Grade 9 / Trigonometry 50 exercises (100% corrected)

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ABCMNPRST ABC ABCD ChingQuizz : 8 exercises available for Quizz assessment : 1. Definition of trigonometric ratios E.6438 Definition: in a right-angled triangle, we have the follow-ing trigonometric ratios : cos R = longueur of the side adjacent to the angle S longueur of the hypothenuse sin W = longueur from the side opposite the W longueur angle of the hypothenuse tan M = longueur of the side opposite the angle M longueur of the side adjacent to the angle M Consider the three triangles ABC , MNP , RST shown below : Using the lengths of the triangles, express the following trigonometric ratios : a cos CAB b sin PNM c tan TSR E.6213 1 a Draw a triangle ABC rectangular in C . b Given the lengths of the sides of the triangle ABC , ex-press the trigonometric ratio of the sine of the angle ∠ ABC . 2 a Draw a triangle DEF rectangular in E . b Given the lengths of the sides of the triangle DEF , express the trigonometric ratio of the tangent of the angle ∠ EDF . E.711 Consider the right triangle ABC at C shown below : 1 Using the measurements taken from the figure above, complete the table with the measurements of the sides of triangle ABC : Angle Adjacent side Opposite side Hypotenuse CAB CBA 2 a By measuring the lengths on the triangle ABC shown above, fill in each cell of the table below with the quotient that defines the value you are looking for, fol-lowed by its value rounded to the nearest hundredth : ¸ cos ¸ sin ¸ tan ¸ CAB ≈ ≈ ≈ CBA ≈ ≈ ≈ b Using a trigonometric table, determine a rounded value for the measures of angles CAB and ABC , rounded to the nearest degree. 3 Using a protractor, verify the accuracy of the results ob-tained in question 2 b . E.11717 1 Dans le triangle ABC , quel est le côté op-posé à l’angle CBA ? 2 Dans le triangle ABD , quel est le côté ad-jacent à l’angle ADB ? 2. Trigonometric ratios https://chingmath.fr chapExoCorrec/6438 sacados/6438 ABCMNPRST chapExoCorrec/6213 sacados/6213 chapExoCorrec/711 sacados/711 ABC chapExoCorrec/11717 sacados/11717 ABCD
ABC4cm52oDEF5cm37o 62oABCx5cm30oEFDx5cm50oHIGx3cm XYZx3cm40oRQPx3cm28oCABx3cm30o ABC6;2m5;8m 3cm4cmABCDE55o35o45o ABCD5cm20o E.5670 We consider the two triangles below : Determine the measures of segments [ AC ] and [ DF ] rounded to the nearest millimeter. E.721 In each case, give the length x of the side indicated. The result should be rounded to the nearest millimetre: E.2261 For each triangle shown below, de-termine the unknown length shown to the nearest millimeter: E.10979 Consider the triangle ABC right-angled at C . Determine the measure of the angle BAC rounded to the nearest tenth of a degree. E.11718 On considère la configuration ci-dessous : 1 Dans le triangle ABC , déterminer la longueur du seg-ment [ BC ] . 2 Dans le triangle ACD , déterminer la longueur du seg-ment [ CD ] . 3 Dans le triangle ADE , déterminer la longueur du seg-ment [ DE ] . Indication : on arrondira les résultats au millimètre près. 3. Trigonometric ratios, problems and models E.4933 Consider the rectangle ABCD be-low : https://chingmath.fr chapExoCorrec/5670 sacados/5670 Fait intervenir le cosinus ABC4cm52oDEF5cm37o chapExoCorrec/721 sacados/721 62oABCx5cm30oEFDx5cm50oHIGx3cm chapExoCorrec/2261 sacados/2261 XYZx3cm40oRQPx3cm28oCABx3cm30o chapExoCorrec/10979 sacados/10979 ABC6;2m5;8m chapExoCorrec/11718 sacados/11718 3cm4cmABCDE55o35o45o chapExoCorrec/4933 sacados/4933 ABCD5cm20o
