

    BPC=BCPBP=BC=20\Rightarrow \angle BPC = \angle BCP \Rightarrow BP = BC = 20 m.

    From question tanα=15,CD=30\tan\alpha = \frac{1}{5}, CD = 30 m and BC=20BC = 20 m.

    In BCD,tanβ=BCCD=2030=23\triangle BCD, \tan\beta = \frac{BC}{CD} = \frac{20}{30} = \frac{2}{3}

    Now tan(α+β)=tanα+tanβ1tanαtanβ=1\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = 1

    α+β=45ADC=DAC=45\Rightarrow \alpha + \beta = 45^\circ \Rightarrow \angle ADC = \angle DAC = 45^\circ

    AC=CD=30\Rightarrow AC = CD = 30 m. AB=ACBC=3020=10\Rightarrow AB = AC - BC = 30 - 20 = 10 m.

    In PAB,h2=PB2AB2=202102h=103\triangle PAB, h^2 = PB^2 - AB^2 = 20^2 - 10^2 \Rightarrow h = 10\sqrt{3} m.

  36. The diagram is given below:

    186th problem

    Let OPOP be the tower having a height of h,Ah, A be the initial position of the man, BB be the second position of the man at a distance aa from AA and CC be the final position of the man at a distance of 5a3\frac{5a}{3} from BB. Given that angles of elevation from A,BA, B and CC of the top of the tower are 30,3030^\circ, 30^\circ and 6060^\circ respectively. OCABOC\perp AB and DNOCDN\perp OC.

    In POA,tan30=13=hOAOA=3h\triangle POA, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{OA} \Rightarrow OA = \sqrt{3}h

    Similarly in POB,OB=3h\triangle POB, OB = \sqrt{3}h and in POD,tan60=hODOD=h3\triangle POD, \tan60^\circ = \frac{h}{OD} \Rightarrow OD = \frac{h}{\sqrt{3}}

    OA=OAAC=BC=a2\because OA = OA \Rightarrow AC = BC = \frac{a}{2}

    OC=OA2AC2=3h2a24,ON=OD2DN2=h23a24OC = \sqrt{OA^2 - AC^2} = \sqrt{3h^2 - \frac{a^2}{4}}, ON = \sqrt{OD^2 - DN^2} = \sqrt{\frac{h^2}{3} - \frac{a^2}{4}}

    BD=CN=OCCN5a3=3h2a24h23a24BD = CN = OC - CN \Rightarrow \frac{5a}{3} = \sqrt{3h^2 - \frac{a^2}{4}} - \sqrt{\frac{h^2}{3} - \frac{a^2}{4}}

    On simplification, we get h=8548ah = \sqrt{\frac{85}{48}}a or h=56ah = \sqrt{\frac{5}{6}}a.

  37. The diagram is given below:

    187th problem

    Let OPOP be the tower having a height of hh. Given ABCABC is an equilateral triangle. Let the angle subtended by OPOP at A,B,CA, B, C be α,β,γ\alpha, \beta, \gamma respectively. According to question tanα=3+1,tanβ=2\tan\alpha = \sqrt{3} + 1, \tan\beta = \sqrt{2} and tanγ=2\tan\gamma = \sqrt{2}. OPOP is perpedicular to the plane of ABC\triangle ABC.

    In AOP,tanα=hOAOA=h3+1\triangle AOP, \tan\alpha = \frac{h}{OA}\Rightarrow OA = \frac{h}{\sqrt{3} + 1}.

    Similarly, OB=h2OB = \frac{h}{\sqrt{2}} and OC=h2OC = \frac{h}{\sqrt{2}}.

    In AOB\triangle AOB and AOC,AB=AC,OB=OCAOC, AB = AC, OB = OC and OAOA is common. So AOB\triangle AOB and AOC\triangle AOC are equal. OAB=OAC\therefore \angle OAB = \angle OAC.

