9. a. DEF UVW. c. UV a. MNO MPQ. b. MP. 11. a. (AC) (AC) (AC)

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1 C H A P T E R 1 3 SIMILAR TRIANGLES AND TRIGONOMETRY LESSON 13-1 pp The literal meaning of trigonometry is triangle measure.. a. No, because no unit was named. Students might use different units to draw triangles. b. Yes. According to the SSS Similarity Theorem, they are similar because their sides are proportional. 3. One way to prove ABC DEF is to find a size-change image of ABC that is congruent to DEF. 4. SSS Similarity Theorem: If three sides of one triangle are proportional to three sides of a second triangle, then the triangles are similar. 5. ST VW, TR WU ; ; True; 5, so the triangles are similar by the SSS Similarity Theorem. 1 9.; 0.55; False; 0 9 similar a. AB ED 0 5 AB ED 9, so the triangles are not 41 AC 4; EF 16 BC 4; 4 DF AC BC EF DF b. Yes, the triangles are similar by the SSS Similarity Theorem. c. me ma 37 md mb 53 mf 180 (me md) 180 (37 53) d. DEF BAC 9. a. DEF UVW b. UW 10 DF 33 ; DF 33 UW 10 ratio of similitude: or c. UV DE VW EF a. MNO MPQ b. MP MN 6 MN 3; MP ratio of similitude: 3 or 1 3 c. QP 3 ON MQ MO OQ MP MN NP a. (AC) 5 65 (AC) (AC) 151 AC (DF) 4 40 (DF) (DF) 104 DF ED 4 AC ; DF 3 CB Yes, the triangles are similar ; FE BA b. Using the Pythagorean Theorem, AC 39 and DF 3. Since , ABC EFD by the SSS Similarity Theorem. 1. QR VU ; QP VT ; PR 11 TU Yes, the triangles are similar; PQR TVU. 16

2 13. Argument: Conclusions Justifications 0. X is the midpoint Given 0. of WY. V is the 0. midpoint of WZ. 1. WX 1 WY; definition of 0. WV 1 midpoint WZ. XV 1 YZ Midpoint Connector Theorem 3. WX WY 1 ; WV WZ 1 ; Multiplication 0. XV YZ 1 4. WX WY WV XV WZ YZ Property of Equality Transitive Property of Equality 5. WXV WYZ SSS Similarity Theorem 14. DEF CAB, and the ratio of similitude is. By the Fundamental Theorem of Similarity, the ratio of areas is the square of the ratio of similitude. Thus, Area( ABC) Area( DEF) V 1 Bh finds volume of a pyramid or a cone 3 with height h and base area B. 17. S.A. lw wh lh finds surface area of a box with sides l, w, and h. 18. p a b c finds perimeter of a triangle with sides a, b, and c. 19. V r h finds volume of a cylinder with height h and base radius r. 0. L.A. rl finds lateral area of a right cone with base radius r and slant height l. 1. a. Area( GHI) 1 bh units b. Area( GHI) 1 HI GJ GJ GJ units 41. mh 180 (mg mi) 180 (90 13) mjgi 180 (mgji mi) 180 (90 13) mhgj 180 (mgjh mh) 180 (90 77) (b); x 10 x 10 x 5 4. (a); a b a b b b b a b x x 0 x 00 x No; for example, the two pentagons shown below have proportional sides, but they are not similar LESSON 13- pp a. SSS Congruence Theorem b. ASA or AAS Congruence Theorem c. SAS Congruence Theorem. A size change is applied to one triangle so that its image is congruent to the other triangle. 3. magnitude of size change: AA Similarity Theorem: If two angles of one triangle are congruent to two angles of another, then the triangles are similar. 5. ASA Congruence Theorem 17

