IB Math Studies - Triangle Trigonometry Practice Key
IB Math Studies - Triangle Trigonometry Practice Key[233 marks]The diagram shows a triangle ABC in which AC 17 cm. M is the midpoint of AC.Triangle ABM is equilateral.1a. Write down the size of angle MCB.[1 mark]Markscheme30 (A1)(C3)[1 mark]Examiners reportPart (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrectassumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule orthe Cosine rule correctly were generally able to substitute correctly and gain at least two marks.1b. Write down the length of BM in cm.[1 mark]Markscheme8.5 (cm)(A1)[1 mark]Examiners reportPart (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrectassumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule orthe Cosine rule correctly were generally able to substitute correctly and gain at least two marks.1c. Write down the size of angle BMC.Markscheme120 (A1)[1 mark][1 mark]
Examiners reportPart (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrectassumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule orthe Cosine rule correctly were generally able to substitute correctly and gain at least two marks.1d.Calculate the length of BC in cm.[3 marks]MarkschemeBCsin 120 8.5sin 30(M1)(A1)(ft)Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.BC 14.7 (17 3)2(A1)(ft)[3 marks]Examiners reportPart (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrectassumptions about triangle BMC being right angled and used Pythagorus theorem incorrectly. Those who used either the Sine rule orthe Cosine rule correctly were generally able to substitute correctly and gain at least two marks.José stands 1.38 kilometres from a vertical cliff.2a.Express this distance in metres.[1 mark]Markscheme1380 (m)(A1)(C1)[1 mark]Examiners reportThis question was well answered by the majority of candidates although it was surprising to find some who could not express thegiven distance in metres. Where working was present, follow through marks could be awarded in the remainder of the question. Mostcandidates could give their answer correct to the nearest metre and find the percentage error correctly, using the formula. A commonerror was to use the calculated value in the denominator.2b. José estimates the angle between the horizontal and the top of the cliff as 28.3 and uses it to find the height of the cliff.Find the height of the cliff according to José’s calculation. Express your answer in metres, to the nearest whole metre.[3 marks]
Markscheme1380 tan 28.3(M1) 743.05 .(A1)(ft) 743 (m)(A1)(ft)(C3)Notes: Award (M1) for correct substitution in tan formula or equivalent, (A1)(ft) for their 743.05 seen, (A1)(ft) for their answercorrect to the nearest m.[3 marks]Examiners reportThis question was well answered by the majority of candidates although it was surprising to find some who could not express thegiven distance in metres. Where working was present, follow through marks could be awarded in the remainder of the question. Mostcandidates could give their answer correct to the nearest metre and find the percentage error correctly, using the formula. A commonerror was to use the calculated value in the denominator.2c.José estimates the angle between the horizontal and the top of the cliff as 28.3 and uses it to find the height of the cliff.[2 marks]The actual height of the cliff is 718 metres. Calculate the percentage error made by José when calculating the height of the cliff.Markschemepercentage error 743.05 718718 100(M1)Note: Award (M1) for correct substitution in formula. 3.49 % (% symbol not required)(A1)(ft)(C2)Notes: Accept 3.48 % for use of 743.Accept negative answer.[2 marks]Examiners reportThis question was well answered by the majority of candidates although it was surprising to find some who could not express thegiven distance in metres. Where working was present, follow through marks could be awarded in the remainder of the question. Mostcandidates could give their answer correct to the nearest metre and find the percentage error correctly, using the formula. A commonerror was to use the calculated value in the denominator.