HCS15o P55o5,8mABAvalAmontSensducourantPortes"busquees"de memelongueur AB12m30o45o324m CABCDEFGHOI¸o˛o˛o4cm AHCB15m17m21o Without using the Pythagorean theorem, determine the perimeter of the rectangle ABCD rounded to the nearest mil-limeter. E.4940 An explorer arrives in front of the pyramid of Kheops. He places his measuring instruments (theodolite) at the point H . Studying the pyramid, he observes that it is a regular pyramid: the foot C of the height from the top S is also the center of the base. It also estimates the distance HC to 511 m . From point H to vertex S , his measuring instruments reveal an angle of 15 o . Determine the measurement, rounded to the nearest meter, of the height SC of the pyramid of Khufu. E.6279 Some locks have doors known as ˇbusquésı which form an angle pointing upstream in order to resist water pressure. Using the diagram above, determine the length of the gates, rounded to the nearest cm . Hint: if the work is not finished, leave a trace of your re-search anyway. It will be taken into account in the grading. E.720 The construction of the Eiffel Tower was completed in 1899 . With a mast bearing the French flag its height was 312 m . In 2005 , the installation of a television antenna increased the size of the Eiffel Tower to 324 m . The figure below shows the Eiffel Tower and 2 houses : 1 Reproduce, in the form of a simplified diagram, the figure below on your sheet. 2 Calculate the distance AB separating the two houses. E.4936 The ABCDEFGH polygon, shown below, is a regular octagon. This means that the circle C with center O passes through all the vertices of this polygon and that the 8 triangles OAB , OBC , OCD , ODE , OEF , OFG , OGH , and OHA are identical. 1 a Determine the measure of the angle ∠ BOC . b Deduce the measures of the angles of triangle OBC . 2 Note I the foot of the height of the triangle OHA from the vertex O . a Determine the measurement to the nearest millimeter of segment [ AI ] . b Give the measure of segment [ AH ] . Justify your state-ment. c Deduce the perimeter of the hexagon ABCDEFGH . 4. Trigonometric ratios, Pythagorean theorem E.9316 Consider the triangle ABC shown below where the point H is the foot of the height from the vertex C : https://chingmath.fr chapExoCorrec/4940 sacados/4940 HCS15o chapExoCorrec/6279 sacados/6279 P55o5,8mABAvalAmontSensducourantPortes"busquees"de memelongueur chapExoCorrec/720 sacados/720 AB12m30o45o324m chapExoCorrec/4936 sacados/4936 CABCDEFGHOI¸o˛o˛o4cm chapExoCorrec/9316 sacados/9316 AHCB15m17m21o
ABDEFG60o ABCDDépartArrivée4;8km5;6km24oSensduvent 45o ¸cos¸¸cos¸¸cos¸¸cos¸¸cos¸¸cos¸199990,99980,99970,99940,99900,99860,99810,99760,99690,99620,99540,99450,99360,99250,99140,99030,98900,98770,98630,98480,98330,98160,97990,97810,97630,97440,97240,97030,96810,96590,96360,96130,95880,95630,95370,95110,94830,94550,94260,93970,93670,93360,93040,92720,92390,92050,91710,91350,91000,90630,90260,89880,89490,89100,88700,88290,87880,87460,87040,86600,86160,85720,85260,84800,84340,83870,83390,82900,82410,81920,81410,80900,80390,79860,79340,78800,78260,77710,77160,76600,76040,75470,74900,74310,73730,73140,72540,71930,71330,70710,70090,69470,68840,68200,67560,66910,66260,65610,64940,64280,63610,62930,62250,61570,60880,60180,59480,58780,58070,57360,56640,55920,55190,54460,53730,52990,52250,51500,50750,50000,49240,48480,47720,46950,46170,45400,44620,43840,43050,42260,41470,40670,39870,39070,38270,37460,36650,35840,35020,34200,33380,32560,31730,30900,30070,29240,28400,27560,26720,25880,25040,24190,23340,22500,21640,20790,19940,19080,18220,17360,16500,15640,14780,13920,13050,12190,11320,10450,09580,08720,07850,06980,06100,05230,04360,03490,02620,01750,00870, whose dimensions are known : AC = 17 m ; AH = 15 m ; ∠ CBH = 21 o 1 Determine the length [ HC ] . 2 Determine, to the nearest decimeter, the length of seg-ment [ BH ] . 3 Give the area of triangle ABC to the nearest square me-ter. E.3547 We know that : EF = 4 cm ; FG = 3 cm ; EG = 5 cm AE = 7 cm ; ∠ DAB = 60 o ; The points A , E and G are aligned ; the points D , E and F are aligned ; ( AB ) is the height originating from A in the triangle AED . Consider the figure above (dimensions not respected) : 1 Demonstrate that EFG is a right-angled triangle. 