32. Height and Distance Solutions Part 5#

  1. The diagram is given below:

    201st problem

    Let BCBC be the tower having a height of hh. According to question AB=hAB = h and BD=h/2BD = h/2.

    In BCD,tanα=hh2=2\triangle BCD, \tan\alpha = \frac{h}{\frac{h}{2}} = 2

    In ABC,tanβ=hhα=45\triangle ABC, \tan\beta = \frac{h}{h} \Rightarrow \alpha = 45^\circ

    Ltanα=10+log2=10.30103L\tan\alpha = 10 + \log 2 = 10.30103 (given)

    LtanαLtan6326=10.3010310.30094L\tan\alpha - L\tan63^\circ26'= 10.30103 - 10.30094

    If the difference is x2x^2, then x2=60×3063152=5.812x^2 = \frac{60\times306}{3152} = 5.81^2

    α=63266\therefore \alpha = 63^\circ26'6''

    Change in sun’s altitude =6326645=18266= 63^\circ26'6'' - 45^\circ = 18^\circ26'6''

  2. This problem is similar to 117117, and has been left as an exercise.

  3. Since the angle of elevation is same from every point it is clear that the balloon is above the circumcenter. From sine rule, sinβ=b2RR=b2sinβ\sin\beta = \frac{b}{2R} \Rightarrow R = \frac{b}{2\sin\beta}.

    If hh is the height of the balloon then tanα=hRh=Rtanα\tan\alpha = \frac{h}{R} \Rightarrow h = R\tan\alpha

    =b2tanα.cosecβ= \frac{b}{2}\tan\alpha.\cosec\beta

  4. The diagram is given below:

    204th problem

    In the diagram AA representes the lighthouse. ABAB is north-east direction and ACAC is north-west direction between which the light is thrown by the lighthouse. Let BCBC is the westward path of the steamer. Given that the distance of steamer is 55 k m from lighthouse i.e. AB=5AB = 5 km and clearly ABC\triangle ABC is a isosceles, right angled triangle so AC=5AC = 5 km and BC=AB2+AC2=52+52=52BC = \sqrt{AB^2 + AC^2} = \sqrt{5^2 + 5^2} = 5\sqrt{2} km.

    Speed of steamer =52302=16= \frac{5\sqrt{2}}{30\sqrt{2}} = \frac{1}{6} km/min.

  5. The diagram is given below:

    205th problem

    Let AA be the initial position of the man and DD be the initial position of the balloon with angle of elevation equal to 6060^\circ. The man moves northward for 400400 yards and balloon moves north-west and they meet at CC where the balloon is vertically above the man. Since BCBC is north-west the ABC\triangle ABC will be isosceles, right angled triangle with AB=AC=400AB = AC = 400 yards.

    \therefore in ABC,tan60=h400h=4003\triangle ABC, \tan60^\circ = \frac{h}{400} \Rightarrow h = 400\sqrt{3} yards.

  6. The diagram is given below:

    206th problem

    Let ABCDABCD be the vertical cross-section of the tower through the middle, let the side of the square is ABAB having length aa and height of the tower be OPOP equal to hh. Let the height of flag-staff PQPQ be bb. MM and NN are points of observation such that AN=MN=100AN = MN = 100 m. Let α\alpha and β\beta be angles of elevation from MM of DD and QQ such that tanα=59\tan\alpha = \frac{5}{9} and tanβ=12\tan\beta = \frac{1}{2}. At NN the man just sees the flag.

    tanβ=12=ADAM=AD100AD=100=OP=h\tan\beta = \frac{1}{2} = \frac{AD}{AM} = \frac{AD}{100} \Rightarrow AD = 100 = OP = h.

    AD=AN=100AND=NDA=45PD=PQa2=ba=2b\therefore AD = AN = 100 \Rightarrow \angle AND = \angle NDA = 45^\circ \Rightarrow PD = PQ \Rightarrow \frac{a}{2} = b \Rightarrow a = 2b

    tanα=59=OQOM=b+h200+a2=b+h200+b\tan\alpha = \frac{5}{9} = \frac{OQ}{OM} = \frac{b + h}{200 + \frac{a}{2}} = \frac{b + h}{200 + b}

    b=25a=50\Rightarrow b = 25 \Rightarrow a = 50.

  7. The diagram is given below:

    207th problem

    Let OCOC be the vertical pole having a height of hh. AA and BB are given points in the question from where anglea of elevations of CC are α\alpha and β\beta respectively. Angle subtended by ABAB at OO is γ\gamma as shown in the diagram. Let OB=xOB = x and OA=yOA = y. Given that AB=dAB = d

    In OAC,tanα=hyy=hcotα\triangle OAC, \tan\alpha = \frac{h}{y}\Rightarrow y = h\cot\alpha

    Similarly x=hcotβx = h\cot\beta

    In OAB,d2=x2+y22xycosγ\triangle OAB, d^2 = x^2 + y^2 - 2xy\cos\gamma

    d2=h2cot2α+h2cot2β2h2cotαcotβcosγd^2 = h^2\cot^2\alpha + h^2\cot^2\beta - 2h^2\cot\alpha\cot\beta\cos\gamma

    h=dcot2α+cot2β2cotαcotβcosγ\Rightarrow h = \frac{d}{\sqrt{\cot^2\alpha + \cot^2\beta - 2\cot\alpha\cot\beta\cos\gamma}}

  8. The diagram is given below:

    208th problem

    Let OPOP be the tree having a height of hh and OABOAB is the hill inclines at angle α\alpha with the horizontal. Let AA be the point from where angle of elevation of the top of the tree be β\beta and BB be the point from where the angle of depression of the top of the tee be γ\gamma. Given AB=mAB = m.