3 6. 56 (5 1 6) 66; 3 (3 1) 36; 134 (13 1 4) h h h 10, or 4 36 The flagpole is 4 feet tall h 5 1.h 1 5 h The height of the garage is about 4.17 meters. 8. a. magnitude of size change: b. SAS Congruence Theorem 9. a. AB AC, and A Y; by the SAS YX YZ Similarity Theorem, ABC YXZ. b. AB XY 6 XY ; 3 AB ratio of similitude: or 1 c. BC XZ a. G I, and GHK IHJ; by the AA Similarity Theorem, GHK IHJ. b. JI 50 KG ; KG JI ratio of similitude: 5 9 or 9 5 c. mk 180 (115 4) mhji mk 41 GH 9 5 HI JH KH 5 9 JH 89 JH 5 9 5(89 JH) 9 JH 445 5JH 9JH 445 4JH JH 11. Argument: Conclusions Justifications 0. WY 3 VY; Given 0. XY 3 YZ 1. mwyx mzyv Vertical Angles Theorem. WY XY 3; VY ZY 3 Multiplication Property of Equality 3. WY XY VY ZY Transitive Property of Equality 4. WXY VZY SAS Similarity Theorem 1. a. ON OM ; OP OQ 8 1 ; O O 3 Yes, they are similar. b. SAS Similarity Theorem c. ONP OMQ 13. a. JL IK 5 7 ; JH 1 IG 4 1 ; KG 7 4 5; LH 13 KG 5 No, they are not similar. b a. R SUT; USR UST Yes, they are similar. b. AA Similarity Theorem c. SRU SUT 15. a. VW ZY 3 XW ; 1.5 AY Yes, they are similar b. SSS Similarity Theorem c. XVW AZY ; VX ZA 4 18

4 16. Argument: Let k XY YZ. Then XY k AB AB BC and YZ k BC. Let ABC be the image of ABC under a size change of magnitude k. So AB k AB and BC k BC. Also, since size transformations preserve angle measure, B B. With transitivity, AB XY, BC YZ, and B Y. So, ABC XYZ by the SAS Congruence Theorem. Thus, XYZ can be mapped onto ABC by a composite of size changes and reflections. So ABC XYZ. 17. Argument: ABC EDC, and BAC DEC from the Lines AIA Theorem. Thus, CED CAB by the AA Similarity Theorem. 18. a. LM PQ ; MN 16 QR ; LN PR No, they are not similar. b AB AB AB 4 3 AB d d (d 1) 5d 3d 3 5d 3 d d n n n 1(n 1) 9n 1n 144 3n 144 n True; x y x y y y y x y 1 3. Argument: Conclusions Justifications 0. a b c d 1. ad bc; d c b a. b a d c 4. front rig h t top Given Multiplication Property of Equality Symmetric Property of Equality 5. Let s be the length of a side of the field. s s 50 s 500 s 150 s meters 6. Yes, because there are HL and SsA Congruence Theorems. These theorems could be deduced using the same techniques used in this lesson to deduce the SAS and AA Similarity Theorems. LESSON 13-3 pp a. AC AB CB or AX AY XY b. AB AC BY CX. Samples: AB 3.4 cm, BC 1.5 cm, AD.9 cm, DE 1.3 cm; AB AD BC DE 19

5 AB AC 3. BD CE 5 3 CE CE CE FG FI FH FJ FJ 8 FJ 3 6 FJ Side-Splitting Converse Theorem: If a line intersects OP and OQ in distinct points X and Y so that OX XP 6. No; because Yes; because OY, then XY PQ. YQ AB AC AD AE 0 AC AC 0 60 AC CE AE AC a. NQ ON b. MP NQ MO ON 1 6 c. MO 6 MP d. OP MO MP e. ON 30 OQ f. OM OP 6 7 g. MN OM PQ OP 6 7 h a. b. MN PQ PQ PQ 14 7 PQ DB BE DA EC DB DB 5 6 DB CE CF EB FA FA 4 FA 6 6 FA c. A and AFE are supplementary angles. d. BD FA DA CF 1. a. Side-Splitting Theorem b. Side-Splitting Theorem c. Transitive Property of Equality x x x 5, m 00 x 100 y y y y 10, m 100 0

