: i’m shantanu and am currently workingas a consultant. my task is to use math to help brian correctlycalculate the dimensions of a ramp he’s required to build, that joins his customer’sliving room to the kitchen. the ramp has to be exactly 13.2Ⱐto the horizontal,with the top being 72 cm vertical distance above the bottom. the ramp will be made from imported wood andthen stained, and the customer doesn’t want any gaps between the ramp and the kitchenfloor. i will now use the data given to create adiagram, not to scale. we know that brian is building a ramp, sowe can assume for now that the cross section
of the ramp is a right angled triangle, butmore on that later. the height is 72 cm and the angle here is13.2â°. now lets get to brian’s calculations. he used trigonometric functions to find x,the horizontal distance from the start of the ramp to the kitchen door. so this is brian's calculation for 'x' the horizontal distance from the start of the ramp to the kitchen door so this is the calculation using trigonometric functions the opposite, is like the height of it
72 cm is equivalent to 0.72 m he divided it by this, and thats what gave him the horizontal distance and that is = 3.1m. he rounded it to one decimal place after that, he used trigonometric functions to findy, the length of the ramp. so this is 'y' it is the hypotenuse basically, of the right angled triangle during his calculation, he used his previousx value which is rounded to 1 decimal place in meters, and he rounded his final answer to one decimal place finally, he found the angle the ramp makes with the kitchen floor wall
based on the 'x' and 'y' values found and he named this theta value to 1 decimal place in meters too. each of brian’s calculations is correctbut is accurate up to 1 decimal place (in metres) only. this means the lengths are only accurate tonearest 10 cm, which is a low accuracy for a room measurement. secondly he has rounded each intermediateanswer to 1 decimal place, due to which the final answer (the angle î¸) has a low degreeof accuracy (to the nearest 2â° as shown by
me later). thirdly he did not have to use a trigonometricmethod to calculate î¸ because the other two angles of the triangle are known. so how could brian improve the accuracy of his calculations? one way would be to take the complete valueof each intermediate calculation from the calculator and only round off to requireddegree of accuracy (doa) for the lengths and the angle î¸ at the end. my suggested approach is a little different. start by computing exact value of î¸ (fromthe fact that the other two angles of the
triangle are precisely known). then starting with brian’s calculation,increase doa of each intermediate calculation by 1 dp at a time and work out calculatedvalue of î¸. compare calculated value of î¸ with exactvalue of î¸. reasonable degree of accuracy will be foundwhen calculated value and exact value of î¸ almost fully agree. i’m using this approach because the investigationis about degree of accuracy and this approach will help me find a suitable degree of accuracy(doa). using the exact answers from the calculatorwill not help me do this.
here are all my calculations with variousdegrees of accuracy, with the respective differences between the exact and calculated values ofî¸. the exact value of î¸ = 180â°, the total angle of a triangle – (13.2â° + 90â°, since it's a right angled triangle). the angle the ramp forms with the wall supportingthe kitchen floor is 13.2â° and we assume that the side surface of the ramp is a rightangle triangle so the second angle will be 90â°. therefore:î¸ = 76.8â° for the calculation of î¸, the change in accuracyof the final answer is significant when the degree of accuracy of intermediate answersis increased.
for each of those calculations i includedthe percentage error compared to the actual answer found above, 76.8â°. therefore the calculation with the answerrounded to 4 decimal places is correct as it has 0% error. in addition, here is a graph showing the changein percentage error for each of my three calculations and brian’s one. my approach and the graph shows that if iuse brian’s calculation with measurements of length up to 4 d.p (in meters) or a doaof tenth of a millimeter, then the final value of î¸ will be accurate up to 3 decimal places,hence the percentage error will be 0%.