2 Deduce that ( FG ) is parallel to ( AB ) . 3 Demonstrate that EB =5.6 cm and AB =4.2 cm . 4 In the DAB triangle, calculate BD , then calculate DE . We’ll give an approximate value of these two numbers to the nearest millimetre. 5 Calculate the area of triangle AED to the nearest 1 cm 2 . E.9313 When a sailboat is upwind, it cannot move forward. If the chosen destination requires heading into the wind, the sail-boat will have to progress by zigzagging. Compare the trajectories of these two sailboats by calculating the distance, in kilometers and rounded to the tenth that each has traveled. E.5779 A city sprawls across the plain and on top of the cliff above it. To facilitate travel through the city, a cable car was built in 1962 . The cable used measured 425 m and, when stretched, formed an angle of 45 o with the level formed by the ground. For safety reasons, the cable car station on the cliff must be moved back 60 m . What will be the new length, rounded to the nearest meter, of the cable connecting the two stations on this beltway? 5. Introduction to reciprocal trigonometric ratios E.6214 Here is a trigonometric table of cosines : https://chingmath.fr chapExoCorrec/3547 sacados/3547 ABDEFG60o chapExoCorrec/9313 sacados/9313 ABCDDépartArrivée4;8km5;6km24oSensduvent chapExoCorrec/5779 sacados/5779 45o chapExoCorrec/6214 sacados/6214 ¸cos¸¸cos¸¸cos¸¸cos¸¸cos¸¸cos¸199990,99980,99970,99940,99900,99860,99810,99760,99690,99620,99540,99450,99360,99250,99140,99030,98900,98770,98630,98480,98330,98160,97990,97810,97630,97440,97240,97030,96810,96590,96360,96130,95880,95630,95370,95110,94830,94550,94260,93970,93670,93360,93040,92720,92390,92050,91710,91350,91000,90630,90260,89880,89490,89100,88700,88290,87880,87460,87040,86600,86160,85720,85260,84800,84340,83870,83390,82900,82410,81920,81410,80900,80390,79860,79340,78800,78260,77710,77160,76600,76040,75470,74900,74310,73730,73140,72540,71930,71330,70710,70090,69470,68840,68200,67560,66910,66260,65610,64940,64280,63610,62930,62250,61570,60880,60180,59480,58780,58070,57360,56640,55920,55190,54460,53730,52990,52250,51500,50750,50000,49240,48480,47720,46950,46170,45400,44620,43840,43050,42260,41470,40670,39870,39070,38270,37460,36650,35840,35020,34200,33380,32560,31730,30900,30070,29240,28400,27560,26720,25880,25040,24190,23340,22500,21640,20790,19940,19080,18220,17360,16500,15640,14780,13920,13050,12190,11320,10450,09580,08720,07850,06980,06100,05230,04360,03490,02620,01750,00870,
ABC9cm9;4cm ¸sin¸¸sin¸¸sin¸¸sin¸¸sin¸¸sin¸00870,02620,04360,06100,07850,09580,11320,13050,14780,16500,18220,19940,21640,23340,25040,26720,28400,30070,31730,33380,35020,36650,38270,39870,41470,43050,44620,46170,47720,49240,50750,52250,53730,55190,56640,58070,59480,60880,62250,63610,64940,66260,67560,68840,70090,71330,72540,73730,74900,76040,77160,78260,79340,80390,81410,82410,83390,84340,85260,86160,87040,87880,88700,89490,90260,91000,91710,92390,93040,93670,94260,94830,95370,95880,96360,96810,97240,97630,97990,98330,98630,98900,99140,99360,99540,99690,99810,99900,99970,99990, ABC3cm2;8cm ¸tan¸¸tan¸¸tan¸¸tan¸¸tan¸¸tan¸00,01750,03490,05240,06990,08750,10510,12280,14050,15840,17630,19440,21260,23090,24930,26790,28670,30570,32490,34430,36400,38390,40400,42450,44520,46630,48770,50950,53170,55430,57740,60090,62490,64940,67450,70020,72650,75360,78130,80980,83910,86930,90040,93250,965711,03551,07241,11061,15041,19181,23491,27991,32701,37641,42811,48261,53991,60031,66431,73211,80401,88071,96262,05032,14452,24602,35592,47512,60512,74752,90423,07773,27093,48743,73214,01084,33154,70465,14465,67136,31387,11548,14439,514411,43014,30019,81128,63657,290 