The diagram shows triangle ABC. Point C has coordinates (4, 7) and the equation of the line AB is x 2y 8.3a. Find the coordinates of A.[1 mark]MarkschemeA(0, 4)Accept x 0, y 4(A1)[1 mark]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.3b.Find the coordinates of B.[1 mark]MarkschemeB(8, 0)Accept x 8, y 0(A1)(ft)Note: Award (A0) if coordinates are reversed in (i) and (A1)(ft) in (ii).[1 mark]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.3c.Show that the distance between A and B is 8.94 correct to 3 significant figures.[2 marks]
Markscheme AB 82 42 80AB 8.944 8.94(M1)(A1)(AG)[2 marks]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.3d.N lies on the line AB. The line CN is perpendicular to the line AB.[3 marks]Find the gradient of CN.Markschemey –0.5x 4 (M1)Gradient AB –0.5 (A1)Note: Award (A2) if –0.5 seen.ORGradient AB 12(A1)(0 4)(8 0)(M1)Note: Award (M1) for correct substitution in the gradient formula. Follow through from their answers to part (a).Gradient CN 2(A1)(ft)(G2)Note: Special case: Follow through for gradient CN from their gradient AB.[3 marks]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.3e.N lies on the line AB. The line CN is perpendicular to the line AB.Find the equation of CN.[2 marks]
MarkschemeCN: y 2x c7 2(4) c(M1)Note: Award (M1)for correct substitution in equation of a line.y 2x – 1(A1)(ft)(G2)Note: Accept alternative forms for the equation of a line including y – 7 2(x – 4) . Follow through from their gradient in (i).Note: If c –1 seen but final answer is not given, award (A1)(d).[2 marks]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.3f.N lies on the line AB. The line CN is perpendicular to the line AB.[3 marks]Calculate the coordinates of N.Markschemex 2(2x – 1) 8or equivalentN(2, 3) (x 2, y 3)(M1)(A1)(A1)(ft)(G3)Note: Award (M1) for attempt to solve simultaneous equations or a sketch of the two lines with an indication of the point ofintersection.[3 marks]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.3g.It is known that AC 5 and BC 8.06.Calculate the size of angle ACB.[3 marks]
Markscheme B) Cosine rule: cos(AC52 8.06 2 8.944 22 5 8.06(M1)(A1)Note: Award (M1) for use of cosine rule with numbers from the problem substituted, (A1) for correct substitution. B 82.9 AC(A1)(G2)Note: If alternative right-angled trigonometry method used award (M1) for use of trig ratio in both triangles, (A1) for correctsubstitution of their values in each ratio, (A1) for answer.Note: Accept 82.8 with use of 8.94.[3 marks]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.3h. It is known that AC 5 and BC 8.06.[3 marks]Calculate the area of triangle ACB.MarkschemeArea ACB 5 8.06 sin(82.9)2(M1)(A1)(ft)Note: Award (M1) for substituted area formula, (A1) for correct substitution. Follow through from their angle in part (e).ORArea ACB AB CN2 8.94 (4 2) 2 (7 3) 22(M1)(M1)(ft)Note: Award (M1) substituted area formula with their values, (M1) for substituted distance formula. Follow through fromcoordinates of N.Area ACB 20.0(A1)(ft)(G2)Note: Accept 20[3 marks]Examiners reportThis question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question tothe end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently fromthe rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.