6 14. Let h the distance from Elm St. to Rasci Rd. along 9th St h 00h h 36, m 00 So, 180 m 360 m of curbs are needed for 9th St. Let t the distance from Elm St. to Rasci Rd. along 10th St t 00t t 45,600 8 m 00 So, 8 m 456 m of curbs are needed for 10th St. 15. No judgment can be made. 16. Yes; PQR TSU by the AA Similarity Theorem. 17. Yes; STO XYZ by the SSS Similarity Theorem. 18. a. Because the two triangles involved here are similar by the AA Similarity Theorem. b. 58 (5 1 8) 68; 43 (4 1 3) 51; 156 (15 1 6) 186 x h e d x x x 1, The building is 08 tall. 19. Argument: Conclusions Justifications 0. BE bisects ABC; Given 0. AE AB; DC CB 1. ABE CBD definition of angle bisector. BAE and BCD are definition of 0.right angles. perpendicular 3. BAE BCD definition of right angle 4. ABE CBD AA Similarity Theorem 0. a. b. c (a); (c); No; the three angle measures are approximately 30, 70, and a. A B 1 B B 3 B4 B5 B 6 C 1 B C C3 C 4 C 5 C6 C7 b. AB 1 AB AC 1 1 from the AA Similarity AC 7 7 Theorem. The other segments have length 1 AB due to the Side-Splitting Theorem. 7 1

7 IN-C LA SS A C T IV IT Y p ,. C A ma 58; mb 3 3. a. mcdb 90 b. mcda 90 c. mdcb 58 d. mdca 3 4. (1) CDA BCA () BCA BDC (3) CDA BDC 5. (1) CD BC DA CA CA BA () BC BA CA BD BC DC (3) CD BD DA CA DC BC 6. (1) DA CA CA BA () BC BA BD BC (3) CD DA BD DC LESSON 13-4 pp Shown above in parts 3. 6., In-Class Activity True B 5. a. mp mqrs b. PRQ RSQ PSR c. QS RQ RQ PQ D 6. d. QS RS RS PS e. RP is the geometric mean of PQ and PS. f. RS is the geometric mean of PS and SQ. g. RQ is the geometric mean of PQ and SQ. PS RS RS QS QS 9 QS 6 6 QS QR a. NC b. c. EC IC IC NC EC EC 5 5 EC IE IN IC NC IE IE 5 1 IE x x 4 4 x y y 3 3 y h xy

8 10. a. True b. True 11. a. part (1) b x 5x 8 8 x height of tower: ft 1. a. Right-Triangle Altitude Theorem b. Geometric Mean Theorem c. Addition Property of Equality d. Distributive Property e. Substitution Property a a 15 a 6 15 a b b 15 b 9 15 b CD CD 9 (CD) 6 9 CD x 8x 3 4 ; 8x 1x 3 No; because 6x 8x 8x 1x 15. Argument: Conclusions Justifications 0. WXYZ is a trapezoid Given 0. with bases WX and YZ. 1. WX ZY definition of trapezoid. XWU ZYU; lines AIA 0. WXU YZU Theorem 3. WXU YZU AA Similarity Theorem 16. a. AC EC ; BC DC ; C C Yes, they are similar. b. SAS Similarity Theorem c. ACB ECD 17. a. mh 180 (40 80) 60; mj 180 (40 50) 90 No, they are not similar. 18. a. LM LN 4 1 ; MN NO ; LN LO Yes, they are similar. b. SSS Similarity Theorem c. LMN LNO 19. a. 3; DB, FH, and EG b. 3; FB, EC, and DH c. 4 positions 0. Let h height of the pyramid. Volume(pyramid) 1 3 Bh Volume(prism) B h Volume(prism) B h 1 Volume(pyramid) 1 3 Bh 3 6 The prism has 6 times the volume of the pyramid. 1. It is circular reasoning to use a theorem to prove itself a. 3 3 b First ten positive integer powers of : 1, 4, 3 8, 4 16, 5 3, 6 64, 7 18, 8 56, 9 51, a. The powers 4, 8, 16, 3, 64, 18, 56, and 51 are geometric means of other powers in the list. b. Sample: a n is a geometric mean of a n1 and a n1. 4. No, it is impossible, because if the two angles at D have different measures, then they must not be corresponding angles, and so their measures are the measures of two angles of each triangle. But these two measures add to 180, leaving no measure for the third angle. 3