ABC3cm4;5cm¸o˛o4cm3;5cmDEF EDF6cm4cm˛oCBA4;5cm3cm¸o EDF3cm5;5cm˛CBA6cm4;5cm¸ ABC6cm3;2cm Consider the triangle ABC below, which is a right triangle at C : 1 Establish that : BC BA ≈ 0 ; 9574 2 Deduce an approximation of the angle ABC to the near-est half degree. E.9314 We give the trigonometric table of the sine : Consider the triangle ABC right-angled C shown below : Determine the measure of the angle ∠ ABC . E.9315 We give the trigonometric table of the sine : Consider the triangle ABC rectangular to C and verifying: CA CB = 3.2 Which of the following is the correct frame : a 17 < ∠ CAB< 18 b 18 < ∠ CAB< 19 c 19 < ∠ CAB< 20 6. Reciprocal trigonometric relations E.5669 Calculate the rounding to the nearest tenth of a degree of the angles ∠ ABC and ∠ EDF shown below : E.714 Calculate the angles ABC and EDF shown below to the nearest tenth of a degree : E.3602 Calculate the rounding to the near-est tenth of a degree of the angles ∠ ABC and ∠ EDF shown below : E.4934 Consider the triangle ABC right-angled B shown below : Determine the measures of the angles ∠ BCA and ∠ CAB rounded to the nearest tenth of a degree. https://chingmath.fr ABC9cm9;4cm chapExoCorrec/9314 sacados/9314 ¸sin¸¸sin¸¸sin¸¸sin¸¸sin¸¸sin¸00870,02620,04360,06100,07850,09580,11320,13050,14780,16500,18220,19940,21640,23340,25040,26720,28400,30070,31730,33380,35020,36650,38270,39870,41470,43050,44620,46170,47720,49240,50750,52250,53730,55190,56640,58070,59480,60880,62250,63610,64940,66260,67560,68840,70090,71330,72540,73730,74900,76040,77160,78260,79340,80390,81410,82410,83390,84340,85260,86160,87040,87880,88700,89490,90260,91000,91710,92390,93040,93670,94260,94830,95370,95880,96360,96810,97240,97630,97990,98330,98630,98900,99140,99360,99540,99690,99810,99900,99970,99990, ABC3cm2;8cm chapExoCorrec/9315 sacados/9315 ¸tan¸¸tan¸¸tan¸¸tan¸¸tan¸¸tan¸00,01750,03490,05240,06990,08750,10510,12280,14050,15840,17630,19440,21260,23090,24930,26790,28670,30570,32490,34430,36400,38390,40400,42450,44520,46630,48770,50950,53170,55430,57740,60090,62490,64940,67450,70020,72650,75360,78130,80980,83910,86930,90040,93250,965711,03551,07241,11061,15041,19181,23491,27991,32701,37641,42811,48261,53991,60031,66431,73211,80401,88071,96262,05032,14452,24602,35592,47512,60512,74752,90423,07773,27093,48743,73214,01084,33154,70465,14465,67136,31387,11548,14439,514411,43014,30019,81128,63657,290 chapExoCorrec/5669 sacados/5669 Fait intervenir le cosinus ABC3cm4;5cm¸o˛o4cm3;5cmDEF chapExoCorrec/714 sacados/714 EDF6cm4cm˛oCBA4;5cm3cm¸o chapExoCorrec/3602 sacados/3602 EDF3cm5;5cm˛CBA6cm4;5cm¸ chapExoCorrec/4934 sacados/4934 ABC6cm3;2cm
EDF6cm4cm˛oCBA4;5cm3cm¸o ABDEF 4;5cm2;1cmABC¸3;6cm5;6cmDEF¸2;7cm4;2cm¸GHI ABC34o5cm ABCD60o¸4cm8;9cm E.708 Consider the two right-angled trian-gles ABC and DEF below : Determine the measure of angles ¸ and ˛ . E.10980 In the figure opposite : points E , A , and F are aligned ; points E , B , and D are aligned ; lines ( FD ) and ( AB ) are parallel; AE = 4.4 cm ; EB = 3.3 cm ; AB = 5.5 cm BD = 6.6 cm 1 Prove that triangle ABE is a right triangle. 2 Calculate the measure of angle ABE , rounded to the nearest degree. 3 Calculate the length FD . E.11719 On considère les trois triangles : 1 Dans chacun des triangles, donner le nom de l’angle codé par la lettre ¸ . 2 Dans chacun des triangles, donner la mesure de l’angle ¸ arrondi au degré près. 7. Direct trigonometric ratio and reciprocal E.5192 The triangle ABC is a right-angled trian-gle in B verifying: AC = 6 cm ; ∠ BAC = 34 o 1 Determine, to the nearest millimetre, the measure of segment [ BC ] . 2 Give, to the nearest square centimeter, the area of the triangle ABC . E.2275 Consider the triangle ABC right-angled B shown below : 1 Determine the length of segment [ BC ] rounded to the nearest millimeter. 2 Derive the measure of the angle ∠ CDB rounded to the nearest degree. https://chingmath.fr chapExoCorrec/708 sacados/708 EDF6cm4cm˛oCBA4;5cm3cm¸o chapExoCorrec/10980 sacados/10980 ABDEF chapExoCorrec/11719 sacados/11719 4;5cm2;1cmABC¸3;6cm5;6cmDEF¸2;7cm4;2cm¸GHI chapExoCorrec/5192 sacados/5192 ABC34o5cm chapExoCorrec/2275 sacados/2275 ABCD60o¸4cm8;9cm