my conclusion is that calculations with doato the nearest tenth of a millimeter will be the desired level of doa. however, though it is possible to round offcalculations of length to the nearest tenth of a millimeter, it is not practical to measuresomething to that degree of accuracy, especially on a 3 meter piece of wood. (i’ll refer to a table of limitations later). therefore i will have to use my second calculationwhere i round my answers to the nearest millimeter. although that is less accurate, it is morepractical, as we are applying math in a real life situation.
however, brian claims his calculations arepretty close to mine. now i want to critique his calculations andgive some suggestions to brian. based on brian’s calculations shown, i wouldsay that they are inaccurate and they can’t be used for the dimensions of the wood. the degree of accuracy used for his calculationsare low. the lengths are rounded to the nearest 10centimeters. lets take a look at brian’s first calculationwhich is x (the horizontal distance from the start of the ramp to the kitchen door). his final answer is 3.1m, which is equivalentto 310 cm.
this means that the lower bound for x is 305cm, the upper bound is 315 cm and the range of inaccuracy is 10 cm which is significant. and similarly for y (the length of the ramp),his final answer is 3.2m, which is equivalent to 320cm. again the range of inaccuracy is 10cm. for the angle î¸, we have already seen thatthe inaccuracy is 1.8â°. brian needs to understand that these are biginaccuracies for measuring furniture and are not acceptable. if there is a mistake with any measurements,it will lead to severe consequences.
for instance, the ramp may not fit or be theright shape which can lead to customer disapproval which can then lead to wastage of the wood. the wood being used is expensive importedwood. therefore, if the wood is wasted, it willbe a huge loss of money for brian. it is also not advisable to build samplesdue to the price of the wood. therefore, there are some things which i wantto recommend to brian before he starts building the ramp. • brian calculated the angle î¸ the rampmakes with the kitchen floor walls. but he actually doesn’t have to worry aboutthat angle because in a right angled triangle,
knowing that the angle of inclination is 13.2â°,we can automatically derive the angle î¸. as we saw earlier it is 76.8 degrees. the only thing that is important is ensuringthat the dimensions and two angles are accurate. • brian should use the following measurementsfor the final dimensions of the ramp. as already justified, the best practical doabrian can use is to the nearest millimeter so: x = 307.0 cm and y = 315.3 cm. • we have to make a three-dimensional ramp. trigonometric functions can help to measurethe dimensions of the triangular cross section. however, he also has to find out the widthof the ramp’s walking surface and the door,
which he has to measure manually to the nearestmillimeter. he has to ensure that the same cross-sectiondimensions are used accurately for the entire ramp and that it accurately matches the cross-sectionmeasurements. • finally i suggest that brian reduces hislength measurements by 1mm before cutting the required wood. therefore he should use x = 306.9 cm and y= 315.2 cm to shape the piece of wood into the required ramp. the angles will remain the same. the reason for this is that brian is goingto use varnish to stain the wood so that it
matches with the wooden ceiling. the varnish itself might increase the thicknessup to approximately 0.5 mm. i will assume that the thickness is not morethan 1 mm. however in case the varnish coating measuresless than 1 mm there will be a tiny gap in between the ramp and the kitchen floor, whichwill not be acceptable. therefore i suggest that after bringing thefinal stained ramp to the customer’s house, brian should place the ramp in the requiredplace and apply additional coats of varnish on the sloping and vertical surfaces to eliminatethe gap. • brian could also consider alternate shapesfor his ramp.