    POA=90α,OAP=α+β,=αγ\angle POA = 90^\circ - \alpha, \angle OAP = \alpha + \beta, \angle = \alpha - \gamma and ABP=(α+β)(αγ)=β+γ\angle ABP = (\alpha + \beta) - (\alpha - \gamma) = \beta + \gamma

    In OAP,APsin(90α)=OPsin(α+β)\triangle OAP, \frac{AP}{\sin(90^\circ - \alpha)} = \frac{OP}{\sin(\alpha + \beta)}

    APcosα=hsin(α+β)\Rightarrow \frac{AP}{\cos\alpha} = \frac{h}{\sin(\alpha + \beta)}

    In PAB,ABsin(β+γ)=APsin(αγ)\triangle PAB, \frac{AB}{\sin(\beta + \gamma) = \frac{AP}{\sin(\alpha - \gamma)}}

    h=msin(α+β)sin(αγ)cosαsin(β+γ)\Rightarrow h = \frac{m\sin(\alpha + \beta)\sin(\alpha - \gamma)}{\cos\alpha\sin(\beta + \gamma)}

  9. The diagram is given below:

    209th problem

    Let ABCDABCDABCDA'B'C'D' be the vertical tower having a height of hh i.e. AA=BB=CC=DD=hAA' = BB' = CC' = DD' = h with side equal to bb and OO be the point of observation on the diagonal ACAC extended at a distance 2a2a from AA. Clearly, OAB=135\angle OAB = 135^\circ. Given that AOA=45\angle AOA' = 45^\circ and BOB=30\angle BOB' = 30^\circ.

    In AOA,tan45=1=AAOA=h2ah=2a\triangle AOA', \tan45^\circ = 1 = \frac{AA'}{OA} = \frac{h}{2a}\Rightarrow h = 2a

    In BOB,tan30=13=BBOBOB=23a\triangle BOB', \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BB'}{OB} \Rightarrow OB = 2\sqrt{3}a

    In OAB,cos135=12=4a2+b212a22.2a.b\triangle OAB, \cos135^\circ = -\frac{1}{\sqrt{2}} = \frac{4a^2 + b^2 - 12a^2}{2.2a.b}

    This is a quadratic equation in b wiht two roots a(2±10)a(-\sqrt{2}\pm\sqrt{10}). Clearly, bb cannot be negaitive so b=a(102)b = a(\sqrt{10} - \sqrt{2}).

  10. The diagram is given below:

    210th problem

    Let ASAS be the steeple having a height of h,Bh, B is the point due south having an angle of elevation of 4545^\circ to the top of the tower and CC is the point due south of BB, at a distance of aa from BB, having an angle of elevation of 1515^\circ to the top of the tower. ASAS is perpendicular to the plane of paper.

    In ABS,tan45=1=ASABAB=h\triangle ABS, \tan45^\circ = 1 = \frac{AS}{AB}\Rightarrow AB = h

    In ACS,tan15=23=ASACAC=h(2+3)\triangle ACS, \tan15^\circ = 2 - \sqrt{3} = \frac{AS}{AC} \Rightarrow AC = h(2 + \sqrt{3})

    In ABC,AC2=AB2+BC2h2(2+3)2=h2+a2\triangle ABC, AC^2 = AB^2 + BC^2 \Rightarrow h^2(2 + \sqrt{3})^2 = h^2 + a^2

    a=h6+23\Rightarrow a = \frac{h}{\sqrt{6 + 2\sqrt{3}}}

  11. The diagram is given below:

    211th problem

    Let CDCD be the given tower with a given height of cc, FGFG be the mountain behind the spire and tower at a distance xx having a height of hh and AA and BB are the points of observation. Let the angle of elevation from AA is α\alpha such that the mountain is just visible behind the tower. Let DEDE be the spire which subtends equal angle of β\beta at AA and BB. Since it subtends equal angles at AA and BB the points A,B,DA, B, D and EE will be concyclic. aa and bb are shown as given in the question.