The diagram shows an office tower of total height 126 metres. It consists of a square based pyramid VABCD on top of a cuboidABCDPQRS.V is directly above the centre of the base of the office tower.The length of the sloping edge VC is 22.5 metres and the angle that VC makes with the base ABCD (angle VCA) is 53.1 .4a. Write down the length of VA in metres.[1 mark]Markscheme22.5 (m)(A1)[1 mark]Examiners reportThis question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in threedimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of thequestion were independent of those preceding them; many candidates did not reach these parts, though for some, these were the onlyparts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.4b.Sketch the triangle VCA showing clearly the length of VC and the size of angle VCA.[1 mark]Markscheme(A1)[1 mark]Examiners reportThis question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in threedimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of thequestion were independent of those preceding them; many candidates did not reach these parts, though for some, these were the onlyparts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.4c.Show that the height of the pyramid is 18.0 metres correct to 3 significant figures.[2 marks]
Markscheme(M1)h 22.5 sin 53.1 17.99 (A1) 18.0 (AG)Note: Unrounded answer must be seen for (A1) to be awarded.Accept 18 as (AG).[2 marks]Examiners reportThis question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in threedimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of thequestion were independent of those preceding them; many candidates did not reach these parts, though for some, these were the onlyparts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.4d.Calculate the length of AC in metres.[3 marks]Markscheme AC 2 22.52 17.99.2(M1)(M1)Note: Award (M1) for multiplying by 2, (M1) for correct substitution into formula.OR(M1)(M1)AC 2(22.5)cos53.1 Notes: Award (M1) for correct use of cosine trig ratio, (M1) for multiplying by 2.ORAC2 22.52 22.52 – 2(22.5)(22.5) cos73.8 (M1)(A1)Note: Award (M1) for substituted cosine formula, (A1) for correct substitutions.ORACsin(73.8 ) 22.5sin(53.1 )(M1)(A1)Note: Award (M1) for substituted sine formula, (A1) for correct substitutions.AC 27.0(A1)(G2)[3 marks]Examiners reportThis question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in threedimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of thequestion were independent of those preceding them; many candidates did not reach these parts, though for some, these were the onlyparts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.
4e.Show that the length of BC is 19.1 metres correct to 3 significant figures.[2 marks]Markscheme BC 13.52 13.52(A1) 19.09 19.1(M1)(AG)ORx2 x2 2722x2 272(A1)BC 19.09 19.1(M1)(A1)(AG)Notes: Unrounded answer must be seen for (A1) to be awarded.[2 marks]Examiners reportThis question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in threedimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of thequestion were independent of those preceding them; many candidates did not reach these parts, though for some, these were the onlyparts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.4f.Calculate the volume of the tower.[4 marks]MarkschemeVolume Pyramid Cuboid 13 (18)(19.12 ) (108)(19.12 )(A1)(M1)(M1)Note: Award (A1) for 108, the height of the cuboid seen. Award (M1) for correctly substituted volume of cuboid and (M1) forcorrectly substituted volume of pyramid. 41 588(41 553 if 2(13.52) is used) 41 600 m3(A1)(ft)(G3)[4 marks]Examiners reportThis question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in threedimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of thequestion were independent of those preceding them; many candidates did not reach these parts, though for some, these were the onlyparts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.4g.To calculate the cost of air conditioning, engineers must estimate the weight of air in the tower. They estimate that 90 % of the [3 marks]volume of the tower is occupied by air and they know that 1 m3 of air weighs 1.2 kg.Calculate the weight of air in the tower.
MarkschemeWeight of air 41 600 1.2 0.9 44 900 kg(M1)(M1)(A1)(ft)(G2)Note: Award (M1) for their part (e) 1.2, (M1) for 0.9.Award at most (M1)(M1)(A0) if the volume of the cuboid is used.[3 marks]Examiners reportThis question also caused many problems for the candidature. There seems to be a lack of ability in visualising a problem in threedimensions – clearly, further exposure to such problems is needed by the students. Further, as in question 2, the final two parts of thequestion were independent of those preceding them; many candidates did not reach these parts, though for some, these were the onlyparts of the question attempted. There is also a lack of awareness of the appropriate volume formula on the formula sheet to use.The base of a prism is a regular hexagon. The centre of the hexagon is O and the length of OA is 15 cm.5a.Write down the size of angle AOB.[1 mark]Markscheme60 (A1)(C1)[1 mark]Examiners reportThis question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many hadproblems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theoremcorrectly to find the height of the triangle AOB. Those who used the formula for the area of a triangle A 12 ab sin C were moresuccessful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism– many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.5b.Find the area of the triangle AOB.[3 marks]
