30. Height and Distance Solutions Part 3#

  1. The diagram is given below:

    101st problem

    Let ABAB and CDCD be the two temples and ACAC be the river. Let the height of temple ABAB be 5050 m. ACAC is the river. The angles are depression are shown as corresponding angles of elevation. Let the height of CDCD be xx m and width of river be ww m. Thus, CD=xCD = x and AC=wAC = w.

    In ABC,tan60=ABACw=503=DE[DEAC]\triangle ABC, \tan60^\circ = \frac{AB}{AC} \Rightarrow w = \frac{50}{\sqrt{3}} = DE[\because DE||AC]

    In BDE,tan30=BDDE13=BD50/3BD=503\triangle BDE, \tan30^\circ = \frac{BD}{DE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BD}{50/\sqrt{3}}\Rightarrow BD = \frac{50}{3}.

    Thus, height of the second temple CD=ABBD=1003CD = AB - BD = \frac{100}{3} m.

  2. This problem is similar to 9595 and has been left as an exercise.

  3. This problem is similar to 9595 and has been left as an exercise.

  4. The diagram is given below:

    104th problem

    Let BEBE be the tower leaning due east where BB is the foot of the tower and EE is the top. ABAB is the vertical height of the tower taken as hh. The angles of elevation are shown from tow points as given in the question.

    In ACE,tanα=hx+a\triangle ACE, \tan\alpha = \frac{h}{x + a} and in ADE,tanβ=hx+b\triangle ADE, \tan\beta = \frac{h}{x + b}.

    bah=1tanβ1tanα\Rightarrow \frac{b - a}{h} = \frac{1}{\tan\beta} - \frac{1}{\tan\alpha}

    h=(ba)tanαtanβtanαtanβ\Rightarrow h = \frac{(b - a)\tan\alpha\tan\beta}{\tan\alpha - \tan\beta}.

  5. The diagram is given below:

    105th problem

    Let ACAC be the lake and BB be the point of observation 25002500 m above lake. Let EE be the cloud and FF be its reflection in the lake. If we take height of the cloud above lake as hh then CD=2500CD = 2500 m where BDACBD||AC . DE=h2500DE = h - 2500 m and CD=2500CD = 2500 m. The angle of elevation and angle of depression of cloud and its reflection are shown as given in the problem.

    In BDF,tan45=DFBDBD=h+2500\triangle BDF, \tan45^\circ = \frac{DF}{BD} \Rightarrow BD = h + 2500

    In BDE,tan15=DEBDDE=1830.6\triangle BDE, \tan15^\circ = \frac{DE}{BD} \Rightarrow DE = 1830.6 m.

    CE=CD+DE=h=2500+1830.6=4330.6\Rightarrow CE = CD + DE = h = 2500 + 1830.6 = 4330.6 m.

  6. The diagram is given below:

    106th problem

    This is a problem similar to previous problem with 25002500 replaced by hh and angles are replaced by α\alpha and β\beta. So the diagram is similar in nature. Let the height of the cloud above lake be hh' m. So DE=hhDE = h' - h and DF=h+hDF = h + h'.

    In BDE,tanα=DEBDBD=(hh)/tanα\triangle BDE, \tan\alpha = \frac{DE}{BD}\Rightarrow BD = (h' - h)/\tan\alpha

    In BDF,tanβ=DFBDBD=(h+h)/tanβ\triangle BDF, \tan\beta = \frac{DF}{BD}\Rightarrow BD = (h' + h)/\tan\beta

    (hh)tanβ=(h+h)tanαh=h(tanα+tanβ)tanβtanα\Rightarrow (h' - h)\tan\beta = (h' + h)\tan\alpha \Rightarrow h' = \frac{h(\tan\alpha + \tan\beta)}{\tan\beta - \tan\alpha}

    BD=2h(tanαtanβ)\therefore BD = \frac{2h}{(\tan\alpha - \tan\beta)}

    Also, secα=BEBDBE=2hsecαtanαtanβ\sec\alpha = \frac{BE}{BD} \Rightarrow BE = \frac{2h\sec\alpha}{\tan\alpha - \tan\beta} which is the distance of the cloud from the point of observation.

  7. The diagram is given below:

    107th problem

    Let ABAB be the height of plane above horizontal ground as hh miles. CC and DD are two consecutive milestones so CD=1CD = 1 mile. Let BC=xBC = x mile. The angles of depression are represented as angles of elevation.

    In ABC,tanα=ABBCh=xtanα\triangle ABC, \tan\alpha = \frac{AB}{BC} \Rightarrow h = x\tan\alpha

    In ABD,tanβ=ABBDh=(x+1)tanβ\triangle ABD, \tan\beta = \frac{AB}{BD} \Rightarrow h = (x + 1)\tan\beta

    x=tanβtanαtanβh=tanαtanαtanβ\Rightarrow x = \frac{\tan\beta}{\tan\alpha - \tan\beta}\Rightarrow h = \frac{\tan\alpha}{\tan\alpha - \tan\beta}.

  8. The diagram is given below:

    108th problem

    Let PQPQ be the post with height hh and ABAB be the tower. Given that the angles of elevation of BB at PP and QQ are α\alpha and β\beta respectively. Draw CQPACQ||PA such that PQ=AC=hPQ = AC = h and A:math:AP = QC = x. Also, let BC=hBC = h' so that AB=AC+BC=h+hAB = AC + BC = h + h'.