    AEC=90(α+β)\angle AEC = 90^\circ - (\alpha + \beta)

    Segment ADAD will also subtend equal angles at BB and EE AED=ABD=90(α+β)CBE=90α\therefore \angle AED = \angle ABD = 90^\circ - (\alpha + \beta) \Rightarrow \angle CBE = 90^\circ - \alpha

    In ACD\triangle ACD and AFG,tanα=ca=hx+ax=ahaccAFG, \tan\alpha = \frac{c}{a} = \frac{h}{x + a} \Rightarrow x = \frac{ah - ac}{c}

    In BFG,tan(90α)=hx+a+b\triangle BFG, \tan(90^\circ - \alpha) = \frac{h}{x + a + b}

    ac=hahacc+a+b\Rightarrow \frac{a}{c} = \frac{h}{\frac{ah - ac}{c} + a + b}

    h=abcc2a2\Rightarrow h = \frac{abc}{c^2 - a^2}

  12. The diagram is given below:

    212th problem

    Let ABAB be the pole having height hh and CDCD be the tower having a height of h+xh + x as shown in the diagram. The angles α\alpha and β\beta are shown as given in the question. Let dd be the distance between the pole and tower. Clearly, ADC=90αBDC=90(αβ)\angle ADC = 90^\circ - \alpha \Rightarrow \angle BDC = 90^\circ - (\alpha - \beta). Let h+x=Hh + x = H

    In ACD,tanα=h+xdd=(h+x)cotα=Hcotα\triangle ACD, \tan\alpha = \frac{h + x}{d} \Rightarrow d = (h + x)\cot\alpha = H\cot\alpha

    In BDE,tan(αβ)=Hhdd=(Hh)cot(αβ)\triangle BDE, \tan(\alpha - \beta) = \frac{H - h}{d} \Rightarrow d = (H - h)\cot(\alpha - \beta)

    Hcotα=(Hh)cot(αβ)\Rightarrow H\cot\alpha = (H - h)\cot(\alpha - \beta)

    H=hcot(αβ)cot(αβ)cotα\Rightarrow H = \frac{h\cot(\alpha - \beta)}{\cot(\alpha- \beta) - \cot\alpha}

  13. The diagram is given below:

    213th problem

    Let A,B,CA, B, C and DD be the points on one bank such that AB=6d,AC=2d,AD=BD=3dAB = 6d, AC = 2d, AD = BD = 3d and PQPQ be the tower on the other bank perpendicular to the plane of the paper having a height of hh. Given that PBQ=PAQ=α\angle PBQ = \angle PAQ = \alpha and PCQ=β\angle PCQ = \beta.

    In PBQ,tanα=PQPBPB=hcotα\triangle PBQ, \tan\alpha = \frac{PQ}{PB} \Rightarrow PB = h\cot\alpha

    Similarly, PA=hcotαPA = h\cot\alpha and PC=hcotβPC = h\cot\beta. Since PA=PAPA = PA the PAB\triangle PAB is an isosceles triangle. As DD is the mid-point of ABAB so PBD,PCD\triangle PBD, \triangle PCD and PAD\triangle PAD will be right angled triangles.

    In PAD,PA2=PD2+AD2\triangle PAD, PA^2 = PD^2 + AD^2 and in PCD,PC2=PD2+CD2\triangle PCD, PC^2 = PD^2 + CD^2

    Subtracting, we get PA2PC2=AD2CD2PA^2 - PC^2 = AD^2 - CD^2

    h2(cot2αcot2β)=9d2d2=8d2\Rightarrow h^2(\cot^2\alpha - \cot^2\beta) = 9d^2 - d^2 = 8d^2

    h=22dcot2αcot2β\Rightarrow h = \frac{2\sqrt{2}d}{\sqrt{\cot^2\alpha - \cot^2\beta}}

    PDPD represents the width of the canal. PD2=PA2AD2=h2cot2α9d2\Rightarrow PD^2 = PA^2 - AD^2 = h^2\cot^2\alpha - 9d^2

    PD=d9cot2βcot2αcot2αcot2β\Rightarrow PD = d\sqrt{\frac{9\cot^2\beta - \cot^2\alpha}{\cot^2\alpha - \cot^2\beta}}

  14. The diagram is given below:

    214th problem

    Let PQPQ be the tower having a height of hh and points A,BA, B are the two stations at a distance of 22 km having angles of elevation of 6060^\circ and 3030^\circ respectively. CC is the mid-point between AA and BB from where the angle of elevation is 4545^\circ.

    In PBQ,tan60=hPBPB=h3\triangle PBQ, \tan60^\circ = \frac{h}{PB}\Rightarrow PB = \frac{h}{\sqrt{3}}

    Similarly, PA=3hPA = \sqrt{3}h and PC=hPC = h.

    Now since CC is the mid-point of ABAB therefore PCPC is the median of the triangle PABPAB.

    PA2+PB2=2(PC2+AC2)\Rightarrow PA^2 + PB^2 = 2(PC^2 + AC^2)

    h23+3h2=2(h2+1)\Rightarrow \frac{h^2}{3} + 3h^2 = 2(h^2 + 1)

    h=32\Rightarrow h = \frac{\sqrt{3}}{\sqrt{2}} km =5006= 500\sqrt{6} m.

