29. Height and Distance Solutions Part 2

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51st problem

Let ABAB be the lamp post having height hh m, and BDBD be the girl having height 1.61.6 m. The distance of the grl from the lamp post is AC=3.2AC = 3.2 m. CECE is the langeth of the shadow given as 4.84.8 m. In the ABE\triangle ABE and CDE,E\triangle CDE, \angle E is common, A=C=90\angle A = \angle C = 90^\circ so third angle will be also equal. This makes the triangles similar.

ABCD=AECEh=83\therefore \frac{AB}{CD} = \frac{AE}{CE} \Rightarrow h = \frac{8}{3} m.

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52nd problem

Let ACAC be the building having a height of 3030 m. Let EE and GG point of observations where angles of elevation are 6060^\circ and 3030^\circ respectively. Let AEFAEF be the line of foot of the building and foot of the observer which is a horizontal line. Let DEDE and FEFE are the heights of the observer. Draw BEGADFBEG\parallel ADF so that AB=DE=FG=1.5AB = DE = FG = 1.5 m. Thus, BC=28.5BC = 28.5 m. We have to find DF=EGDF = EG.

In BCE,tan60=3=BCCECE=28.53\triangle BCE, \tan60^\circ = \sqrt{3} = \frac{BC}{CE} \Rightarrow CE = \frac{28.5}{\sqrt{3}} m.

In BCG,tan30=13=BCCGCG=28.53\triangle BCG, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{CG} \Rightarrow CG = 28.5\sqrt{3} m.

Thus, DF=EG=CGCF=573DF = EG = CG - CF = \frac{57}{\sqrt{3}} m, which is the distance walked by the observer.

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53rd problem

Let the height of the tower ABAB is hh m. When the altitude of the sun is 6060^\circ let the length of the shadown be AC=xAC = x m. Then according to question length of shadow when the sun’s altitude i 3030^\circ the length of shadow will be AD,40AD, 40 m longer i.e. AD=x+40AD = x + 40.

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

In ABC,tan30=13=ABAC=hx+40x=20\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AC} = \frac{h}{x + 40} \Rightarrow x = 20 m and h=203h = 20\sqrt{3} m.

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54th problem

Let ABAB be the building with 2020 m height. Let the height of tower be hh m represented by BCBC in the figure. Let DD be the point of observation at a distance xx from the foot of the building ABAB.

In ABD,tan30=13=ABAD=20xx=203\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{20}{x} \Rightarrow x = 20\sqrt{3} m.

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

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55th problem

Let DEDE be the building having a height of 88 m. Let ACAC be the multistoried building having height h+8h + 8 m. Foot of both the buildings are joined on horizontal plane i.e. ADAD. Draw a line parallel to ADAD which is BEBE. So BEBE is equal to ADAD which we have let as xx m. Clearly, AB=8AB = 8 m. Let height of BCBC to be hh m.

In CBE,tan30=13=BCBE=hx3h=x\triangle CBE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{BE} = \frac{h}{x} \Rightarrow \sqrt{3}h = x.

In ACD,tan30=13=ACAD=h+8xh=831h+8=8331\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.

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56th problem

Let ABAB be the pedestal having height hh m and BCBC be the statue having height 1.61.6 m on top of pedestal. Let DD be the point of observation from where the angles of elevation as given in the question are 4545^\circ and 6060^\circ.

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

In ACD,tan60=3=ACCD=h+1.6xh=1.631\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{CD} = \frac{h + 1.6}{x} \Rightarrow h = \frac{1.6}{\sqrt{3} - 1} m.

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

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58th problem

Let ABAB be the tower having height 7575 m. Let CC and DD 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=xAC = x m and distance between the ships as dd m.

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

In ABC,tan30=13=75x+dd=75(31)\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{75}{x + d} \Rightarrow d = 75(\sqrt{3} - 1) m.

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59th problem

Let ABAB be the building and CDCD be thw tower having height 5050 m. The angles of elevation are shown as given in the question. Let distance between the foot of the tower and the building be dd m and height of the building be hh m.

In ABC,tan30=13=hdd=3h\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{d} \Rightarrow d = \sqrt{3}h.

In ABD,tan60=3=50d3h=50h=503\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{50}{d} \Rightarrow 3h = 50 \Rightarrow h = \frac{50}{3} m.

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60th problem

Let DEDE represent the banks of river and BCBC the bridge. Given that height of the bridge is 3030 m. BD=CE=30\therefore BD = CE = 30 m. The angles of depression from point AA is shown as given in the question. We have to find DE=BCDE = BC i.e. width of the river.

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

In ABD,tan30=13=BDABAB=303\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BD}{AB} \Rightarrow AB = 30\sqrt{3} m.

Thus, width of river =30+303=30(3+1)= 30 + 30\sqrt{3} = 30(\sqrt{3} + 1) m

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61st problem

Let BCBC and DEDE be the poles, both having the same height hh m. Let BDBD be the road having a width of 8080 m. Also, let AA be the point of observation from which angles of observation are shown as given in the question. Let AB=xAB = x m and thus AD=80xAD = 80 - x m.

In ABC,tan60=3=BCABBC=h=x3\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{BC}{AB} \Rightarrow BC = h = x\sqrt{3} m.

In ADE,tan30=13=DEADDE=h=80x3\triangle ADE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{DE}{AD} \Rightarrow DE = h = \frac{80 - x}{\sqrt{3}} m.

Since the poles have equal height, therefore 3x=80x3x=20\sqrt{3}x = \frac{80 - x}{\sqrt{3}}\Rightarrow x = 20 m.

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

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62nd problem

Let ABAB be the tree having height 2020 m, CDCD the width of river. Let CC and DD be the foot of the poles on opposite banks of the river such that foot of the poles and the tree are in the same horizontal plane. Let AC=xAC = x and AD=yAD = y m. The angles of depression are shown as given in the question.

In ABC,tan60=3=ABACAC=203\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} \Rightarrow AC = \frac{20}{\sqrt{3}} m.

In ABD,tan30=13=ABADAD=203\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} \Rightarrow AD = 20\sqrt{3} m.

Thus, CD=AC+AD=803CD = AC + AD = \frac{80}{\sqrt{3}} m.

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63rd problem

Let ABAB be the tower having height hh m, BCBC be the flag staff having height 77 m. Also, let point of observation at DD having angles of observation as 3030^\circ and 4545^\circ as given in the question such that AD=xAD = x m.

In ABD,tan30=13=ABAD=hx3h=x\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{h}{x} \Rightarrow \sqrt{3}h = x.

In ACD,tan45=1=ACAD=h+7xh=731\triangle ACD, \tan45^\circ = 1 = \frac{AC}{AD} = \frac{h + 7}{x} \Rightarrow h = \frac{7}{\sqrt{3} - 1} m.

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64th problem

Let ACAC and ADAD be the length of shadows with angles of depression 4545^\circ and 3030^\circ respectively such that CD=2xCD = 2x m. Also let ABAB as the tower having height hh m and ACAC as the length of shadow with angle of depression 4545^\circ having length dd m.

In ABC,tan45=1=ABAC=hdh=d\triangle ABC, \tan45^\circ = 1 = \frac{AB}{AC} = \frac{h}{d} \Rightarrow h = d m.

In ABD,tan30=13=ABAD=h2x+dh=2x31=2x(3+1)31=x(3+1)\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{h}{2x + d} \Rightarrow h = \frac{2x}{\sqrt{3} - 1} = \frac{2x(\sqrt{3} + 1)}{3 - 1} = x(\sqrt{3} + 1) m.