    In ABP,tanα=h+hxx=h+htanα\triangle ABP, \tan\alpha = \frac{h + h'}{x}\Rightarrow x = \frac{h + h'}{\tan\alpha}

    In BCQ,tanβ=hxx=htanβ\triangle BCQ, \tan\beta = \frac{h'}{x}\Rightarrow x = \frac{h'}{\tan\beta}

    h+htanα=htanβ\Rightarrow \frac{h + h'}{\tan\alpha} = \frac{h'}{\tan\beta}

    h=htanβtanαtanβx=htanαtanβh' = \frac{h\tan\beta}{\tan\alpha - \tan\beta} \Rightarrow x = \frac{h}{\tan\alpha - \tan\beta}

    PQ=h+h=htanαtanαtanβ\therefore PQ = h + h' = \frac{h\tan\alpha}{\tan\alpha - \tan\beta}.

  9. The diagram is given below:

    109th problem

    Let ADAD be the wall, BDBD and CECE are two positions of the ladder. Then according to question BC=a,DE=bBC = a, DE = b and angles of elevations at BB and CC are α\alpha and β\beta. Let AB=xAB = x and AE=yAE = y. Also, let length of ladder be ll i.e BD=CE=lBD = CE = l.

    In ABD,sinα=ADBD=y+bl,cosα=ABBD=xl\triangle ABD, \sin\alpha = \frac{AD}{BD} = \frac{y + b}{l}, \cos\alpha = \frac{AB}{BD} = \frac{x}{l}

    In ACE,sinα=AECE=yl,cosβ=ACCE=a+xl\triangle ACE, \sin\alpha = \frac{AE}{CE} = \frac{y}{l}, \cos\beta = \frac{AC}{CE} = \frac{a + x}{l}.

    cosβcosα=al\Rightarrow \cos\beta - \cos\alpha = \frac{a}{l} and sinαsinβ=bl\sin\alpha - \sin\beta = \frac{b}{l}

    ab=cosαcosβsinβsinα\Rightarrow \frac{a}{b} = \frac{\cos\alpha - \cos\beta}{\sin\beta - \sin\alpha}.

  10. The diagram is given below:

    110th problem

    Let CDCD be the tower subtending angle α\alpha at AA. Let BB be the point bb m above AA from which angle of depression to foot of tower at CC is β\beta which is shown as angle of elevation. Let AC=xAC = x and CD=hCD = h.

    In ACD,tanα=hxx=hcotα\triangle ACD, \tan\alpha = \frac{h}{x} \Rightarrow x = h\cot\alpha

    In ABC,tanβ=bxx=bcotβ\triangle ABC, \tan\beta = \frac{b}{x} \Rightarrow x = b\cot\beta

    hcotα=bcotβh=btanαcotβ\Rightarrow h\cot\alpha = b\cot\beta \Rightarrow h = b\tan\alpha\cot\beta

  11. The diagram is given below:

    111th problem

    Let ABAB be the observer with a height of 1.51.5 m, 28.528.5 m i.e. ADAD from tower DE,30DE, 30 m high. Draw BCADBC||AD such that AB=CD=1.5AB = CD = 1.5 m and thus CE=28.5CE = 28.5 m. Let the angle of elevation from observer’s eye to the top of the tower be α\alpha.

    In BCE,tanα=CEBC=28.528.5=1α=45\triangle BCE, \tan\alpha = \frac{CE}{BC} = \frac{28.5}{28.5} = 1\Rightarrow \alpha = 45^\circ.

  12. The diagram is given below:

    112th problem

    Let ABAB be the tower havin a height of hh and CC and DD are two objects at a distance of xx and x+yx + y such that angles of depression shown as angles of elevatin are β\beta and α\alpha respectively.

    In ABC,tanβ=hxx=hcotβ\triangle ABC, \tan\beta = \frac{h}{x} \Rightarrow x = h\cot\beta

    In ABD,tanα=hx+yx+y=hcotα\triangle ABD, \tan\alpha = \frac{h}{x + y} \Rightarrow x + y = h\cot\alpha

    Distance between CC and D=y=h(cotαcotβ)D = y = h(\cot\alpha - \cot\beta).

  13. The diagram is given below:

    113th problem

    Let ABAB be the height of the window at a height hh and DEDE be the house opposite to it. Let the distance between the houses be AD=xAD = x. Draw BCADBC||AD such that BC=xBC = x and CD=hCD = h. The angles are shown as given in the problem. Let CE=yCE = y

    In BCD,tanβ=hxx=hcotβ\triangle BCD, \tan\beta = \frac{h}{x} \Rightarrow x = h\cot\beta

    In BCE,tanα=yxx=ycotα\triangle BCE, \tan\alpha = \frac{y}{x} \Rightarrow x = y\cot\alpha

    y=htanαcotβ\Rightarrow y = h\tan\alpha\cot\beta

    Total height of the second house DE=CD+DE=y+h=h(1+tanαcotβ)DE = CD + DE = y + h = h(1 + \tan\alpha\cot\beta)

  14. The diagram is given below:

    114th problem

    Let ADAD be the ground, BB be the lower window at a height of 22 m, CC be the upper window at a height of 44 m above lower window and GG be the balloon at a height of x+2+4x + 2 + 4 m above ground. Draw DGAC,BEADDG||AC, BE||AD and CFADCF||AD so that DE=2DE = 2 m, EF=4EF = 4 m and FG=xFG = x m. Also, let BE=CF=dBE = CF = d m. The angles of elevation are shown as given in the problem.

    In BEG,tan60 circ=3=EGBE=x+4dd=x+43\triangle BEG, \tan60^\ circ = \sqrt{3} = \frac{EG}{BE} = \frac{x + 4}{d}\Rightarrow d = \frac{x + 4}{\sqrt{3}}

    In CFG,tan30=13=FGCF=xdd=3x\triangle CFG, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{FG}{CF} = \frac{x}{d}\Rightarrow d = \sqrt{3}x

    3x=x+43x=2\Rightarrow \sqrt{3}x = \frac{x + 4}{\sqrt{3}}\Rightarrow x = 2

    \therefore the height of the balloon =2+4+2=8= 2 + 4 + 2 = 8 m.