  15. The diagram is given below:

    215th problem

    Let PQPQ be the flag-staff standing inside equilateral ABC\triangle ABC and since all sides subtend an angle of 6060^\circ it is guaranteed that PP will be centroid of the ABC\triangle ABC. Given that the height of the flag-staff is 1010 m. Also, according to question AQB=BQC=CQA=60AQ=BQ\angle AQB = \angle BQC = \angle CQA = 60^\circ \therefore AQ = BQ. Let each side of the triangle has length of 2a2a m.

    Thus, AQB\triangle AQB is an equilateral triangle. AQ=BQ=AB=2a=CQ\therefore AQ = BQ = AB = 2a = CQ

    We know from geometry that AP=23ADAP = \frac{2}{3}AD. We also know that median of an equilateral triangle is perpendicular bisector. ABD\therefore \triangle ABD is a right-angle triangle where DD is the point where APAP would meet BCBC.

    sin60=ADABAD=2asin60\Rightarrow \sin60^\circ = \frac{AD}{AB} \Rightarrow AD = 2a\sin60^\circ

    AP=2a3\Rightarrow AP = \frac{2a}{\sqrt{3}}

    APQ\triangle APQ is also a right angle triangle.

    AQ2=AP2+PQ24a24a23=10\Rightarrow AQ^2 = AP^2 + PQ^2 \Rightarrow 4a^2 - \frac{4a^2}{3} = 10

    a=532\Rightarrow a = 5\sqrt{\frac{3}{2}}

    2a=56\Rightarrow 2a = 5\sqrt{6} m.

  16. The diagram is given below:

    216th problem

    Let CDCD be the cliff having a height of H,DEH, DE be the tower on the cliff having a height of hh and A,BA, B are two points on horizontal level where the tower subtends the equal angle β\beta at a distance of a,ba, b from the cliff’s foot. Let α\alpha be the angle of elevation from AA of the cliff’s top.

    Since DEDE subtends equal angles at A,BA, B therefore a circle will pass through these four points and thus chord ADAD will also subtend equal angles AEC\angle AEC and ABD\angle ABD equal to 90(α+β)90^\circ - (\alpha + \beta).

    In ACD,tanα=Ha\triangle ACD, \tan\alpha = \frac{H}{a} and tan(α+β)=H+ha=bH\tan(\alpha + \beta) = \frac{H + h}{a} = \frac{b}{H}

    In BCE,tan(90α)=cotα=H+hb\triangle BCE, \tan(90^\circ - \alpha) = \cot\alpha = \frac{H + h}{b}

    We have H+ha=bHabH2=Hh\frac{H + h}{a} = \frac{b}{H} \Rightarrow ab - H^2 = Hh

    We have tan(α+β)=bH\tan(\alpha + \beta) = \frac{b}{H}

    tanα+tanβ1tanαtanβ=bH\Rightarrow \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{b}{H}

    Ha+tanβ1Hatanβ=bH\Rightarrow \frac{\frac{H}{a} + \tan\beta}{1 - \frac{H}{a}\tan\beta} = \frac{b}{H}

    (H+bHa)tanβ=abH2a\Rightarrow \left(H + \frac{bH}{a}\right)\tan\beta = \frac{ab - H^2}{a}

    h=(a+b)tanβ\Rightarrow h = (a + b)\tan\beta

  17. The diagram is given below:

    217th problem

    Let ABAB be the tower and BCBC be the flag-staff having heights of xx and yy respectively. According to question BCBC makes an angle of α\alpha at EE which is cc distance from the tower. Let the angle of elevation from EE to the top of tower BB is β\beta.

    In ABE,tanβ=xc\triangle ABE, \tan\beta = \frac{x}{c}

    In ACE,tan(α+β)=x+yc\triangle ACE, \tan(\alpha + \beta) = \frac{x + y}{c}

    tanα+tanβ1tanαtanβ=x+yc\Rightarrow \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{x + y}{c}

    x+ctanαcxtanα=x+yc\Rightarrow \frac{x + c\tan\alpha}{c - x\tan\alpha} = \frac{x + y}{c}

    tanα=cyx2+c2+xy\Rightarrow \tan\alpha = \frac{cy}{x^2 + c^2 + xy}

    Given that α\alpha is the greatest angle made which means tanα\tan\alpha will be greatest. So equating the derivative w.r.t to cc to zero, we get

    ddc[cyx2+c2+xy]=c[x(x+c)c2][x2+c2+xy]2=0\frac{d}{dc}\left[\frac{cy}{x^2 + c^2 + xy}\right] = \frac{c[x(x + c) - c^2]}{[x^2 + c^2 + xy]^2} = 0

    c2=x(x+y)\Rightarrow c^2 = x(x + y)

    tanα=cy2c2=y2cy=2ctanα\Rightarrow \tan\alpha = \frac{cy}{2c^2} = \frac{y}{2c} \Rightarrow y = 2c\tan\alpha

