29. Height and Distance Solutions Part 2

51. The diagram is given below:

    

    Let $AB$ be the lamp post having height $h$ m, and $BD$ be the girl having height $1.6$ m. The distance of the grl from the lamp post is $AC = 3.2$ m. $CE$ is the langeth of the shadow given as $4.8$ m. In the $\triangle ABE$ and $\triangle CDE, \angle E$ is common, $\angle A = \angle C = 90^\circ$ so third angle will be also equal. This makes the triangles similar.

    $\therefore \frac{AB}{CD} = \frac{AE}{CE} \Rightarrow h = \frac{8}{3}$ m.

52. The diagram is given below:

    

    Let $AC$ be the building having a height of $30$ m. Let $E$ and $G$ point of observations where angles of elevation are $60^\circ$ and $30^\circ$ respectively. Let $AEF$ be the line of foot of the building and foot of the observer which is a horizontal line. Let $DE$ and $FE$ are the heights of the observer. Draw $BEG\parallel ADF$ so that $AB = DE = FG = 1.5$ m. Thus, $BC = 28.5$ m. We have to find $DF = EG$.

    In $\triangle BCE, \tan60^\circ = \sqrt{3} = \frac{BC}{CE} \Rightarrow CE = \frac{28.5}{\sqrt{3}}$ m.

    In $\triangle BCG, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{CG} \Rightarrow CG = 28.5\sqrt{3}$ m.

    Thus, $DF = EG = CG - CF = \frac{57}{\sqrt{3}}$ m, which is the distance walked by the observer.

53. The diagram is given below:

    

    Let the height of the tower $AB$ is $h$ m. When the altitude of the sun is $60^\circ$ let the length of the shadown be $AC = x$ m. Then according to question length of shadow when the sun’s altitude i $30^\circ$ the length of shadow will be $AD, 40$ m longer i.e. $AD = x + 40$.

    In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} = \frac{h}{x} \Rightarrow x = \sqrt{3}h$ m.

    In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AC} = \frac{h}{x + 40} \Rightarrow x = 20$ m and $h = 20\sqrt{3}$ m.

54. The diagram is given below:

    

    Let $AB$ be the building with $20$ m height. Let the height of tower be $h$ m represented by $BC$ in the figure. Let $D$ be the point of observation at a distance $x$ from the foot of the building $AB$.

    In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{20}{x} \Rightarrow x = 20\sqrt{3}$ m.

    In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{AD} = \frac{h + 20}{x}\Rightarrow h = 40$ m.

55. The diagram is given below:

    

    Let $DE$ be the building having a height of $8$ m. Let $AC$ be the multistoried building having height $h + 8$ m. Foot of both the buildings are joined on horizontal plane i.e. $AD$. Draw a line parallel to $AD$ which is $BE$. So $BE$ is equal to $AD$ which we have let as $x$ m. Clearly, $AB = 8$ m. Let height of $BC$ to be $h$ m.

    In $\triangle CBE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{BE} = \frac{h}{x} \Rightarrow \sqrt{3}h = x$.

    In $\triangle ACD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AC}{AD} = \frac{h + 8}{x}\Rightarrow h = \frac{8}{\sqrt{3} - 1} \Rightarrow h + 8 = \frac{8\sqrt{3}}{\sqrt{3} - 1}$ m.

56. The diagram is given below:

    

    Let $AB$ be the pedestal having height $h$ m and $BC$ be the statue having height $1.6$ m on top of pedestal. Let $D$ be the point of observation from where the angles of elevation as given in the question are $45^\circ$ and $60^\circ$.

    In $\triangle ABD, \tan45^\circ = 1 = \frac{AB}{BD} = \frac{h}{x} \Rightarrow h = x$.

    In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{CD} = \frac{h + 1.6}{x} \Rightarrow h = \frac{1.6}{\sqrt{3} - 1}$ m.

57. This problem is similar to 55 and has been left as an exercise.

58. The diagram is given below:

    

    Let $AB$ be the tower having height $75$ m. Let $C$ and $D$ be the position of two ships and angles of elevation are as given in the question. Let foor of the tower be in line with ships such that $AC = x$ m and distance between the ships as $d$ m.

    In $\triangle ABC, \tan45^\circ = 1 = \frac{AB}{AC} = \frac{75}{x} \Rightarrow x = 75$ m.

    In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{75}{x + d} \Rightarrow d = 75(\sqrt{3} - 1)$ m.

59. The diagram is given below:

    

    Let $AB$ be the building and $CD$ be thw tower having height $50$ m. The angles of elevation are shown as given in the question. Let distance between the foot of the tower and the building be $d$ m and height of the building be $h$ m.

    In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{d} \Rightarrow d = \sqrt{3}h$.

    In $\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{50}{d} \Rightarrow 3h = 50 \Rightarrow h = \frac{50}{3}$ m.