9 LESSON 13-5 pp In an isosceles right triangle, the acute angles each measure 45.. If one leg of an isosceles right triangle has length 10 cm, the hypotenuse has length cm. 3. From first base to third base is feet. 4. a. False b. True c. True 5. In a right triangle with a 30 angle, the hypotenuse has double the length of the short leg. 6. a b d c d f e f longer leg: cm hypotenuse: cm 11. longer leg: 5 3 height of cell: mm 1. mzxy Perimeter( XYZ) XZ YZ XY 4 (4 4 3) units 14. The length of each diagonal is s units. 15. length of altitude: Perimeter( BLT) BL LT TO OB h h h h 3 3h h h h triangles 18. r 1 6; h r S.A. L.A. B 1 lp B 1 1 ( 6) units V 1 3 Bh 1 3 ( 6 ) units OP OP 3 3 OP MO y y 6 3 y 18 9 x ( 6) (3 9) x 8 1 x x 80 x XA XB XC XD 5 XB XB 5 7 XB BD XD XB XC XD CE DF 8 CE CE 8 4 CE ADC CBA BPA CPB 4

10 4. Area(square) s 400 s s Perimeter(square) 4(0) 80 meters Area(circle) r 400 r r r 11.3 Circumference(circle) meters The square has the larger perimeter. 5. a. Check students drawings. b. 6 c. The angles have measures x, x, and 4x, where x IN-CLASS ACTIVITY p B 1 C 1 7 mm; AC 1 15 mm; B 1 C AC B C 11 mm; AC 3 mm; B C AC 3 B 3 C 3 15 mm; AC 3 3 mm; B 3 C AC 3 3 BC 6 mm; AC 55 mm; BC AC BC AC 0.47, B 1C , B C 0.48, AC 1 AC B 3 C ; they should be very close. AC 3 4. about about They should be nearly the same. 7. Sample: BC AC They should be nearly the same. LESSON 13-6 pp a. about 0.47 b. They are nearly the same.. Sample: tan tan tan 3 h 5 h 5 tan The wall is about 15.6 meters high. 5. tan E tan F tan 30 x x tan 45 x x 1 9. Sample: A 30 m m 4 B 6 C m m 6 a. about 34 b. about 56 ( ) ( ) 3 c. tan ; tan

11 10. tan 40 x 0 x 0 tan height of the pole: feet 11. tan E tan E ABC and XYZ are similar because of the AA Similarity Theorem; so a x c z (corresponding sides are proportional). By multiplying both sides by x c, a x x c c z x c, and a c x z. 13. a. largest tangent: 4 b. smallest tangent: Sample: In the figure below, tan 75 a b and tan 74 a. Since c b, tan 75 tan 74. c 15. a. perimeter 3 ( units ) b. area , units 16. height of trapezoid b c 75 a area (9 4) units AC EC BC DC DC 00 DC DC 43, DE EC DC units 18. No; because a. No. Suppose a trapezoid is as the one below, where no two sides are congruent. Therefore, no two of the triangles formed are congruent. B C b. Yes. Use the trapezoid in part a. Because AD BC, DAO BCO and ADO CBO ( Lines AIA Theorem). Therefore, by the AA Similarity Theorem, ADO CBO. 0. d m A. m A B B O A D B' D = A' B' D' A' D' It is a translation in the direction from m to l perpendicular to m and l, and twice the distance from m to l. 3. Sample: Let ma 7, mb 97, and mc 56. a. tan 7 tan 97 tan b. tan 7 tan 97 tan c. Let ma 41, mb 4, and mc 115. The sum of the tangents is about The product of the tangents is also about D 6