    We had x2+xyc2=0x2+2cxtanαc2=0x^2 + xy - c^2 = 0 \Rightarrow x^2 + 2cx\tan\alpha - c^2 = 0

    Neglecting the negative root we have x=ctanα+csecα\Rightarrow x = -c\tan\alpha + c\sec\alpha

    =c(1sinαcosα)=2d(1+tan2α22tanα21tan2α2)\Rightarrow = c\left(\frac{1 - \sin\alpha}{\cos\alpha}\right) = 2d\left(\frac{1 + \tan^2\frac{\alpha}{2} - 2\tan\frac{\alpha}{2}}{1 - \tan^2\frac{\alpha}{2}}\right)

    =c(1tanα21+tanα2)= c\left(\frac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}}\right)

    =ctan(π4α2)= c\tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)

  18. The diagram is given below:

    218th problem

    We know that BB is due north of DD at a distance of 22 km and DD is due west of CC such that BCD=25\angle BCD = 25^\circ we can plot B,C,DB, C, D as shown in the diagram. It is given that BB lies on ACAC such that BDA=40\angle BDA = 40^\circ. From figure it is clear that ACD=CAD=25\angle ACD = \angle CAD = 25^\circ thus ACD\triangle ACD is an isoscelels triangle. Let AD=CD=xAD = CD = x.

    In BCD,tan25=2xx=2cot25=4.28\triangle BCD, \tan25^\circ = \frac{2}{x} \Rightarrow x = 2\cot25^\circ = 4.28 km.

  19. The diagram is given below:

    219th problem

    Let the train move along the line PQPQ. The train is at OO at some instant. AA is the observation point. Ten minutes earlier let the train position be PP and ten minutes afterwards let the train be at QQ.

    According to question, OAP=α1,OAQ=α2,NOQ=θ\angle OAP = \alpha_1, \angle OAQ = \alpha_2, \angle NOQ = \theta. Hence POA=θ\angle POA = \theta.

    Since the speed of the train is constant OP=OQ\therefore OP = OQ

    Applying m:nm:n rule in PAQ,(1+1)cotθ=cotα2cotα1\triangle PAQ, (1 + 1)\cot\theta = \cot\alpha_2 - \cot\alpha_1

    cotθ=cotα2cotα12\Rightarrow \cot\theta = \frac{\cot\alpha_2 - \cot\alpha_1}{2}

    tanθ=2sinα1sinα2sin(α1α2)\Rightarrow \tan\theta = \frac{2\sin\alpha_1\sin\alpha_2}{\sin(\alpha_1 - \alpha_2)}

  20. The diagram is given below:

    220th problem

    Let OPOP be the flag-staff and that the man walk along the horizontal circle. Clearly, the flag-staff will subtend the greatest and least angles when the man is at AA and BB respectively. Let CC be the mid-point of the arc ACBACB. According to question, PAO=α,PBO=β,PCO=θ\angle PAO = \alpha, \angle PBO = \beta, \angle PCO = \theta. Clearly, POC=90\angle POC = 90^\circ. Let OP=h,POD=ϕOP = h, \angle POD = \phi.

    Also, OA=OB=OC=rOA = OB = OC = r where rr is the radius of the circle.

    In PDO,sinϕ=PDOPPD=hsinϕ\triangle PDO, \sin\phi = \frac{PD}{OP} \Rightarrow PD = h\sin\phi

    cosϕ=ODOPOD=hcosϕ\cos\phi = \frac{OD}{OP} \Rightarrow OD = h\cos\phi

    BD=r+hcosϕ\therefore BD = r + h\cos\phi and AD=rhcosϕAD = r - h\cos\phi

    In POC,tanθ=hrh=rtanθ\triangle POC, \tan\theta = \frac{h}{r} \Rightarrow h = r\tan\theta

    In PDA,tanα=PDAD=hsinϕrhcosϕ=rtanθsinϕrrtanθcosϕ\triangle PDA, \tan\alpha = \frac{PD}{AD} = \frac{h\sin\phi}{r - h\cos\phi} = \frac{r\tan\theta\sin\phi}{r - r\tan\theta\cos\phi}

    tanα=tanθsinϕ1tanθcosϕ\Rightarrow \tan\alpha = \frac{\tan\theta\sin\phi}{1 - \tan\theta\cos\phi}

    tanθsinϕ=tanαtanαtanθcosϕ\Rightarrow \tan\theta\sin\phi = \tan\alpha - \tan\alpha\tan\theta\cos\phi

    In PDB,tanβ=hsinϕr+hcosϕ=rtanθsinϕr+rtanθcosϕ\triangle PDB, \tan\beta = \frac{h\sin\phi}{r + h\cos\phi} = \frac{r\tan\theta\sin\phi}{r + r\tan\theta\cos\phi}

    tanβ=tanθsinϕ1+tanθcosϕ\Rightarrow \tan\beta = \frac{\tan\theta\sin\phi}{1 + \tan\theta\cos\phi}

    tanθsinϕ=tanβ+tanβtanθcosϕ\Rightarrow \tan\theta\sin\phi = \tan\beta + \tan\beta\tan\theta\cos\phi

    tanαtanαtanθcosϕ=tanβ+tanβtanθcosϕ\Rightarrow \tan\alpha - \tan\alpha\tan\theta\cos\phi = \tan\beta + \tan\beta\tan\theta\cos\phi

    tanαtanβ=tanθcosϕ(tanα+tanβ)\Rightarrow \tan\alpha - \tan\beta = \tan\theta\cos\phi(\tan\alpha + \tan\beta)