60. The diagram is given below:

    

    Let $DE$ represent the banks of river and $BC$ the bridge. Given that height of the bridge is $30$ m. $\therefore BD = CE = 30$ m. The angles of depression from point $A$ is shown as given in the question. We have to find $DE = BC$ i.e. width of the river.

    In $\triangle ACE, \tan45^\circ = 1 = \frac{CE}{AC} \Rightarrow AC = 30$ m.

    In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BD}{AB} \Rightarrow AB = 30\sqrt{3}$ m.

    Thus, width of river $= 30 + 30\sqrt{3} = 30(\sqrt{3} + 1)$ m

61. The diagram is given below:

    

    Let $BC$ and $DE$ be the two poles. Let $A$ be the point between them such that $AB = x$ m and, thus $AD = 80 - x$ m. Let the elevation from $A$ to $C$ is $60^\circ$ and to $E$ is $30^\circ$. Let the height of poles be $h$ m.

    In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{BC}{AB} = \frac{h}{x} \Rightarrow h = \sqrt{3}x$ m.

    In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{DE}{AD} = \frac{h}{80 - x} \Rightarrow 3x = 80 -x \Rightarrow x = 20$ m. $\Rightarrow h = 20\sqrt{3}$ m.

62. The diagram is given below:

    

    Let $BD$ and $CE$ be the poles and $AJ$ be the tree. Given, $AJ = 20$ m and angles of depression to base of poles are $60^\circ$ and $30^\circ$. Let $\angle DAB = 6-00^\circ$ and $\angle EAC = 30^\circ$.

    Clearly, $AB = DJ = y$ m(say) and $AC = EJ = x$ m(say).

    In $\triangle AEJ, \tan60^\circ = \sqrt{3} = \frac{AJ}{EJ} \Rightarrow x = \frac{20}{\sqrt{3}}$ m.

    Similarly, $y = 20\sqrt{3}$ m.

    Thus, width of river $x + y = \frac{80}{\sqrt{3}}$ m.

63. This problem is similar to $56$ and has been left as an exercise.

64. This problem is similar to $58$ and has been left as an exercise.

65. This problem is similar to $49$ and has been left as an exercise.

66. The diagram is given below:

    

    Let $A$ be the point on the ground, $AC$ be the string and $BC$ the height of balloon. Then given, angle of elevation $\angle BAC = 60^\circ$.

    In $\triangle ABC, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{BC}{AC} = \frac{BC}{215} = 107.5\sqrt{3}$ m.

67. The diagram is given below:

    

    Let $AB$ be the cliff having a height of $80$ m. Let $C$ and $D$ be two points on eihter side of the cliff from where angle of elevations are $60^\circ$ and $30^\circ$ respectively.

    In $\triangle ABC, \tan60^\circ = \frac{AB}{AC} \Rightarrow AC = \frac{80}{\sqrt{3}}$ m.

    In $\triangle ABD, \tan30^\circ = \frac{AB}{AD}\Rightarrow AD = 80\sqrt{3}$ m.

    Distance bettwen points of observation $CD = AC + AD = \frac{320}{\sqrt{3}}$ m.

68. Since the length of shadow is equal to height of pole the angle of elevation would be $45^\circ$ as $\tan45^\circ = 1$.

69. This problem is similar to $62$ and has been left as an exercise.

70. This problem is similar to $25$ and has been left as an exercise.

71. The diagram is given below:

    

    Let $AB$ be the lighthouse having a height of $200$ m. Let $C$ and $D$ be the ships. The angles of depression are converted to angles of elevation.

    In $\triangle ABC, \tan45^\circ = 1 = \frac{AB}{AC}\Rightarrow AC = 200$ m.

    In $\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{AB}{AD} \Rightarrow AD = \frac{200}{\sqrt{3}}$ m.

    Thus distance between ships $CD = AC + AD = \frac{200(\sqrt{3} + 1)}{\sqrt{3}}$ m.

72. The diagram is given below:

    

    Let $AB$ be the first pole and $CD$ be the second pole. Given, $CD = 24$ m and $AC = 15$ m. Draw $BE || AC \Rightarrow BE = 15$ m. Angle of depression is converted to angle of elevation.

    In $\triangle BDE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{ED}{BE} \Rightarrow ED = \frac{15}{\sqrt{3}} = 5\sqrt{3}$ m.

    $\Rightarrow CE = BD - ED = 24 - 5\sqrt{3} = AB$ which is height of the first pole.

73. This problem is similar to $71$ and has been left as an exercise.

74. The diagram is given below:

    

    xLet $AB$ be the tower and $C$ and $D$ are two points at a distance of $4$ m and $9$ m respectively. Because it is given that angles of elevations are complementary we have chosen and angle of $\theta$ for $C$ and $90^\circ - \theta$ for $D$.