  15. The diagram is given below:

    115th problem

    Let ABAB be the lamp post, EFEF and GHGH be the two positions of the man having height 66 ft. Let the shdows be ECEC and GDGD of lengths 2424 ft. and 3030 ft. for initial and final position. Since the man moves eastward from his initial position ACD=90\therefore \angle ACD = 90^\circ.

    Let AB=h,AE=xAB = h, AE = x and AG=yAG = y.

    From similar triangles CEFCEF and ABCABC

    h6=24+x24\frac{h}{6} = \frac{24 + x}{24}

    From similar triangles ABDABD and DGHDGH

    h6=30+y30\frac{h}{6} = \frac{30 + y}{30}

    Thus, 1+x24=1+y30y=5x41 + \frac{x}{24} = 1 + \frac{y}{30} \Rightarrow y = \frac{5x}{4}.

    From right angles ACD\triangle ACD,

    y2=x2+3002x=400y^2 = x^2 + 300^2 \Rightarrow x = 400 ft.

    h=106\Rightarrow h = 106 ft.

  16. The diagram is given below:

    116th problem

    Let ABAB be the tower having a height of hh m, ACAC be the final length of shadow taken as xx m, ADAD is the initial length of shadow which is 55 m more than finla length i.e. CD=5CD = 5 m. The angles of elevation are shown as given in the problem.

    In ABC,tan60=3=hxx=h3\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}

    In ABD,tan30=13=hx+52h3=5\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{x + 5} \Rightarrow \frac{2h}{\sqrt{3}} = 5

    h=532=4.33\Rightarrow h = \frac{5\sqrt{3}}{2} = 4.33 m.

  17. The diagram is given below:

    117th problem

    Let AA be the initial position of the man and DD and EE be the objects in the west. Let DE=x,AD=y,ADB=θ,AEB=ϕDE = x, AD = y, \angle ADB=\theta, \angle AEB = \phi and ADC=ψ\angle ADC=\psi. α\alpha and β\beta are the angles made by objects on the two positions of the man as given in the problem.

    tanθ=cy\Rightarrow \tan\theta = \frac{c}{y} and tanϕ=cx+y\tan\phi = \frac{c}{x + y}

    Now θϕ=αtan(θϕ)=tanα\theta - \phi = \alpha \Rightarrow \tan(\theta - \phi) = \tan\alpha

    tanθtanϕ1+tanθtanϕ=tanα\Rightarrow \frac{\tan\theta - \tan\phi}{1 + \tan\theta\tan\phi} = \tan\alpha

    cycx+y1+cycx+y=tanα\Rightarrow \frac{\frac{c}{y} - \frac{c}{x + y}}{1 + \frac{c}{y}\frac{c}{x + y}} = \tan\alpha

    cxcotα=xy+y2+c2\Rightarrow cx\cot\alpha = xy + y^2 + c^2

    Similarly, substituting 2c2c for xx and ψ\psi for ϕ\phi, we get

    2cxcotβ=xy+y2+4c2x=3c2cotβcotα2cx\cot\beta = xy + y^2 + 4c^2 \Rightarrow x= \frac{3c}{2\cot\beta - \cot\alpha}.

  18. The diagram is given below:

    118th problem

    Let PP be the object and OAOA be the straight line on which BB and CC lie underneath the object. Let OP=hOP = h. According to question the angles of elevation made are α,2α\alpha, 2\alpha and 3α3\alpha from A,BA, B and CC i.e. PCO=3α,PBO=2α\angle PCO = 3\alpha, \angle PBO = 2\alpha and PAO=α\angle PAO = \alpha. Given that AB=αAB = \alpha and BC=bBC = b.

    APB=2αα=α\angle APB = 2\alpha - \alpha = \alpha and BPC=3α2α=α\angle BPC = 3\alpha - 2\alpha = \alpha

    AB=BP=a\therefore AB = BP = a

    In PBC,BCsinα=PBsin(1803α)bsinα=a sin3α\triangle PBC, \frac{BC}{\sin\alpha} = \frac{PB}{\sin(180^\circ - 3\alpha)} \Rightarrow \frac{b}{\sin\alpha} = \frac{a}{\ sin3\alpha}

    ab=sin3αsinα=34sin2αsinα=3ba4b\Rightarrow \frac{a}{b} = \frac{\sin3\alpha}{\sin\alpha} = 3 - 4\sin^2\alpha \Rightarrow \sin\alpha = \sqrt{\frac{3b - a}{4b}}

    In OPB,OP=BPsin2α=2asinαcosα=a2b(a+b)(3ba)\triangle OPB, OP = BP\sin2\alpha = 2a\sin\alpha\cos\alpha = \frac{a}{2b}\sqrt{(a + b)(3b - a)}

  19. This problem is similar to 9292 and has been left as an exercise.

  20. The diagram is given below:

    120th problem

    Let θ\theta be the angle of inclination of the inclines plane ACAC. Let AB=cAB = c and BC=cBC = c. Let the object be at DD. Now DBA=θα\angle DBA = \theta - \alpha and DCA=θβ\angle DCA = \theta - \beta.