for instance he could build a ramp where thedimensions are the same but the bottom part of the ramp is supported by stands insteadof having the full wood cross section covered. that would actually save the customer a lotof money and wood. however, i would recommend the solid rampwith a right angle triangle cross section as that will be the strongest and most stable. he could also build a ramp with a flat surfacejust before the kitchen door (quadrilateral cross section). now lets get to the assumptions and limitationsrelated to my work:the sources used for
finding out about paint or varnish thicknessare giving different results from different people (from 0.1mm up to 0.5mm). however their estimates of thickness of paintare all less than 1 mm. i have decided to assume maximum thicknessof 1 mm for the coat of varnish, in order to give brian a clear recommendation to makethe dimensions 1 mm less. this leads to the limitation that the paint/varnishwill not be exactly 1 mm thick. the chosen assumption isn’t fully validas the thickness of paint given by those sources are just rough estimations. in this investigation, we don’t actuallyknow what type of varnish he is using and
how thick it will be. two sources aren’t necessarily enough togive us the needed data. though my assumption is approximate, i canbe confident that the normal thickness of paint cannot be more than 1 mm. secondly my recommendation to brian is toapply additional varnish at the end to make the ramp the right size. however i have decided to assume a right angledtriangle cross section. it is the simplest in terms of measurements.
it doesn’t require any additional measurementsto be made. also it will give the strongest possible supportas a ramp. however in my recommendations to brian i willsuggest that he discusses other possible shapes with the customer. in this project, it would have been idealto measure dimensions to the nearest tenth of a millimeter. in real life this is theoretically possiblebut practically difficult. according to the first source, there are someinstruments like vernier calipers and micrometer gauges that can measure tiny lengths likethat, but they can be used only for very small
or thin objects, not a 3 meter long ramp. the second source mentions some laser measuringequipment but these must be expensive scientific equipment, not practical for cutting a pieceof wood. so we will have to use measurements to thenearest millimeter which reduces the accuracy. there will continue to be an unavoidable limitationin brian being able to make the ramp with the dimensions and angle to the required degreeof accuracy (nearest millimeter for lengths). we do know there is a requirement to varnishthe wood but we don’t know how thick it will be and how much it will affect the measurements. due to this we are forced to assume up to1mm thickness and to reduce this from the
lengths of the ramp. but there is a risk that this will lead togaps where the ramp touches the kitchen floor. the sources used is the same as in assumption#1. it is difficult to ensure that the same cross-sectiondimensions will accurately be used for the entire width of the ramp. if the cross-section shapes aren’t accurate,it can affect the shape of the whole ramp and lead to gaps, which will not be acceptable. despite all the assumptions and limitations,i helped brian as much as i could and this also shows how all the math we learn can beused in real life.
although nothing is 100% accurate, math isa great help in solving real life problems like brian’s ramp building.
of the ramp is a right angled triangle, butmore on that later. the height is 72 cm and the angle here is13.2â°. now lets get to brian’s calculations. he used trigonometric functions to find x,the horizontal distance from the start of the ramp to the kitchen door. so this is brian's calculation for 'x' the horizontal distance from the start of the ramp to the kitchen door so this is the calculation using trigonometric functions the opposite, is like the height of it
72 cm is equivalent to 0.72 m he divided it by this, and thats what gave him the horizontal distance and that is = 3.1m. he rounded it to one decimal place after that, he used trigonometric functions to findy, the length of the ramp. so this is 'y' it is the hypotenuse basically, of the right angled triangle during his calculation, he used his previousx value which is rounded to 1 decimal place in meters, and he rounded his final answer to one decimal place finally, he found the angle the ramp makes with the kitchen floor wall
based on the 'x' and 'y' values found and he named this theta value to 1 decimal place in meters too. each of brian’s calculations is correctbut is accurate up to 1 decimal place (in metres) only. this means the lengths are only accurate tonearest 10 cm, which is a low accuracy for a room measurement. secondly he has rounded each intermediateanswer to 1 decimal place, due to which the final answer (the angle î¸) has a low degreeof accuracy (to the nearest 2â° as shown by
me later). thirdly he did not have to use a trigonometricmethod to calculate î¸ because the other two angles of the triangle are known. so how could brian improve the accuracy of his calculations? one way would be to take the complete valueof each intermediate calculation from the calculator and only round off to requireddegree of accuracy (doa) for the lengths and the angle î¸ at the end. my suggested approach is a little different. start by computing exact value of î¸ (fromthe fact that the other two angles of the
triangle are precisely known). then starting with brian’s calculation,increase doa of each intermediate calculation by 1 dp at a time and work out calculatedvalue of î¸. compare calculated value of î¸ with exactvalue of î¸. reasonable degree of accuracy will be foundwhen calculated value and exact value of î¸ almost fully agree. i’m using this approach because the investigationis about degree of accuracy and this approach will help me find a suitable degree of accuracy(doa). using the exact answers from the calculatorwill not help me do this.