    In $\triangle ABC, \tan\theta = \frac{AB}{AC} = \frac{h}{4}$

    In $\triangle ABD, \tan(90^\circ - \theta) = \cot\theta = \frac{AB}{AD} = \frac{h}{9}$

    Substituting for $\cot\theta$, we get

    $\frac{4}{h} = \frac{h}{9}\Rightarrow h^2 = 36 \Rightarrow h = 6$ m.

75. This problem is similar to $72$ and has been left as an exercise.

76. This problem is similar to $56$ and has been left as an exercise.

77. This problem is similar to $55$ and has been left as an exercise.

78. This problem is similar to $71$ and has been left as an exercise

79. This problem is similar to $55$ and has been left as an exercise.

80. This problem is similar to $58$ and has been left as an exercise.

81. This problem is similar to $26$ annd has been left as an exercise.

82. This problem is similar to $71$ and has been left as an exercise.

83. This problem is similar to $26$ annd has been left as an exercise.

84. This problem is similar to $28$ annd has been left as an exercise.

85. This problem is similar to $71$ and has been left as an exercise

86. This problem is similar to $23$ and has been left as an exercise

87. The diagram is given below:

    

    Let $AB$ be the tower and $BC$ be the flag-staff having a height of $h$ m. Let $D$ be the point of observation having angle of elevations $\alpha$ and $\beta$ as given in the question.

    In $\triangle ABC, \tan\alpha = \frac{AB}{AD} \Rightarrow AB = AD\tan\alpha$

    In $\triangle ABD, \tan\beta = \frac{AC}{AD} = \frac{AB + BC}{AD}$

    $\Rightarrow \frac{AB\tan\beta}{\tan\alpha} = AB + h \Rightarrow AB = \frac{h\tan\alpha}{\tan\beta - \tan\alpha}.$

88. This proble is similar to $74$ and has been left as an exercise.

89. The diagram is given below:

    

    Let $BE$ be the tower leaning northwards and $AB$ be the vertical height of tower taken as $h$. Let $C$ and $D$ be the points of observation. Given that angle of leaning is $\theta$ and angles of elevation are $\alpha$ at $C$ and $\beta$ at $D$. Let $AB = x$. Given $BC = a$ and $BD = b$.

    In $\triangle ABE, \cot\theta = \frac{x}{h}$, in $\triangle ACE, \cot\alpha = \frac{x + a}{h}$ and in $\triangle ADE, \cot\beta = \frac{x + b}{h}$.

    $\Rightarrow b\cot\alpha = \frac{bx + ab}{h}, a\cot\beta = \frac{ax + ab}{h}$

    $\Rightarrow b\cot\alpha - a\cot\beta = \frac{bx - ax}{h}\Rightarrow \frac{x}{h} = \cot\theta = \frac{b\cot\alpha - a\cot\beta}{b - a}$.

90. The diagram is given below:

    

    Let $AE$ be the plane of lake and $AC$ be the height of the cloud. $F$ is the point of observation at a height $h$ from lake. $AD$ is the reflection of cloud in the lake. Clearly, $AC = AD$. Draw $AE || BF$ and let $BF = x$. $\alpha$ and $\beta$ are angles of elevation and depression as given.

    In $\triangle BCF, \tan\alpha = \frac{BC}{BF} = \frac{BC}{x}\Rightarrow BC = x\tan\alpha$

    $AC = AD = AB + BC = h + x\tan\alpha$

    In $\triangle BDF, \tan\beta = \frac{AB + AD}{BF} = \frac{h + h + x\tan\alpha}{x} \Rightarrow x = \frac{2h}{\tan\beta - \tan\alpha}$

    $AC = AB + BC = h + x\tan\alpha = \frac{h(\tan\alpha + \tan\beta)}{\tan\beta - \tan\alpha}$.

91. The diagram is given below:

    

    Let the cicle represent round balloon centered at $O$ having radius $r$. $B$ is the point of observation from where angle of elevation to the center of the balloon is given as $\beta$. $BL$ and $BM$ are tangents to the balloon and $OL$ and $OM$ are perpendiculars. Clearly $OL = OM = r$. GIven $\angle LBM = \alpha$ and $\angle OBL = \angle OBM = \alpha/2$.

    In $\triangle OBL, \sin\alpha/2 = \frac{OL}{OB} \Rightarrow OB = r\cosec\alpha/2$.

    In $\triangle ABO, \sin\beta = \frac{AO}{OB}\Rightarrow AO = r\sin\beta\cosec\alpha/2$.

92. The diagram is given below:

    

    Let $AB$ be the cliff having a height $h$ and $F$ be the initial point of observation from where the angle of elevation is $\theta$. Let $D$ be the point reached after walking a distance $k$ towards the top at an angle $\phi$. The angle of elevation at $D$ is $\alpha$.