HRJ ABCH MIJKL CDBA30o¸o3cm2cm E.719 The unit of length is the meter. The drawing is not to scale. 1 Romeo ( R ) wants to join Juliet ( J ) at her window. To do this, he places a ladder [ JR ] against the wall [ JH ] . The wall and floor are perpendicular. We give : HR =3 ; JH =4 . a Calculate JR . b Calculate cos HJR then the value of the angle HJR rounded to the degree. 2 The ladder slides : it changes position (we then note J the point of support of the ladder against the wall) . We give : JR =5 and HJR =40 o . a Calculate HR (give value rounded to the tenth) b Write the expression for tan HJR then calculate JH (give the value rounded to the tenth) . E.723 The figure is not made full size. It is not to be reproduced AHC is a right-angled triangle in H . The line through A is perpendicular to the line ( AC ) inter-sects the line ( HC ) at B . We know that : AH =4.8 cm ; HC =6.4 cm 1 a Justify equality: ACH =90 o − HAC b Justify equality: BAH =90 o − HAC c What can be deduced for the angles ACH and BAH ? 2 a Show that : tan( ACH ) = 3 4 b Using the triangle BAH , express tan( BAH ) as a func- tion of BH . 3 Deduce from questions 1 and 2 that : BH =3.6 cm 4 Calculate the measure in degrees, rounded to the degree, of the angle ACH . E.2281 Consider the figure opposite which is not full size : The segments [ KL ] and [ JM ] intersect at point I ; IK =4 cm ; JK =2.4 cm ; LM =4.2 cm . triangle IJK is right-angled K ; triangle LIM is right-angled at M . 1 Calculate the exact value of the tangent of the angle ∠ KIJ . 2 Why are the angles ∠ KIJ and ∠ LIM equal? 3 Give the expression for the tangent of the angle ∠ LIM as a function of IM 4 Using the answers to the previous questions, prove that the length IM in centimeters is a whole number. 5 Determine the rounding to the degree of the angle ∠ KIJ . E.725 The figure opposite is composed of the triangles ABC and BDC rectangle in B and D respectively. Give the value of angle ¸ to the near-est tenth. 8. Trigonometry and Pythagorean theorem E.722 1 Draw the triangle REC such that : RE =7.5 cm ; RC =10 cm ; EC =12.5 cm 2 Show that the triangle REC is rectangular to R . 3 Give the values, rounded to the nearest degree, of the angles of this triangle. https://chingmath.fr chapExoCorrec/719 sacados/719 HRJ chapExoCorrec/723 sacados/723 Moyen Orient ou Groupement Est - Juin 2005 - ? points ABCH chapExoCorrec/2281 sacados/2281 MIJKL chapExoCorrec/725 sacados/725 CDBA30o¸o3cm2cm chapExoCorrec/722 sacados/722
ABC¸o ABCD4cm6cm60o UOMAI E.6415 We consider the car represented be-low : It is assumed that the light emitted by its headlight can be considered to be emitted from a single point A and that with the current setting the headlight illuminates horizontally. It is desired to lower the headlight by an angle ¸ so that the emitted light reaches but does not exceed the point C . Here are some measurements obtained : The lighthouse is at a height of 1.1 m from the ground. The point C in front of the car at a distance of 6 m 1 Using the points A , B , and C , denote the lengths with values 1.1 m , and 6 m . 2 Determine the measure of the angle ¸ of tilt of the head-light so that it reaches the point C , rounded to the near-est tenth of a degree. E.3585 Given : BD =4 cm ; BA =6 cm ; DBC =60 o You are not required to draw a full-scale diagram. 1 Show that : BC =8 cm . 2 Calculate the length CD , rounded to the nearest millime-ter. 3 Calculate AC . 