    Also, 1tanθcosϕ=tanθsinϕtanα1 - \tan\theta\cos\phi = \frac{\tan\theta\sin\phi}{\tan\alpha}

    and 1+tanθcosϕ=tanθsinϕtanβ1 + \tan\theta\cos\phi = \frac{\tan\theta\sin\phi}{\tan\beta}

    2=tanθsinϕ(1tanα+1tanβ)\Rightarrow 2 = \tan\theta\sin\phi\left(\frac{1}{\tan\alpha} + \frac{1}{\tan\beta}\right)

    2tanαtanβ=tanθsinϕ(tanα+tanβ)\Rightarrow 2\tan\alpha\tan\beta = \tan\theta\sin\phi(\tan\alpha + \tan\beta)

    (tanαtanβ)2+4tan2αtan2β=tan2θ(tanα+tanβ)2\Rightarrow (\tan\alpha - \tan\beta)^2 + 4\tan^2\alpha\tan^2\beta = \tan^2\theta(\tan\alpha + \tan\beta)^2

    tan2θ[sinαcosα+sinβcosβ]2=(sinαcosαsinβcosβ)2+4sin2αsin2βcos2αcos2β\Rightarrow \tan^2\theta\left[\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}\right]^2 = \left(\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}\right)^2 + \frac{4\sin^2\alpha\sin^2\beta}{\cos^2\alpha\cos^2\beta}

    tan2θsin2(α+β)cos2αcos2β=sin2(αβ)+4sin2αsin2βcos2αcos2β\Rightarrow \tan^2\theta\frac{\sin^2(\alpha + \beta)}{\cos^2\alpha\cos^2\beta} = \frac{\sin^2(\alpha - \beta) + 4\sin^2\alpha\sin^2\beta}{\cos^2\alpha\cos^2\beta}

    tanθ=sin2(αβ)+4sin2αsin2βsin(α+β)\Rightarrow \tan\theta = \frac{\sqrt{\sin^2(\alpha - \beta) + 4\sin^2\alpha\sin^2\beta}}{\sin(\alpha + \beta)}

  21. The diagram is given below:

    221st problem

    Let OO be the position of the observer. PRQSPRQS is the horizontal circle in which the bird is flying. PP and QQ are the two extremes and RR is the mid point of the arc of the circle. PRQSP'R'Q'S is the vertical projection of the ground. CC is the center of the circle PRQSPRQS.

    According to the question, POP=60,QOQ=30,ROP=θ\angle POP'' = 60^\circ, \angle QOQ' = 30^\circ, \angle ROP'= \theta.

    Also, let PP=QQ=RR=h,rPP' = QQ' = RR' = h, r be the radius of the horizontal circle and OP=zOP' = z.

    In PPO,tan60=3=PPOP=hzh=3z\triangle PP'O, \tan60^\circ = \sqrt{3} = \frac{PP'}{OP'} = \frac{h}{z} \Rightarrow h = \sqrt{3}z

    In QOQ,tan30=13=hz+2rz+2r=3h\triangle QOQ', \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{z + 2r} \Rightarrow z + 2r = \sqrt{3}h

    z+2r=33zz=r\Rightarrow z + 2r = \sqrt{3}\sqrt{3}z \Rightarrow z = r

    In ROR,tanθ=hOR=hOC2+CR2=h(z+r)2+r2=3r(r+r)2+r2=35\triangle ROR', \tan\theta = \frac{h}{OR'} = \frac{h}{\sqrt{OC^2 + C'R'^2}} = \frac{h}{\sqrt{(z + r)^2 + r^2}} = \frac{\sqrt{3}r}{\sqrt{(r + r)^2 + r^2}} = \sqrt{\frac{3}{5}}

    tan2θ=35\Rightarrow \tan^2\theta = \frac{3}{5}

  22. The diagram is given below:

    222nd problem

    Let OO be the position of the observer and OPQOPQ be the horizontal line through OO meeting the hill at PP and the vertical through the center CC of the sphere at QQ.

    Let OAOA be the tangent to the sphere from OO touching it at AA. According to question, AOQ=β,QPC=90α,ACN=β\angle AOQ = \beta, \angle QPC = 90^\circ - \alpha, \angle ACN = \beta. Let rr be the radius of the hill. Draw AMOQAM\perp OQ and ANCRAN\perp CR.