here are all my calculations with variousdegrees of accuracy, with the respective differences between the exact and calculated values ofî¸. the exact value of î¸ = 180â°, the total angle of a triangle – (13.2â° + 90â°, since it's a right angled triangle). the angle the ramp forms with the wall supportingthe kitchen floor is 13.2â° and we assume that the side surface of the ramp is a rightangle triangle so the second angle will be 90â°. therefore:î¸ = 76.8â° for the calculation of î¸, the change in accuracyof the final answer is significant when the degree of accuracy of intermediate answersis increased.
for each of those calculations i includedthe percentage error compared to the actual answer found above, 76.8â°. therefore the calculation with the answerrounded to 4 decimal places is correct as it has 0% error. in addition, here is a graph showing the changein percentage error for each of my three calculations and brian’s one. my approach and the graph shows that if iuse brian’s calculation with measurements of length up to 4 d.p (in meters) or a doaof tenth of a millimeter, then the final value of î¸ will be accurate up to 3 decimal places,hence the percentage error will be 0%.
my conclusion is that calculations with doato the nearest tenth of a millimeter will be the desired level of doa. however, though it is possible to round offcalculations of length to the nearest tenth of a millimeter, it is not practical to measuresomething to that degree of accuracy, especially on a 3 meter piece of wood. (i’ll refer to a table of limitations later). therefore i will have to use my second calculationwhere i round my answers to the nearest millimeter. although that is less accurate, it is morepractical, as we are applying math in a real life situation.
however, brian claims his calculations arepretty close to mine. now i want to critique his calculations andgive some suggestions to brian. based on brian’s calculations shown, i wouldsay that they are inaccurate and they can’t be used for the dimensions of the wood. the degree of accuracy used for his calculationsare low. the lengths are rounded to the nearest 10centimeters. lets take a look at brian’s first calculationwhich is x (the horizontal distance from the start of the ramp to the kitchen door). his final answer is 3.1m, which is equivalentto 310 cm.
this means that the lower bound for x is 305cm, the upper bound is 315 cm and the range of inaccuracy is 10 cm which is significant. and similarly for y (the length of the ramp),his final answer is 3.2m, which is equivalent to 320cm. again the range of inaccuracy is 10cm. for the angle î¸, we have already seen thatthe inaccuracy is 1.8â°. brian needs to understand that these are biginaccuracies for measuring furniture and are not acceptable. if there is a mistake with any measurements,it will lead to severe consequences.