12 3. d. Sample: If ma mb mc 180, with ma 90, mb 90, and mc 90, then tan A tan B tan C tan A tan B tan C. LESSON 13-7 pp a. leg opposite N: MO b. hypotenuse: MN c. leg adjacent to M: MO d. leg adjacent to N: NO. cos A AC BC ; sin A AB AB 3. a. sin F b. cos F c. tan G 4 d. sin G a. sin 60 x 3 3 x b. cos 60 x x 1 c. tan 60 x 3 x 3 5. a. sin 45 x x 1 b. cos 45 x x 1 c. tan 45 x x 1 6. a. sin b. cos c. tan d. cos a. sin b. cos a. cos 65 AC 30 AC 30 cos feet b. sin 65 BC 30 BC 30 sin feet 9. sin 80 BC 0 BC 0 sin feet 10. cos 80 AC 0 AC 0 cos feet 11. cos 65 AC 0 AC 0 cos feet Yes, since 6 feet is between 3.5 and 8.5 feet. 1. a. sin b. A right triangle with an 89 angle is close to isosceles (because the 89 angle is nearly the measure of the right angle), so the length of the leg opposite the 89 angle is almost the same as the hypotenuse. 14 mm 13. a. Sample: sin B 0.41; 34 mm 10 mm sin B mm b. Yes; they probably will not be equal, due to measurement error, but very close. 14. a. 5 has the largest sine. b. 1 has the largest cosine. 15. sin 85 0 x x sin 85 0 x feet sin cos x x cos x feet cos

13 17. tan 5 t 14.5 t 14.5 tan feet 18. tan 30 x 440 x 440 tan m 160 cm 1.6 m height of building: m 19. From isosceles right triangle ACD, AC DC. Then AO 1 AC 1. Diameter of O is, so the radius OE 1. Then AE AO OE AE AB 3 4 ; AC AD ; BAE DAC 4 Yes, ABE ADC by the SAS Similarity Theorem. 1. a. Converse: If mabc mdef, then ABC DEF. Inverse: If ABC is not similar to DEF, then mabc mdef. Contrapositive: If mabc mdef, then ABC is not similar to DEF. b. The contrapositive is true.. Let h be the length of the altitude from A. 6 h 7 36 h 49 h 13 h units 3. D' D A' A B' C' B C 4. a. x sin x cos x b. all c. For any value of x, (sin x) (cos x) 1. LESSON 13-8 pp y 340 mph 15 x 0. Because XYZ is a right triangle and mx 90, sin Z XY YZ. Therefore, XY YZ sin Z. 3. sin 63 x 4 x 4 sin cos 0 y 10 y 10 cos cos 6 z 150 z 150 cos

14 6. east component 0 cos mph; north component 0 sin mph 7. a. tan A ; tan1 (5) 79 The plane should fly about 79 south of west or 11 west of south. b , The plane will travel about km. 8. Using the SAS Triangle Area Formula, Area( ABC) sin units 9. Using the SAS Triangle Area Formula, Area( DEF) sin units 10. Suppose BD is the altitude from B to side AC. B C a D b A Area( ABC) 1 AC BD 1 AC (BC sin C) 1 AC BC sin C 1 ba sin C 1 ab sin C 11. a. Divide the pentagon into 5 congruent isosceles triangles, as shown in the diagram. D E 5 A M B maob maom 7 36 mmao 180 (90 36) 54 sin 54 OM 5 OM 5 sin cos 54 AM 5 AM 5 cos AB AM Area( AOB) 1 AB OM Area(pentagon ABCDE) 5 Area( AOB) cm b. Area(circle) r cm material cut off: cm 1. Draw the altitude from T to SU, intersecting SU at M. Since STU is isosceles, M is also the midpoint of SU. cos 7 SM 15 SM 15 cos SU SM Using the SAS Triangle Area Formula, Area( STU) sin 7 91 units O C 9