    In AMO,tanβ=AMOM=QNOP+PM=QNOP+PQMQ\triangle AMO, \tan\beta = \frac{AM}{OM} = \frac{QN}{OP + PM} = \frac{QN}{OP + PQ - MQ}

    =CNCQOP+PQMQ= \frac{CN - CQ}{OP + PQ - MQ}

    sinβcosβ=rcosβrcosαa+rsinαrsinβ[MQ=AN]\Rightarrow \frac{\sin\beta}{\cos\beta} = \frac{r\cos\beta - r\cos\alpha}{a + r\sin\alpha - r\sin\beta} [\because MQ = AN]

    asinβ+rsinβ(sinαsinβ)=rcosβ(cosβcosα)\Rightarrow a\sin\beta + r\sin\beta(\sin\alpha - \sin\beta) = r\cos\beta(\cos\beta - \cos\alpha)

    asinβ=r[1cos(αβ)]\Rightarrow a\sin\beta = r[1 - \cos(\alpha - \beta)]

    r=asinβ2sin2αβ2\Rightarrow r = \frac{a\sin\beta}{2\sin^2\frac{\alpha - \beta}{2}}

    Height of the hill above the plane =OR=CRCQ=rrcosα=2rsin2α2= OR = CR _ CQ = r - r\cos\alpha = 2r\sin^2\frac{\alpha}{2}

    =asinβsin2α2sin2αβ2= \frac{a\sin\beta\sin^2\frac{\alpha}{2}}{\sin^2\frac{\alpha - \beta}{2}}

  23. The diagram is given below:

    223rd problem

    Let OO be the center of the sphere and rr be its radius. Given, PAM=θ,PBM=ϕ,CA=a,CB=b\angle PAM = \theta, \angle PBM = \phi, CA = a, CB = b

    Let DOC=β\angle DOC = \beta

    In PMA,tanθ=PMAM=DNAC+CM=ONODAC+DCDM\triangle PMA, \tan\theta = \frac{PM}{AM} = \frac{DN}{AC + CM} = \frac{ON _ OD}{AC + DC - DM}

    sinθcosθ=rcosθrcosβa+rsinβrsinθ[DM=NP]\Rightarrow \frac{\sin\theta}{\cos\theta} = \frac{r\cos\theta - r\cos\beta}{a + r\sin\beta - r\sin\theta} [\because DM = NP]

    Proceeding like previous problem asinβ=r[1cos(θβ)]a\sin\beta = r[1 - \cos(\theta - \beta)]

    2rsin2θβ2=2asinθ2cosθ2\Rightarrow 2r\sin^2\frac{\theta - \beta}{2} = 2a\sin\frac{\theta}{2}\cos\frac{\theta}{2}

    rsinθβ2=asinθ2cosθ2\Rightarrow \sqrt{r}\sin\frac{\theta - \beta}{2} = \sqrt{a\sin\frac{\theta}{2}\cos\frac{\theta}{2}}

    r[sinθ2cosβ2cosθ2sinβ2]sinθ2=asinθ2cosθ2sinθ2\Rightarrow \sqrt{r}\frac{\left[\sin\frac{\theta}{2}\cos\frac{\beta}{2} - \cos\frac{\theta}{2}\sin\frac{\beta}{2}\right]}{\sin\frac{\theta}{2}} = \frac{\sqrt{a\sin\frac{\theta}{2}\cos\frac{\theta}{2}}}{\sin\frac{\theta}{2}}

    r[cosβ2cotθ2sinβ2]=acotθ2\Rightarrow \sqrt{r}\left[\cos\frac{\beta}{2} - \cot\frac{\theta}{2}\sin\frac{\beta}{2}\right] = \sqrt{a\cot\frac{\theta}{2}}

    Similarly, r[cosβ2cotϕ2sinβ2]=bcosϕ2\sqrt{r}\left[\cos\frac{\beta}{2} - \cot\frac{\phi}{2}\sin\frac{\beta}{2}\right] = \sqrt{b\cos\frac{\phi}{2}}

    Subtracting, we get rsinβ2[cotθ2cotϕ2]=bcosϕ2acotθ2\sqrt{r}\sin\frac{\beta}{2}\left[\cot\frac{\theta}{2} - \cot\frac{\phi}{2}\right] = \sqrt{b\cos\frac{\phi}{2}} - \sqrt{a\cot\frac{\theta}{2}}

    Height of the hill DR=OROD=rrcosβ=2rsin2β2DR = OR - OD = r - r\cos\beta = 2r\sin^2\frac{\beta}{2}

    =2[bcosϕ2acotθ2cotθ2cotϕ2]2= 2\left[\frac{\sqrt{b\cos\frac{\phi}{2}} - \sqrt{a\cot\frac{\theta}{2}}}{\cot\frac{\theta}{2} - \cot\frac{\phi}{2}}\right]^2

  24. The diagram is given below:

    224th problem

    Let OO be the center of the hemisphere and PQPQ is the flag-staff. Given, OP=OR=r,AB=dOP = OR = r, AB = d.