4 What is the value of tan BAC ? 5 Deduce the measure of angle BAC , rounded to the near-est degree. 9. Trigonometry, Pythagoras and/or Thales theorem E.712 Questions are independent of each other MNP is a right triangle at P such that : MP = 5 cm ; MN = 7 cm 1 Calculate the measure, rounded to the degree, of the an-gle ∠ MNP . 2 Calculate the exact value of NP ; Give its value rounded to the mm . 3 Let I be the point on segment [ MP ] such that PI =2 cm . The parallel to ( MN ) passing through I intersects [ PN ] at J . Compute IJ . E.716 Segments [ OA ] and [ UI ] inter-sect in M . We have : MO =21 ; MA =27 ; MU =28 ; MI =36 ; AI =45 (the unit of length is the millimetre) 1 Prove that the straight lines ( OU ) and ( AI ) are parallel 2 Calculate the length OU . 3 Prove that the triangle AMI is a right-angled triangle. 4 Determine, to the nearest degree, the measure of the an-gle AIM 5 Show that the angles MAI and MOU have the same measure. https://chingmath.fr chapExoCorrec/6415 sacados/6415 ABC¸o chapExoCorrec/3585 sacados/3585 Amerique du Nord Juin 2009 ABCD4cm6cm60o chapExoCorrec/712 sacados/712 chapExoCorrec/716 sacados/716 Groupe Est - 2004 - 5 points UOMAI
AOSIMN ABCDE 70m50m30m60oABCDEM E.713 The unit of length is the meter The drawing opposite shows a cross-section of a house. The triangle MAI is isosceles, with vertex M . The line perpendicular to the line ( AI ) , passing through M , intersects ( AI ) at S . We know that : MS =2.5 and AI =11 . 1 a Calculate AS . (justify) b Calculate the measure of angle AMS , rounded to the nearest 0.1 degrees. 2 There is a leak in the roof at N that causes a stain at O on the ceiling. The line ( NO ) is perpendicular to the line ( AI ) . AO =4.5 To perform the calculations, we will assume : OAN = 24 o . Calculate the length AN , rounded to the nearest decime-ter. E.718 The unit of length is the cen-timeter. RST is a triangle such that : RS = 6.4 ; ST = 8 ; RT = 4.8 1 Construct the figure to full size. 2 Demonstrate that the triangle RST is right-angled at R . 3 Calculate the value, rounded to the nearest degree, of the measure of the angle RST . 4 M is the point on segment [ SR ] such that SM =4 ; and N is the point on segment [ ST ] such that SN =5 . a Demonstrate that the straight lines ( MN ) and ( RT ) are parallel. b Calculate the distance MN . E.2280 On this figure, we have the following lengths : AB = 5.4 cm ; BC = 7.2 cm AC = 9 cm ; AD = 2.6 cm The straight lines ( AE ) and ( BC ) are parallel. the figure is not to be redone. It is not given in full size. 1 Show that the triangle ABC is a right-angled triangle in B . 2 Calculate the tangent of the angle ACB , then deduce the measure of the angle ACB (value rounded to the nearest degree) . 3 Calculate AE . E.11720 Indication : la figure ci-dessous n’est pas en vraie grandeur. On a les données suivantes : Les points A , B , E et M sont alignés Les points A , C et D sont alignés ADE est un triangle rectangle en E ABC est un triangle rectangle en B AD = 70 m BC = 30 m AC = 50 m DME = 60 ◦ 1 Calculer la longueur AB . 2 Montrer que les droites ( DE ) et ( BC ) sont parallèles. 3 Montrer que la longueur DE est égale à 42 m . 4 Montrer que la longueur EM est environ égale à 24 ; 2 m . 5 En déduire l’aire du triangle AMD . 10. Open problems, problems with initiative, complex tasks https://chingmath.fr chapExoCorrec/713 sacados/713 Groupe Ouest - 2002 - 4 points AOSIMN chapExoCorrec/718 sacados/718 Groupe Nord - Juin 2003 - 5,5 points chapExoCorrec/2280 sacados/2280 Antilles-Guyane - Septembre 2006 - 5,5 points ABCDE chapExoCorrec/11720 sacados/11720 70m50m30m60oABCDEM