    Using sine rule in DAB,csinα=ADsin(θα)\triangle DAB, \frac{c}{\sin\alpha} = \frac{AD}{\sin(\theta - \alpha)}

    AD=csin(θα)sinα\Rightarrow AD = \frac{c\sin(\theta - \alpha)}{\sin\alpha}

    Applying sine rule in DAC,2csinβ=ADsin(θβ)\triangle DAC, \frac{2c}{\sin\beta} = \frac{AD}{\sin(\theta - \beta)}

    AD=2sin(θβ)sinβ\Rightarrow AD = \frac{2\sin(\theta - \beta)}{\sin\beta}

    csin(θα)sinα=2csin(ββ)sinβ\Rightarrow \frac{c\sin(\theta - \alpha)}{\sin\alpha} = \frac{2c\sin(\beta - \beta)}{\sin\beta}

    sinθcosαcosθsinαsinθsinα=2[sinθcosβcosθsinβ]sinθsinβ\Rightarrow \frac{\sin\theta\cos\alpha - \cos\theta\sin\alpha}{\sin\theta\sin\alpha} = \frac{2[\sin\theta\cos\beta - \cos\theta\sin\beta]}{\sin\theta\sin\beta}

    cotαcotθ=2(cotβcotβ)\Rightarrow \cot\alpha - \cot\theta = 2(\cot\beta - \cot\beta)

    cotθ=2cotβcotα\Rightarrow \cot\theta = 2\cot\beta - \cot\alpha.

  21. The question is same as 109109 just that we have a different relation to prove. From 109109 we have

    ab=cosβcosαsinαsinβ\frac{a}{b} = \frac{\cos\beta - \cos\alpha}{\sin\alpha - \sin\beta}

    =2sinα+β2sinαβ22cosα+β2sinαβ2= \frac{2\sin\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}{2\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}

    =tanα+β2a=btanα+β2= \tan\frac{\alpha + \beta}{2}\Rightarrow a = b\tan\frac{\alpha + \beta}{2}.

  22. The diagram is given below:

    122nd problem

    Given that AA and BB are two points of observation on ground 10001000 m apart. Let CC be the point where the balloon will hit the ground at a distance xx m from BB. Also, let DD and EE be the points above AA and BB respectively such that BAE=30\angle BAE= 30^\circ and DBA=60\angle DBA = 60^\circ.

    In ABD,tan60=3=ADABAD=10003\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{AD}{AB}\Rightarrow AD = 1000\sqrt{3} m.

    In ABE,tan30=13=BEABBE=10003\triangle ABE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{AB}\Rightarrow BE = \frac{1000}{\sqrt{3}} m.

    Clearly, BCE\triangle BCE and ACDACD are similar. Therefore,

    BCAC=BEADxx+1000=100010003.3\frac{BC}{AC} = \frac{BE}{AD} \Rightarrow \frac{x}{x + 1000} = \frac{1000}{1000\sqrt{3}.\sqrt{3}}

    x=500AC=1500\Rightarrow x = 500 \Rightarrow AC = 1500 m.

  23. The diagram is given below:

    123rd problem

    Let ABAB be tree having height hh m and BCBC be the width of the river having width ww m. According to question angle of elevation of the tree from the opposite bank is 6060^\circ. Also, let DD be the point when the man retires 4040 m from where the angle of elevation of the tree is 3030^\circ.

    In ABC,tan60=3=hwh=w3\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{h}{w} \Rightarrow h = w\sqrt{3} m.

    In ABD,tan30=hw+403w=w+40w=20\triangle ABD, \tan30^\circ = \frac{h}{w + 40}\Rightarrow 3w = w + 40 \Rightarrow w = 20 m.

    h=203\Rightarrow h = 20\sqrt{3} m.

    Thus, width of the river is 2020 m and height of the tree is 20320\sqrt{3} m.

  24. The diagram is given below:

    124th problem

    Let OO be the point of observation. The bird is flying in the horizontal line WXYZWXYZ. The angles of elevation of the bird is given at equal intervals of time. Since the speed of the bird is constant WX=XY=YZ=yWX = XY = YZ = y (let). From question AOW=α,BOX=β,COY=γ\angle AOW = \alpha, \angle BOX = \beta, \angle COY = \gamma and DOZ=δ\angle DOZ = \delta. Let OA=xOA = x and AW=hAW = h.

    In AOW,cotα=xh\triangle AOW, \cot\alpha = \frac{x}{h}

    In BOX,cotβ=x+yh\triangle BOX, \cot\beta = \frac{x + y}{h}

    In COY,cotγ=x+2yh\triangle COY, \cot\gamma = \frac{x + 2y}{h}

    In DOZ,cotδ=x+3yh\triangle DOZ, \cot\delta = \frac{x + 3y}{h}

    L.H.S. =cot2αcot2δ=6xy9y2h2= \cot^2\alpha - \cot^2\delta = \frac{-6xy - 9y^2}{h^2}

    R.H.S. =cot2βcot2γ=6xy9y2h2= \cot^2\beta - \cot^2\gamma = \frac{-6xy - 9y^2}{h^2}

    \therefore L.H.S. = R.H.S.

  25. The diagram is given below:

    125th problem

    Let ABAB be the tower, BCBC be the pole and DD be the point of observation where the tower and the pole make angles α\alpha and β\beta respectively. Let the height of the tower be hh' and AD=dAD = d. Given that the height of the pole is hh.

    In ABD,cotα=ADAB=dhd=hcotα\triangle ABD, \cot\alpha = \frac{AD}{AB} = \frac{d}{h'} \Rightarrow d = h'\cot\alpha

    In ACD,tan(α+β)=ACAB=h+hd\triangle ACD, \tan(\alpha + \beta) = \frac{AC}{AB} = \frac{h + h'}{d}

    h+h=hcotαtan(α+β)\Rightarrow h + h' = h'\cot\alpha\tan(\alpha + \beta)

    h=hcotαtan(α+β)1=hsinαcos(α+β)sin(α+β)cosαcos(α+β)sinαh' = \frac{h}{\cot\alpha\tan(\alpha + \beta) - 1} = \frac{h\sin\alpha\cos(\alpha + \beta)}{\sin(\alpha + \beta)\cos\alpha - \cos(\alpha + \beta)\sin\alpha}

    =hsinαcos(α+beta)sin(α+βα)=hsinαcosecβcos(α+β)= \frac{h\sin\alpha\cos(\alpha + beta)}{\sin(\alpha + \beta - \alpha)} = h\sin\alpha\cosec\beta\cos(\alpha + \beta)

  26. The diagram is given below:

    126th problem

    Given AC=BC=xAC = BC = x (let) and BPC=β\angle BPC = \beta.