for instance, the ramp may not fit or be theright shape which can lead to customer disapproval which can then lead to wastage of the wood. the wood being used is expensive importedwood. therefore, if the wood is wasted, it willbe a huge loss of money for brian. it is also not advisable to build samplesdue to the price of the wood. therefore, there are some things which i wantto recommend to brian before he starts building the ramp. • brian calculated the angle î¸ the rampmakes with the kitchen floor walls. but he actually doesn’t have to worry aboutthat angle because in a right angled triangle,
knowing that the angle of inclination is 13.2â°,we can automatically derive the angle î¸. as we saw earlier it is 76.8 degrees. the only thing that is important is ensuringthat the dimensions and two angles are accurate. • brian should use the following measurementsfor the final dimensions of the ramp. as already justified, the best practical doabrian can use is to the nearest millimeter so: x = 307.0 cm and y = 315.3 cm. • we have to make a three-dimensional ramp. trigonometric functions can help to measurethe dimensions of the triangular cross section. however, he also has to find out the widthof the ramp’s walking surface and the door,
which he has to measure manually to the nearestmillimeter. he has to ensure that the same cross-sectiondimensions are used accurately for the entire ramp and that it accurately matches the cross-sectionmeasurements. • finally i suggest that brian reduces hislength measurements by 1mm before cutting the required wood. therefore he should use x = 306.9 cm and y= 315.2 cm to shape the piece of wood into the required ramp. the angles will remain the same. the reason for this is that brian is goingto use varnish to stain the wood so that it
matches with the wooden ceiling. the varnish itself might increase the thicknessup to approximately 0.5 mm. i will assume that the thickness is not morethan 1 mm. however in case the varnish coating measuresless than 1 mm there will be a tiny gap in between the ramp and the kitchen floor, whichwill not be acceptable. therefore i suggest that after bringing thefinal stained ramp to the customer’s house, brian should place the ramp in the requiredplace and apply additional coats of varnish on the sloping and vertical surfaces to eliminatethe gap. • brian could also consider alternate shapesfor his ramp.
for instance he could build a ramp where thedimensions are the same but the bottom part of the ramp is supported by stands insteadof having the full wood cross section covered. that would actually save the customer a lotof money and wood. however, i would recommend the solid rampwith a right angle triangle cross section as that will be the strongest and most stable. he could also build a ramp with a flat surfacejust before the kitchen door (quadrilateral cross section). now lets get to the assumptions and limitationsrelated to my work:the sources used for
finding out about paint or varnish thicknessare giving different results from different people (from 0.1mm up to 0.5mm). however their estimates of thickness of paintare all less than 1 mm. i have decided to assume maximum thicknessof 1 mm for the coat of varnish, in order to give brian a clear recommendation to makethe dimensions 1 mm less. this leads to the limitation that the paint/varnishwill not be exactly 1 mm thick. the chosen assumption isn’t fully validas the thickness of paint given by those sources are just rough estimations. in this investigation, we don’t actuallyknow what type of varnish he is using and
how thick it will be. two sources aren’t necessarily enough togive us the needed data. though my assumption is approximate, i canbe confident that the normal thickness of paint cannot be more than 1 mm. secondly my recommendation to brian is toapply additional varnish at the end to make the ramp the right size. however i have decided to assume a right angledtriangle cross section. it is the simplest in terms of measurements.
it doesn’t require any additional measurementsto be made. also it will give the strongest possible supportas a ramp. however in my recommendations to brian i willsuggest that he discusses other possible shapes with the customer. in this project, it would have been idealto measure dimensions to the nearest tenth of a millimeter. in real life this is theoretically possiblebut practically difficult. according to the first source, there are someinstruments like vernier calipers and micrometer gauges that can measure tiny lengths likethat, but they can be used only for very small
or thin objects, not a 3 meter long ramp. the second source mentions some laser measuringequipment but these must be expensive scientific equipment, not practical for cutting a pieceof wood. so we will have to use measurements to thenearest millimeter which reduces the accuracy. there will continue to be an unavoidable limitationin brian being able to make the ramp with the dimensions and angle to the required degreeof accuracy (nearest millimeter for lengths). we do know there is a requirement to varnishthe wood but we don’t know how thick it will be and how much it will affect the measurements. due to this we are forced to assume up to1mm thickness and to reduce this from the
lengths of the ramp. but there is a risk that this will lead togaps where the ramp touches the kitchen floor. the sources used is the same as in assumption#1. it is difficult to ensure that the same cross-sectiondimensions will accurately be used for the entire width of the ramp. if the cross-section shapes aren’t accurate,it can affect the shape of the whole ramp and lead to gaps, which will not be acceptable. despite all the assumptions and limitations,i helped brian as much as i could and this also shows how all the math we learn can beused in real life.
although nothing is 100% accurate, math isa great help in solving real life problems like brian’s ramp building.
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