ABCD96cm150cm55cmprofondeurd’unemarchehauteurd’unemarcheSchéma 20cm12cm10cm ABCH12cm E.5924 We wish to build a structure for a skatepark, consisting of a staircase with six identical steps providing access to yan inclined plane whose height is equal to 96 cm . The design for this structure is shown below : Stair construction standards : 60 2 h + p 65 où h is the height of a step and p the depth of a step in cm . Requests from skatepark regulars : Length of the inclined plane (i.e., the length AD ) be-tween 2.20 m and 2.50 m . Angle formed by the inclined plane with the ground (here the angle ∠ BDA ) between 20 o and 30 o . 1 Are the construction standards for the staircase being met? 2 Are the requests of skatepark regulars for the incline plane being met? E.5721 In this exercise, any record of research, even if incomplete, or initiative, even if unsuccessful, will be considered in the assessment. In billiards, a player wants to hit the black ball with the white ball by making a stripe (by touching only one edge of the bil-liard table) . The diagram shows the situation in which the player finds himself : Since the pool table is brand new, the white ball leaves the cushion with the same angle with which it arrived. What angle does it have to arrive at to hit the black ball? E.6414 Consider the triangle ABC right-angled C and the point H foot of the height from the vertex C . We have the following information : segment [ CH ] measures 12 cm ; triangle ABC has area 150 cm 2 The measure of segment [ AC ] is assumed to be less than that of segment [ BC ] . Determine the measure of the angle ∠ CBH rounded to the nearest tenth of a degree. Leave any trace of research, even if it is not successful. Hint: we can establish the factorization : 2 x 2 − 50 x + 288 = 2 x − 16 x − 9 https://chingmath.fr chapExoCorrec/5924 sacados/5924 ABCD96cm150cm55cmprofondeurd’unemarchehauteurd’unemarcheSchéma chapExoCorrec/5721 sacados/5721 Correction de Sylvain C. 20cm12cm10cm chapExoCorrec/6414 sacados/6414 ABCH12cm
4mBarreslatérales PSHUT90cm140cm HRJ E.10982 Olivia decided to install four so-lar panels on the flat ground in her garden to generate some of the electricity she consumes. Description A solar panel is a device that generates elec-tricity from light energy. Panel characteristics : Length 1700 mm Width 1000 mm Thickness 40 mm Optimal operation: angle of inclination from the hori-zontal between 30 o and 35 o . Orientation : South. To tilt her panels and achieve optimal performance, Olivia decides to make her own support structure. To do this, she draws up the following diagrams for a support structure con-sisting of three identical brackets, connected by three side bars 4 m long. Each support is designed to hold four panels. General plan of the support, one panel is shown : General plan of the support, one panel is shown : 1 a Check that the distance HS rounded to the nearest millimeter is equal to 166 ; 4 cm . b To ensure that the panel is securely held in place, the manufacturer recommends that the distance HS from the support be at least 95 % of the length of the panel Remember that this length measures 1 700 mm . Will this support comply with the manufacturer’s rec-ommendations? 2 Will the angle of inclination, HSP , allow the panels to function optimally? 3 To reinforce the structure, Olivia attaches a 50 cm long reinforcement bar inside her brackets. On the detailed plan of a bracket, this bar is represented by the segment [ UT ] perpendicular to the segment [ PS ] . Calculate the length ST . Round to the nearest millime-ter. 4 Olivia buys stainless steel tubes ranging in length from 4 ; 5 m to 37 e each to make the support consisting of three brackets and three side bars. Show that she must budget a minimum of 222 e for the purchase of stainless steel tubes. 11. Unclassified exercises E.966 Note: The unit of length is the meter. The drawing is not to scale. 