    Let BPA=θ\angle BPA = \theta then CPA=θβ\angle CPA = \theta - \beta

    In APC,tan(θβ)=xAP\triangle APC, \tan(\theta - \beta) = \frac{x}{AP}

    In APB,tanθ=2xAP\triangle APB, \tan\theta = \frac{2x}{AP}

    tanθ=2tan(θβ)=2(tanθtanβ)1+tanθtanβ\Rightarrow \tan\theta = 2\tan(\theta - \beta) = \frac{2(\tan\theta - \tan\beta)}{1 + \tan\theta\tan\beta}

    tanθ=ABAP=1n\tan\theta = \frac{AB}{AP} = \frac{1}{n} (from question)

    1n=2(1ntanβ)1+tanβn\Rightarrow \frac{1}{n} = \frac{2\left(\frac{1}{n} - \tan\beta\right)}{1 + \frac{\tan\beta}{n}}

    tanβ=n2n2+1\Rightarrow \tan\beta = \frac{n}{2n^2 + 1}.

  27. The diagram is given below:

    127th problem

    Let ABAB be the first chimney and CDCD be the second chimney. The angles of elevation are shown as angles of elevation as given in the problem. Draw BEACBE||AC and let AC=BE=dAC = BE = d m and AB=CE=hAB = CE = h m. Given CD=150CD = 150 m. Clearly, DE=150hDE = 150 - h m.

    In BED,tanθ=43=150hd4d=4503h\triangle BED, \tan\theta = \frac{4}{3} = \frac{150 - h}{d} \Rightarrow 4d = 450 - 3h

    In ACD,tanϕ=52=150dd=60\triangle ACD, \tan\phi = \frac{5}{2} = \frac{150}{d} \Rightarrow d = 60 m.

    h=70\Rightarrow h = 70 m. BE=60\Rightarrow BE = 60 m and ED=15070=80ED = 150 - 70 = 80 m.

    BD2=BE2+DE2=802+602BD=100BD^2 = BE^2 + DE^2 = 80^2 + 60^2 \Rightarrow BD = 100 m, which, is the distance between the tops of two chimneys.

  28. The diagram is given below:

    128th problem

    Let CDCD be the tower of height hh having an elevation of 3030^\circ from AA which is southward of it. Let BB be eastward of AA at a distance of aa from it from where the angle of elevation is 1818^\circ. Since BB is eastward of ACAB=90A \angle CAB = 90^\circ.

    In ACD,tan30=hACAC=h3\triangle ACD, \tan 30^\circ = \frac{h}{AC} \Rightarrow AC = h\sqrt{3}

    In BCD,,tan18=hBCBC=hcot18\triangle BCD,, \tan18^\circ = \frac{h}{BC} \Rightarrow BC = h\cot18^\circ

    In ABC,BC2=a2+AC2h2cot218=a2+3h2\triangle ABC, BC^2 = a^2 + AC^2 \Rightarrow h^2\cot^218^\circ = a^2 + 3h^2

    h=acot2183\therefore h = \frac{a}{\sqrt{\cot^218^\circ - 3}}

    Now cot218=5+25h=a2+25\cot^218^\circ = 5 + 2\sqrt{5} \therefore h = \frac{a}{\sqrt{2 + 2\sqrt{5}}}.

  29. The diagram is given below:

    129th problem

    Let ABAB be the tower having height hh. Given that PP is north of the tower and QQ is due west of PAPQ=90P\therefore \angle APQ= 90^\circ.

    In ABP,tanθ=hAPAP=hcotθ\triangle ABP, \tan\theta = \frac{h}{AP} \Rightarrow AP = h\cot\theta

    In ABQ,tanϕ=hAQAQ=hcotϕ\triangle ABQ, \tan\phi = \frac{h}{AQ} \Rightarrow AQ = h\cot\phi

    In APQ,AQ2=AP2+PQ2\triangle APQ, AQ^2 = AP^2 + PQ^2

    PQ2=h2[cot2ϕcot2θ]\Rightarrow PQ^2 = h^2[\cot^2\phi - \cot^2\theta]

    h=PQcot2ϕcot2θ\Rightarrow h = \frac{PQ}{\sqrt{\cot^2\phi - \cot^2\theta}}

    =PQsinθsinϕsin2θcos2ϕsin2ϕcos2θ= \frac{PQ\sin\theta\sin\phi}{\sqrt{\sin^2\theta\cos^2\phi - \sin^2\phi\cos^2\theta}}

    =PQsinθsinϕ(sin2θ(1sin2ϕ))sin2ϕ(1sin2θ)= \frac{PQ\sin\theta\sin\phi}{\sqrt{(\sin^2\theta(1 - \sin^2\phi)) - \sin^2\phi(1 - \sin^2\theta)}}

    =PQsinθsinϕsin2θsin2ϕ= \frac{PQ\sin\theta\sin\phi}{\sqrt{\sin^2\theta - \sin^2\phi}}.

  30. The diagram is given below:

    130th problem

    Let BB be the peak having a height of hh with base AA. Let PQPQ is the horizontal base having a length 2a2a making angle of elevation of θ\theta from each end. Let RR be the mid-point of PQPQ from where the angle of elevation of BB is ϕ\phi as given in the question.