    In ORB,cos45=rOBOB=2r\triangle ORB, \cos45^\circ = \frac{r}{OB} \Rightarrow OB = \sqrt{2}r

    In QOA,tan30=13=OQOA=h+r2r+d\triangle QOA, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{OQ}{OA} = \frac{h + r}{\sqrt{2}r + d}

    h+r=2r+d3\Rightarrow h + r = \frac{\sqrt{2}r + d}{\sqrt{3}}

    In QOB,tan45=1=OQOBh+r=2r\triangle QOB, \tan45^\circ = 1 = \frac{OQ}{OB} \Rightarrow h + r = \sqrt{2}r

    2r=2r+d3\Rightarrow \sqrt{2}r = \frac{\sqrt{2}r + d}{\sqrt{3}}

    (62)r=dr=3+122d\Rightarrow (\sqrt{6} - \sqrt{2})r = d \Rightarrow r = \frac{\sqrt{3} + 1}{2\sqrt{2}}d

    h=(21)r=(21)(3+1)22d\Rightarrow h = (\sqrt{2} - 1)r = \frac{(\sqrt{2} - 1)(\sqrt{3} + 1)}{2\sqrt{2}}d

  25. The diagram is given below:

    225th problem

    Let the direction in which man starts walking be the xx-axis. From question OA=AB=BC=aOA = AB = BC = a.

    Let coordinate of last point be (X,Y)(X, Y) then X=a+acosα+acos2α+X = a + a\cos\alpha + a\cos2\alpha + \cdots up to nn terms

    =a.cos(n1)α2sinnα2sinα2= a.\frac{\cos(n - 1)\frac{\alpha}{2}\sin\frac{n\alpha}{2}}{\sin\frac{\alpha}{2}}

    Y=0+asinα+asin2α+Y = 0 + a\sin\alpha + a\sin2\alpha + \cdots up to nn terms

    =a[sin(n1)α2sinnα2sinα2]= a\left[\frac{\sin(n - 1)\frac{\alpha}{2}\sin\frac{n\alpha}{2}}{\sin\frac{\alpha}{2}}\right]

    Distance from the starting point =X2+Y2=asinnα2sinα2= \sqrt{X^2 + Y^2} = \frac{a\sin\frac{n\alpha}{2}}{\sin\frac{\alpha}{2}}

    Let θ\theta be the angle which this distance makes with xx-axis. Then

    tanθ=YX=tan(n1)α2θ=(n1)α2\tan\theta = \frac{Y}{X} = \tan(n - 1)\frac{\alpha}{2} \Rightarrow \theta = (n - 1)\frac{\alpha}{2}

  26. The diagram is given below:

    226th problem

    Let ABCABC be the horizontal triangle. ABA'B' represents the stratum of coal. Suppose this startus meets the horizontal plane in line DDDD'. Let θ\theta be the angle between the horizontal and the stratum of the coal.

    Clearly, ADA=θ\angle ADA' = \theta. According to question, AA=x,BB=x+yAA' = x, BB' = x + y and CC=x+zCC' = x + z.

    In AAD,tanθ=AAADx=ADtanθ\triangle AA'D, \tan\theta = \frac{AA'}{AD} \Rightarrow x = AD\tan\theta

    In BBD,tanθ=BBBDx+y=(AD+AB)tanθ\triangle BB'D, \tan\theta = \frac{BB'}{BD} \Rightarrow x + y = (AD + AB)\tan\theta

    x+z=(AD+c)tanθ\Rightarrow x + z = (AD + c)\tan\theta

    In CCD,tanθ=CCCDx+z=(AD+bcosA)tan]θ\triangle CC'D, \tan\theta = \frac{CC'}{CD'} \Rightarrow x + z = (AD + b\cos A)\tan]\theta

    y=ctanθyc=tanθ\Rightarrow y = c\tan\theta \Rightarrow \frac{y}{c} = \tan\theta and z=bcosAtanθzb=cosAtanθz = b\cos A\tan\theta \Rightarrow \frac{z}{b} = \cos A\tan\theta

    Now, y2c2+z2b22yzbccosA=y2c2sin2A+y2c2cos2A+z2b22yzbccosA\frac{y^2}{c^2} + \frac{z^2}{b^2} - \frac{2yz}{bc}\cos A = \frac{y^2}{c_2}\sin^2A + \frac{y^2}{c^2}\cos^2A + \frac{z^2}{b^2} - \frac{2yz}{bc}\cos A

    =y2c2sin2A+(yccosAzb)2= \frac{y^2}{c^2}\sin^2A + \left(\frac{y}{c}\cos A - \frac{z}{b}\right)^2

    =tan2θsin2A+(tanθcosAcosAtanθ)2=tan2θsin2A= \tan^2\theta\sin^2A + (\tan\theta\cos A - \cos A\tan\theta)^2 = \tan^2\theta\sin^2A

    tanθsinA=y2c2+z2b22yzbccosA\Rightarrow \tan\theta\sin A = \sqrt{\frac{y^2}{c^2} + \frac{z^2}{b^2} - \frac{2yz}{bc}\cos A}