1 Romeo ( R ) wants to join Juliet ( J ) at her window. To do this, he places a ladder [ JR ] against the wall [ JH ] . The wall and the ground are perpen-dicular. Given : HR =3 ; JH =4 . a Calculate JR . b Calculate cos HJR then the value of the angle HJR rounded to the nearest degree. 1 The scale slides. We are given : JR =5 ; HJR =40 o . a Calculate HR , rounded to the nearest tenth. b Write the expression for tan HJR then calculate JH rounded to the nearest decimeter. https://chingmath.fr chapExoCorrec/10982 sacados/10982 4mBarreslatérales PSHUT90cm140cm chapExoCorrec/966 sacados/966 HRJ
ABCDEFJHGradins NordPiscine olympiqueGradins Sud 1;7m1m ABCDE E.10974 Construction of the Olympic Aquatic Center in Saint-Denis began in 2021 to host the artis-tic swimming events of the Paris Olympic Games 2024 . Alyssa and Jules visit the Olympic Aquatic Center and take their seats in the stands. Their positions in relation to the Olympic pool are shown in the diagram below, which models the situation : Alyssa is seated in the north stands at point A and Jules is seated in the south stands at point J . The diagram is not to scale. Given : AC = FJ =15 m ; BC =27 m ; FH =7 m ; EF =18 m Points F , J , and D are aligned. Points F , H , and E are aligned. Points C , B , D , and E are aligned. Lines ( HJ ) and ( ED ) are parallel. 1 Jules and Alyssa discuss who is in the best position to attend the event. a Calculate the distance between Alyssa and the edge of the pool, i.e., calculate the length AB . Round the result to the nearest meter. b Check that the distance between Jules and the edge of the pool, i.e., the length JD , is 24 m , rounded to the nearest meter. c Deduce which of the two friends is closest to the edge of the pool 2 To comply with safety standards, the angle of inclination ABC of the north stands must not exceed 35 o . Do the north bleachers comply with this standard? 3 The roof of the Olympic Aquatic Center has a surface area of 5 000 m 2 . It is estimated that 4 678.4 m 2 of this roof is covered with photovoltaic panels. Here are the characteristics of a standard photovoltaic panel provided by the manufacturer : Dimensions: 1 m wide and 1.7 m long. Energy produced : approximately 350 kWh per year. Show that the annual amount of energy produced by all the photovoltaic panels on the roof of the Olympic Aquatic Center is 963 200 kilowatt hours. (kWh) . 4 The regulatory temperature of the water in the pool dur-ing the Olympic Games must be between 25 o and 28 o . To comply with this regulation, the water in the Olympic pool in Saint-Denis should be at a temperature of 26 o . It is assumed that the water in this pool occupies a rectan-gular block with the following dimensions : Length : 50 m ; Width : 25 m ; Depth : 3 m It is assumed that before the Olympic pool is heated for the first time, the water temperature is 18 o . It is estimated that it takes approximately 9.3 kWh to heat 1 m 3 of water from 18 o to 26 o . How much energy, in kWh, will be needed to heat all the water in the Olympic swimming pool to 26 o ? E.3584 the following figure is not full-scale. It is not required to be reproduced. The unit is centimeters. The point B belongs to the segment [ DE ] and the point A to the segment [ CE ] . We give : ED = 9 ; EB = 5.4 ; EC = 12 EA = 7.2 ; CD = 15 1 Show that the lines ( AB ) and ( CD ) are parallel. 2 Calculate the length of segment [ AB ] . 3 Show that the lines ( CE ) and ( DE ) are perpendicular. 4 a Calculate the value rounded to the nearest degree of the angle ∠ ECD . b Deduce, without doing any calculations, that of the angle ∠ EAB . Justify. https://chingmath.fr chapExoCorrec/10974 sacados/10974 ABCDEFJHGradins NordPiscine olympiqueGradins Sud 1;7m1m chapExoCorrec/3584 sacados/3584 Nouvelle-Caledonie Mars 2009 ABCDE