    Thus, APB=AQB=θ\angle APB = \angle AQB = \theta and ARB=ϕ\angle ARB = \phi.

    In APB,tanθ=hAPAP=hcotθ\triangle APB, \tan\theta = \frac{h}{AP} \Rightarrow AP = h\cot\theta

    Similarly, AQ=hcotθAQ = h\cot\theta and AR=hcotϕAR = h\cot\phi

    AR\because AR is the median of the APQ\triangle APQ

    AP2+AQ2=2PR2+2AR2\therefore AP^2 + AQ^2 = 2PR^2 + 2AR^2

    2h2cot2θ=2a2+2h2cot2ϕ\Rightarrow 2h^2\cot^2\theta = 2a^2 + 2h^2\cot^2\phi

    h(cot2θcot2ϕ)=a2\Rightarrow h^(\cot^2\theta - \cot^2\phi) = a^2

    h=acos2θsin2θcos2ϕsin2ϕ\Rightarrow h = \frac{a}{\sqrt{\frac{\cos^2\theta}{\sin^2\theta} - \frac{\cos^2\phi}{\sin^2\phi}}}

    =asinθsinϕ(sinϕcosθ+cosϕsinθ)(sinϕcosθcosϕsinθ)= \frac{a\sin\theta\sin\phi}{\sqrt{(\sin\phi\cos\theta + \cos\phi\sin\theta)(\sin\phi\cos\theta - \cos\phi\sin\theta)}}

    =asinθsinϕsin(θ+ϕ)sin(ϕθ)= \frac{a\sin\theta\sin\phi}{\sqrt{\sin(\theta + \phi)\sin(\phi - \theta)}}

  31. The diagram is given below:

    131st problem

    Let BB be the top of the hill such that height of the hill ABAB is hh and P,R,QP, R, Q be the three consecutive milestones. Given, APB=α,ARB=β,AQB=γ\angle APB = \alpha, \angle ARB = \beta, \angle AQB = \gamma.

    In APB,tanα=hAPAP=hcotα\triangle APB, \tan\alpha = \frac{h}{AP}\Rightarrow AP = h\cot\alpha

    Similalrly, AR=hcotβAR = h\cot\beta and AQ=hcotγAQ = h\cot\gamma

    Also, PR=QR=1PR = QR = 1 mile.

    PR=QR,AR\because PR = QR, AR is the median of the triangle APRAPR.

    AP2+AR2=2PR2+2AQ2h2(cot2α+cot2γ)=2+2h2cot2β\therefore AP^2 + AR^2 = 2PR^2 + 2AQ^2 \Rightarrow h^2(\cot^2\alpha + \cot^2\gamma) = 2 + 2h^2\cot^2\beta

    h=2cot2α+cot2γ2cot2β\Rightarrow h = \sqrt{\frac{2}{\cot^2\alpha + \cot^2\gamma - 2\cot^2\beta}} miles.

  32. The diagram is given below:

    132nd problem

    Let OPOP be the tower haing a height of hh which is to be found. Let ABCABC be the equilateral triangle. Given that OPOP subtends angles of α,β,γ\alpha, \beta, \gamma at A,B,CA, B, C respectively. Given that tanα=3+1\tan\alpha = \sqrt{3} + 1 and tanβ=tanγ=2\tan\beta = \tan\gamma = \sqrt{2}. It is given that OPOP is perpendicular to the plane of ABC\triangle ABC.

    In AOP,tanα=hOA3+1=hOAOA=h3+1\triangle AOP, \tan\alpha = \frac{h}{OA} \Rightarrow \sqrt{3} + 1 = \frac{h}{OA} \Rightarrow OA = \frac{h}{\sqrt{3} + 1}

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

    In AOB\triangle AOB and AOCAOC, AB=AC,OB=OC,OAAB = AC, OB = OC, OA is common. AOB\therefore \triangle AOB and AOC\triangle AOC are equal.

    OAB=OAC\therefore \angle OAB = \angle OAC but BAC=60\angle BAC = 60^\circ

    OAB=OAC=30\Rightarrow \angle OAB = \angle OAC = 30^\circ

    Using sine rule in the OAB,OBsin30=OAsinθ\triangle OAB, \frac{OB}{\sin30^\circ} = \frac{OA}{\sin\theta} (let ABO=θ\angle ABO = \theta)

    h212=h3+1sinθ\Rightarrow \frac{\frac{h}{\sqrt{2}}}{\frac{1}{2}} = \frac{\frac{h}{\sqrt{3} + 1}}{\sin\theta}

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

    θ=15\Rightarrow \theta = 15^\circ

    OBD=ABCθ=45\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 OBD,cos45=BDOB12=40h2h=80\triangle OBD, \cos45^\circ = \frac{BD}{OB} \Rightarrow \frac{1}{\sqrt{2}} = \frac{40}{h\sqrt{2}}\Rightarrow h = 80'

  33. The diagram is given below:

    133rd problem

    In the diagram we have shown only one tower instead of three. We will apply cyclic formula to this one tower relationships. Let PP be the position of the eye and height of PQ=xPQ = x. Let ABAB be the tower having a height of aa as given in the question and let the angle subtended by ABAB at PP is θ\theta.