    In $\triangle DEF, \sin\phi = \frac{DE}{DF} \Rightarrow DE = k\sin\phi, \cos\phi = \frac{EF}{DF} \Rightarrow EF = k\cos\phi$.

    In $\triangle ABF, \tan\theta = \frac{AB}{BF} \Rightarrow \frac{x}{k\cos\phi + (x - k\sin\phi)\cot\alpha}$

    $\Rightarrow x\cot\theta = k\cos\phi + x\cot\alpha - k\sin\phi\cot\alpha \Rightarrow x(\cot\theta - \cot\alpha) = k(\cos\phi - \sin\phi\cot\alpha)$

    $\Rightarrow x = \frac{k(\cos\phi - \sin\phi\cot\alpha)}{\cot\theta - \cot\alpha}$.

93. The diagram is given below:

    

    Let $CD$ be the tower having a height $h$. Point $A$ is due south of $A$ making an angle of elevation $\alpha$ and $B$ is due east of tower making an angle of elevation $\beta$. Clearly, $\angle ACB = 90^\circ$. Given that $AB = d$.

    In $\triangle ACD, \tan\alpha = \frac{CD}{AC} \Rightarrow AC = h\cot\alpha$ and in $\triangle BCD, \tan\beta = \frac{CD}{BC} \Rightarrow BC = h\cot\beta$.

    In $\triangle ABC, AB^2 = AC^2 + AD^2 \Rightarrow d^2 = h^2\cot^2\alpha + h^2\cot^2\beta \Rightarrow h = \frac{d}{\sqrt{\cot^2\alpha + \cot^2\beta}}$.

94. This problem is similar to $93$ and has been left as an exercise.

95. The diagram is given below:

    

    Let $AB$ be the girl having a height of $1.2$ m, $C$ and $F$ be the two places of balloon for which angle of elevations are $60^\circ$ and $30^\circ$ respectively. Height of ballon above ground level is given as $88.2$ m and thus height of balloon above the girl’s eye-level is $88.2 - 1.2 = 87$ m.

    In $\triangle ACD, \tan60^\circ = \frac{CD}{AD} \Rightarrow AD = 87/\sqrt{3}$ m.

    In $\triangle AFG, \tan30^\circ = \frac{FG}{AG} \Rightarrow AG = 87\sqrt{3}$ m.

    Thus distance trarvelled by the ballon $= 87\sqrt{3} - 87/\sqrt{3} = 174/\sqrt{3}$

96. The diagram is given below:

    

    Let $AB$ represent the tower with a height $h$. Let $C$ and $D$ be the points to which angles of depression are given as $60^\circ$ and $30^\circ$ which are shown as angles of elevation at these points.

    In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} \Rightarrow AC = h/\sqrt{3}$

    In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} \Rightarrow AD = h\sqrt{3}$

    $CD = AD - AC = 2h/\sqrt{3}$

    The car covers the distance $CD$ in six seconds. Thus speed of the car if $2h/(6\sqrt{3}) = h/3\sqrt{3}$

    Time taken to cover $AC$ to reach the foot of the tower is $\frac{h}{\sqrt{3}}\times\frac{3\sqrt{3}}{h} = 3$ seconds.

97. Proceeding like previous problem the answer would be three minutes.

98. This problem is similar to $96$ and has been left as an exercise.

99. The diagram is given below:

    

    Let $AB$ be the building having height $h$ m. Let $C$ and $D$ be the fire stations from which the angles of elevation are $60^\circ$ and $45^\circ$ separated by $20,000$ m.

    In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC}\Rightarrow AC = h/\sqrt{3}$ m.

    In $\triangle ABD, \tan45^\circ = h = \frac{AB}{AD}\Rightarrow AD = h$ m.

    Since $AD < AD$ so the fire station at $C$ will reach the building faster.

    $AD = AC + CD \Rightarrow h = h/\sqrt{3} + 20000 \Rightarrow h = \frac{20000\sqrt{3}}{\sqrt{3} - 1}$

    $\therefore AC = \frac{2000}{\sqrt{3} - 1}$ m.

100. The diagram is given below:

     

     Let $AB$ be the deck of the ship with given height of $10$ m. Let $CE$ be the cliff with base at $C$. Let the height of portion $DE$ be $x$ m. The angles of elevation of the top and of the bottom of the cliff are shown as given in the question.

     In $\triangle BDE, \tan45^\circ = DE/BD \Rightarrow BD = x$ m.

     In $\triangle, \tan30^circ = CD/BD \Rightarrow BD = 10\sqrt{3} = x$

     Thus, $CE = 10 + 10\sqrt{3} = 27.32$ m.

     So height of the cliff is $27.32$ m and distance of cliff from the ship is $10$ m.