    But BAC=60OAB=OAC=30\angle BAC = 60^\circ \therefore \angle OAB = \angle OAC = 30^\circ

    Let OBA=θ\angle OBA = \theta

    Using sine rule in OAB,OBsin30=OAsinθ\triangle OAB, \frac{OB}{\sin30^\circ} = \frac{OA}{\sin\theta}

    sinθ=3122=sin15\Rightarrow \sin\theta = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \sin15^\circ

    θ=15.OBD=ABCθ=45\Rightarrow \theta = 15^\circ. \Rightarrow \angle OBD = \angle ABC - \theta = 45^\circ

    In BOC,OB=OC,ODBCBD=DC=40\triangle BOC, OB = OC, OD\perp BC \therefore BD = DC = 40'

    In BOD,cos45=BDOB=40h/2h=80\triangle BOD, \cos45^\circ = \frac{BD}{OB} = \frac{40}{h/\sqrt{2}}\Rightarrow h = 80'

  38. The diagram is given below:

    188th problem

    Let OPOP be the tower having a height of hh and PQPQ be the flag-staff having a height of xx. Since PQPQ subtends equal angle α\alpha at AA and BB so a circle will pass through A,B,PA, B, P and QQ. Since CC is the mid-point of ABAC=BC=aAB \therefore AC = BC = a.

    Let OA=dOA = d and PAO=θ\angle PAO = \theta. In AOP,tanθ=hd\triangle AOP, \tan\theta = \frac{h}{d}

    In AOQ,tan(θ+α)=h+xdtanθ+tanα1tanθtanα=hd+tanα1hdtanα\triangle AOQ, \tan(\theta + \alpha) = \frac{h + x}{d} \Rightarrow \frac{\tan\theta + \tan\alpha}{1 - \tan\theta\tan\alpha} = \frac{\frac{h}{d} + \tan\alpha}{1 - \frac{h}{d}\tan\alpha}

    h+dtanαdhtanα=h+xd\Rightarrow \frac{h + d\tan\alpha}{d - h\tan\alpha} = \frac{h + x}{d}

    d2+h(x+h)=xdcotα\Rightarrow d^2 + h(x + h) = xd\cot\alpha

    Similarly, (d+a)2+h(x+h)=x(d+a)cotβ(d + a)^2 + h(x + h) = x(d + a)\cot\beta

    As the points A,B,PA, B, P and QQ are concyclic OA.OB=OP.OQ\therefore OA.OB = OP.OQ

    d(d+2a)=h(h+x)d(d + 2a) = h(h + x)

    d2+d(d+2a)=xdcotαd+a=x2cotα\Rightarrow d^2 + d(d + 2a) = xd\cot\alpha \Rightarrow d + a = \frac{x}{2}\cot\alpha

    Similarly, (d+a)2+(d+a)2a2=x(d+a)cotβ(d + a)^2 + (d + a)^2 - a^2 = x(d + a)\cot\beta

    Solving the above two equations

    x24cot2α+x24cot2αa2=x.x2cotαcotβ\frac{x^2}{4}\cot^2\alpha + \frac{x^2}{4}\cot^2\alpha - a^2 = x.\frac{x}{2}\cot\alpha\cot\beta

    x22(cot2αcotαcotβ)=a2\Rightarrow \frac{x^2}{2}(\cot^2\alpha - \cot\alpha\cot\beta) = a^2

    x=asinα2sinβcosαsin(βα)x = a\sin\alpha\sqrt{\frac{2\sin\beta}{\cos\alpha\sin(\beta - \alpha)}}

  39. The diagram is given below:

    189th problem

    Let A1,A2,,A10,,A17A_1, A_2, \ldots, A_{10}, \ldots, A_{17} be the feet of the first, second, …, tenth, and seventeenth pillars respectively and hh be the height of each of these pillars. Given that these pillars are equidistant, therefore A1A2=A2A3==A16A17=xA_1A_2 = A_2A_3 = \cdots = A_{16}A_{17} = x (let).

    Clearly, A1A10=9xA_1A_{10} = 9x and A1A17=16xA_1A_{17} = 16x. We have let OO as the position of the observer and A2A1O=θ\angle A_2A_1O = \theta.

    In A10OP,tanα=hOA10OA10=hcotα\triangle A_{10}OP, \tan\alpha = \frac{h}{OA_{10}}\Rightarrow OA_{10} = h\cot\alpha