    Thus, APB=θ,PAQ=αABP=180θ(90α)=90+(αθ)\angle APB = \theta, \angle PAQ = \alpha \Rightarrow \angle ABP = 180^\circ - \theta - (90^\circ - \alpha) = 90^\circ + (\alpha - \theta)

    By sine rule in APB\triangle APB,

    asinθ=APsin[90+(αθ)]=APcos(αθ)\frac{a}{\sin\theta} = \frac{AP}{\sin[90^\circ + (\alpha - \theta)]} = \frac{AP}{\cos(\alpha - \theta)}

    In APQ,sinα=xAPAP=xsinα\triangle APQ, \sin\alpha = \frac{x}{AP} \Rightarrow AP = \frac{x}{\sin\alpha}

    asinθ=xsinαcos(αθ)\Rightarrow \frac{a}{\sin\theta} = \frac{x}{\sin\alpha\cos(\alpha - \theta)}

    xsinθ=asinαcos(αθ)cos(αθ)=xsinθasinα\Rightarrow x\sin\theta = a\sin\alpha\cos(\alpha - \theta)\Rightarrow \cos(\alpha - \theta) = \frac{x\sin\theta}{a\sin\alpha}

    If we consider the other two towers we will have similar relations i.e.

    cos(βθ)=xsinθbsinβ\Rightarrow \cos(\beta - \theta) = \frac{x\sin\theta}{b\sin\beta} and cos(γθ)=xsinθcsinγ\cos(\gamma - \theta) = \frac{x\sin\theta}{c\sin\gamma}

    Now, sin(βγ)asinα+sin(γα)bsinβ+sin(αβ)csinγ\frac{\sin(\beta - \gamma)}{a\sin\alpha} + \frac{\sin(\gamma - \alpha)}{b\sin\beta} + \frac{\sin(\alpha - \beta)}{c\sin\gamma}

    =sin(αβ)csinγ=sin(αθ+θβ)csinγ=\displaystyle\sum\frac{\sin(\alpha - \beta)}{c\sin\gamma} = \displaystyle\sum\frac{\sin(\alpha - \theta + \theta - \beta)}{c\sin\gamma}

    =sin(αθ)cos(θβ)+cos(αθ)sin(θβ)csinγ= \displaystyle\sum\frac{\sin(\alpha - \theta)\cos(\theta - \beta) + \cos(\alpha - \theta)\sin(\theta - \beta)}{c\sin\gamma}

    =1csinγ[sin(αθ)xsinθbsinβ+sin(θbeta)xsinθasinα]=\displaystyle\sum\frac{1}{c\sin\gamma}\left[\frac{\sin(\alpha - \theta)x\sin\theta}{b\sin\beta} + \frac{\sin(\theta - beta)x\sin\theta}{a\sin\alpha}\right]

    =xsinθabcsinαsinβsinγ[asin(αθ)sinαbsin(βθ)sinβ]= \displaystyle\sum\frac{x\sin\theta}{abc\sin\alpha\sin\beta\sin\gamma}\left[a\sin(\alpha - \theta)\sin\alpha - b\sin(\beta - \theta)sin\beta\right]

    =xsinθabcsinαsinβsinγ.0=0= \frac{x\sin\theta}{abc\sin\alpha\sin\beta\sin\gamma}.0 = 0

  34. The diagram is given below:

    134th problem

    Let SS be the initial position of the man and PP and QQ be the poosition of the objects. Since PQPQ subtends greatest angle at RR, a circle will pass through P,QP, Q and RR and RSRS will be a tangent to this circle at RR.

    Also, PQR=PRS=θ\angle PQR = \angle PRS = \theta (let). Let PQ=xPQ = x.

    Clearly SRQ=θ+β\angle SRQ = \theta + \beta

    Using sine law in PRQ,xsinβ=PRsinθx=PRsinβsinθ\triangle PRQ, \frac{x}{\sin\beta} = \frac{PR}{\sin\theta} \Rightarrow x = \frac{PR\sin\beta}{\sin\theta}

    Using sine law in PRS,PRsinα=csin(θ+β)PR=csinαsin(θ+β)\triangle PRS, \frac{PR}{\sin\alpha} = \frac{c}{\sin(\theta + \beta)} \Rightarrow PR = \frac{c\sin\alpha}{\sin(\theta + \beta)}

    x=csinαsinβsin(θ+β)sin(θ)=2csinαsinβ2sin(θ+β)sin(θ)\Rightarrow x = \frac{c\sin\alpha\sin\beta}{\sin(\theta + \beta)\sin(\theta)} = \frac{2c\sin\alpha\sin\beta}{2\sin(\theta + \beta)\sin(\theta)}

    =2csinαsinβcosβcos(2θ+β)= \frac{2c\sin\alpha\sin\beta}{\cos\beta - \cos(2\theta + \beta)}

    In QRS,α+β+2θ=1802θ+β=180α\triangle QRS, \alpha + \beta + 2\theta = 180^\circ \Rightarrow 2\theta + \beta = 180^\circ - \alpha

    cos(2θ+β)=cosα\Rightarrow \cos(2\theta + \beta) = -\cos\alpha

    x=2csinαsinβcosα+cosβ\Rightarrow x = \frac{2c\sin\alpha\sin\beta}{\cos\alpha + \cos\beta}.

  35. The diagram is given below:

    135th problem

    Let OPOP be the tower having a height of hh and PQPQ be the flag-staff having a height of xx. AA and BB are the two points on the horizontal line OAOA. Let OB=yOB = y. Given, AB=d,QAP=QBP=αAB = d, \angle QAP = \angle QBP = \alpha.

    Since QAP=QBP\angle QAP = \angle QBP, a circle will pass through the points A,B,PA, B, P and QQ because angles in the same segment of a circle are equal.