Markscheme15 152 7.522 97.4 cm2(97.5 cm2)(A1)(M1)(A1)Notes: Award (A1) for correct height, (M1) for substitution in the area formula, (A1) for correct answer.Accept 97.5 cm2 from taking the height to be 13 cm.OR12 152 sin 60 97.4 cm2(M1)(A1)(A1)(ft)(C3)Notes: Award (M1) for substituted formula of the area of a triangle, (A1) for correct substitution, (A1)(ft) for answer.Follow through from their answer to part (a).If radians used award at most (M1)(A1)(A0).[3 marks]Examiners reportThis question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many hadproblems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theoremcorrectly to find the height of the triangle AOB. Those who used the formula for the area of a triangle A 12 ab sin C were moresuccessful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism– many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.5c.The height of the prism is 20 cm.[2 marks]Find the volume of the prism.Markscheme97.4 120 11700 cm3(M1)(A1)(ft)(C2)Notes: Award (M1) for multiplying their part (b) by 120.[2 marks]
Examiners reportThis question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many hadproblems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theoremcorrectly to find the height of the triangle AOB. Those who used the formula for the area of a triangle A 12 ab sin C were moresuccessful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism– many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.In the diagram below A, B and C represent three villages and the line segments AB, BC and CA represent the roads joining them.The lengths of AC and CB are 10 km and 8 km respectively and the size of the angle between them is 150 .6a.Find the length of the road AB.[3 marks]MarkschemeAB2 102 82 – 2 10 8 cos150 AB 17.4 km(M1)(A1)(A1)(G2)Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer.[3 marks]Examiners reportThe weak students answered parts (a) and (b) using right-angled trigo
IB Math Studies - Triangle Trigonometry Practice Key [233 marks] 1a. [1 mark] The diagram shows a triangle ABC in which AC 17 cm. M is the midpoint of AC. Triangle ABM is equilateral. Write down the size of angle MCB. Markscheme 30 (A1) (C3) [1 mark] Examiners report Part (a) was generally well answered with many candidates gaining full marks. Some candidates went on to make incorrect .
So, ADÆis a median of the triangle. The three medians of a triangle are concurrent. The point of concurrency is called the The centroid, labeled Pin the diagrams below, is always inside the triangle. acute triangle right triangle obtuse triangle The medians of a triangle have a special concurrency property, as described in Theorem 5.7.File Size: 695KBPage Count: 7
The triangle has three acute angles that are not all equal. It is an acute triangle. 62/87,21 One angle of the triangle measures 90, so it is a right angle. Since the triangle has a right angle, it is a right triangle. 62/87,21 Since all the angles are congruent, it is a equiangular triangle. 62/87,21 The triangle
acute triangle equiangular triangle right triangle obtuse triangle equilateral triangle isosceles triangle scalene triangle Vocabulary. Holt McDougal Geometry 4-2 Classifying Triangles Recall that a triangle ( ) is a polygon with three sides. Triangles can be classified in
13 Section 2.3 – Solving Right Triangle Trigonometry Example In the right triangle ABC, A 40 and c 12 cm. Find a, b, and B. Solution 12 0 aa c a 0 7cm 0 b c 12 b b 0 2cm B 90 - A 90 - 40 50 Example A circle has its center at C and a radius of 18 inches. If triangle ADC is a right triangle and A 35 . Find x, the distance from A to B.
1 Right Triangle Trigonometry Trigonometry is the study of the relations between the sides and angles of triangles. The word “trigonometry” is derived from the Greek words trigono (τρ ιγων o), meaning “triangle”, and metro (µǫτρω ), meaning “measure”. Though the ancient Greeks, such as Hipparchus
Glencoe/McGraw-Hill 184 Glencoe Geometry Classify Triangles by Sides You can classify a triangle by the measures of its sides. Equal numbers of hash marks indicate congruent sides. If all three sides of a triangle are congruent, then the triangle is an equilateral triangle. If at least two sides of a triangle are congruent, then the triangle is an isosceles triangle.
Name _ 1 Geometry 1 Chapter 4 – Triangle Congruence Terms, Postulates and Theorems 4.1 Scalene triangle - A triangle with all three sides having different lengths. Equilateral triangle - All sides of a triangle are congruent. Isosceles triangle - A tri
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