    Thus, BAP=BQP=β\angle BAP = \angle BQP = \beta (angles on the same segment BPBP)

    BPO=QAO=α+β\Rightarrow \angle BPO = \angle QAO = \alpha + \beta

    In AOP,tanβ=hy+d\triangle AOP, \tan\beta = \frac{h}{y + d}

    In BOP,tan(α+β)=yhy=htan(α+β)\triangle BOP, \tan(\alpha + \beta) = \frac{y}{h} \Rightarrow y = h\tan(\alpha + \beta)

    h=ytanβ+dtanβh=dtanβ1tan(α+β)tanβ\Rightarrow h = y\tan\beta + d\tan\beta\Rightarrow h = \frac{d\tan\beta}{1 - \tan(\alpha + \beta)\tan\beta}

    In BOQ,tanβ=yx+hxtanβ+htanβ=htan(α+β)\triangle BOQ, \tan\beta = \frac{y}{x + h}\Rightarrow x\tan\beta + h\tan\beta = h\tan(\alpha + \beta)

    x=d[tan(α+β)tanβ]1tan(α+β)+tanβ\Rightarrow x = \frac{d[\tan(\alpha + \beta) - \tan\beta]}{1 - \tan(\alpha + \beta) + \tan\beta}

  36. This question is same as 9292 with α\alpha replaced by θ\theta and β\beta replaced by ϕ\phi.

    Referring to diagram of 92,AC=h(tanθ+tanϕ)tanϕtanθ92, AC = \frac{h(\tan\theta + \tan\phi)}{\tan\phi - \tan\theta}

    =h(sinθcosϕ+sinϕcosθ)sinϕcosθsinθcosϕ= \frac{h(\sin\theta\cos\phi + \sin\phi\cos\theta)}{\sin\phi\cos\theta - \sin\theta\cos\phi}

    =hsin(θ+ϕ)sin(ϕθ)= \frac{h\sin(\theta + \phi)}{\sin(\phi - \theta)}.

  37. The diagram is given below:

    137th problem

    Let BCBC represent the road inclined at 1010^\circ to the vertical towards sun and AB=2.05AB = 2.05 m represents the shadow where the elevation of the sun is BAC=38\angle BAC = 38^\circ. Thus, BCA=180(10+90+38)=42\angle BCA = 180^\circ - (10^\circ + 90^\circ + 38^\circ) = 42^\circ.

    Using sine rule in ABC\triangle ABC,

    BCsin38=ABsin426BC=2.05sin38sin42\frac{BC}{\sin38^\circ} = \frac{AB}{\sin426\circ}\Rightarrow BC = \frac{2.05\sin38^\circ}{\sin42^\circ}.

  38. The diagram is given below:

    138th problem

    Let CDCD be the tower having a height of hh m. Let BCBC be its shadow when altitude of the sun is 6060^\circ and ACAC be its shadow when altitude of the sun is 3030^\circ.

    Given that shadow decreases by 3030 m when altitude changes from 3030^\circ to 6060^\circ i.e. AB=30AB = 30 m. Let BC=xBC = x m.

    In BCD,tan60=hxh=3x\triangle BCD, \tan60^\circ = \frac{h}{x} \Rightarrow h = \sqrt{3}x

    In ACD,tan30=hx+30h=153\triangle ACD, \tan30^\circ = \frac{h}{x + 30} \Rightarrow h = 15\sqrt{3} m.

  39. This problem is similar to 138138 and has been left as an exercise.

  40. This problem is similar to 9696 and has been left as an exercise. The answer is 9090 seconds.

  41. The diagram is given below:

    141st problem

    Let CC be the position of the aeroplane flying 30003000 m above ground and DD be the aeroplane below it. Given that the angles of elevation of these aeroplanes are 4545^\circ and 6060^\circ respectively. Let the height of DD is hh m and AB=dAB = d m.

    In ABC,tan60=3=3000dd=10003\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{3000}{d} \Rightarrow d = 1000\sqrt{3} m.

    In ABD,tan45=1=hdh=10003\triangle ABD, \tan45^\circ = 1 = \frac{h}{d} \Rightarrow h = 1000\sqrt{3} m.

    \therefore Distance between heights of the aeroplanes =CD=300010003=1268= CD = 3000 - 1000\sqrt{3} = 1268 m.

  42. The diagram is given below:

    142nd problem

    Let CC and DD be two consecutive milestones so that CD=1CD = 1 mile. Let DD be position of aeroplane having a height hh above AA, to which angles of elevation are α\alpha and β\beta from CC and DD respectively. Let AC=xAD=1xAC = x \Rightarrow AD = 1 - x.

    In ABC,tanα=hxh=xtanα\triangle ABC, \tan\alpha = \frac{h}{x} \Rightarrow h = x\tan\alpha

    In ABD,tanβ=h1xh=tanαtanβtanα+tanβ\triangle ABD, \tan\beta = \frac{h}{1 - x} \Rightarrow h = \frac{\tan\alpha\tan\beta}{\tan\alpha + \tan\beta}

  43. This problem is similar to 119119 and has been left as an exercise.

  44. The diagram is given below:

    144th problem

    Using m:nm:n theorem from section 16.9,

    2ccot(θ30)=ccot15ccot302c\cot(\theta - 30^\circ) = c\cot15^\circ - c\cot30^\circ

    cot(θ30)=1=cot45\Rightarrow \cot(\theta - 30^\circ) = 1 = \cot45^\circ

    θ=75\Rightarrow \theta = 75^\circ.

  45. This problem is simmilar to 138138, and has been left as an exercise.

  46. The diagram is given below:

    146th problem

    Let ABAB be the height of air-pilot which has height of hh. Let CDCD be the tower whose angles of depression of top and bottom of tower be 3030^\circ and 6060^\circ respectively. Draw DEACDE||AC such that DE=ACDE = AC. Let the height of tower CDCD be xx.

    In ABC,tan60=3=hxh=x3\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}

    In ADE,tan30=13=BExBE=x3=h3\triangle ADE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{x}\Rightarrow BE = \frac{x}{\sqrt{3}} = \frac{h}{3}

    \therefore Height of the tower CD=hh3=2h3CD = h - \frac{h}{3} = \frac{2h}{3}