15 13. Let J Johor, S Singapore, and B Batam Island. To find JB, draw JYB as shown in the diagram below, find JY and YB, then use the Pythagorean Theorem. N W 9 J X Y 0 Z S Sou th Using SZB, sin 66 ZB 50 ZB 50 sin cos 66 SZ 50 SZ 50 cos Using JXS, sin 9 XS 0 XS 0 sin cos 9 JX 0 JX 0 cos JY JX XY JX SZ YB YZ ZB XS ZB JB (40.09) (48.81) 63 km 14. AB sin A 7 ; cos A 11 ; tan A sin B ; cos B ; tan B 11 7 E B 16. mghi (8 ) Draw the altitude from H to GI, intersecting GI at M. Since GHI is isosceles, mghm 1 mghi sin 60 GM 10 GM 10 sin GI GM units h altitude: h 5 1 h segments x and y: x x 1 1 x y y 5 5 y a. Yes, the two triangles are congruent by the SSS Congruence Theorem, so they are similar. b. No; c. No; Rudolph is both a y and a z, but not an x. 0. a. It is a sphere with the same center and radius as circle C. b. It is a doughnut (torus). 1. Sample: If 0 90, then the area of a triangle with fixed sides 7 and 10 increases as increases. If , then the area of the triangle with fixed sides 7 and 10 decreases as increases. 30

16 C H A P T E R 1 3 P R OG R ESS SELF -TEST pp AB AB 6 6 AB WY WY WY WZ YZ WY VW VX VY VZ 11 VX VX VX WX VW 4. YZ VY 8 VW 0 VW 15 0 VW 8(VW 15) 0VW 8VW VW 10 VW a. BC AB 7 b. AC AB 7 6. A right triangle is formed when the altitude is drawn. The shorter leg is 15, and the longer leg (altitude) is ; ; Yes, the triangles are similar by the SSS Similarity Theorem. 10. cos sin tan B AC BC has the smallest tangent. 14. cos A AC 9 sin B AB sin E x x x 33,000 0 m sin 80 x 15 x 15 sin ft a. x x 1 67 x m y tan north of west b km 0. tan 35 x 40 x 40 tan 35 8 ft height of tree: ft 0 x 8. Area( TRI) sin units 9. Argument: Since AB DE, A E and B D because of the Lines AIA Theorem. Therefore, ABC EDC by the AA Similarity Theorem. 31

17 C H A P T E R 1 3 REVIEW pp a.. b. JK KL JM MN MN 4 MN 6 7 MN JK JL KM LN 4 5 LN 4 LN 5 LN MN MP MQ MS MN MN MN No; since Yes; since AB BE CD DE 4 0 DE 1 DE 4 DE 1(0 DE) 4DE 40 1DE 30DE 40 DE 8 SW WV SU UT SW 11 SW SW 11(SW 18) 0SW 11SW 198 9SW 198 SW 7. CD AD DB AD AC 8. AC AB AD AD 7 7 AD AD CD 9. CD DB DB 18 DB 1 1 DB BC (DC) (DB) a QT RT RT TS QT QT 4 4 QT 16 8 b. QR (QT) (RT) c. RS TS QS d. QS QT TS AC AB BC AB DE DF 16 EF DF altitude: diagonal: q 15. a. TU ST b. US ST 7 c. SK ST d. TK ST

18 16. a. ST b. SU ST 7.51 c. TK ST Sample: ma 35, tan A.69, sin A.60, and cos A has the largest tangent has the largest sine. 0. sin A cos B tan B Area( DEF) sin units 4. Area( GHI) sin units 5. sin tan sin 30 x x 1 8. tan 60 x 3 x 3 9. tan 45 x x cos 45 x x a ; ; Yes, the triangles are similar. b. SSS Similarity Theorem 3. True; all angles in equilateral triangles measure 60; thus all equilateral triangles are similar by the AA Similarity Theorem (59 60) 61 Yes, by the AA Similarity Theorem ; Yes, by the SAS Similarity Theorem. 35. Argument: Conclusions Justifications 0. AB CD Given 1. BAE DCE; Lines AIA 0. ABE CDE Theorem. ABE CDE AA Similarity Theorem 36. Argument: Conclusions Justifications 0. AC CD, BC x, Given 0. AC x, and DC 4x. 1. BC AC x x 1 ; definition of ratio 0. AC x DC 4x 1. BC AC AC DC Transitive Property of Equality 3. ABC DAC SAS Similarity Theorem 37. cos A AC AB 38. sin A BC AB 39. tan A BC AC 40. MP is the tangent of angle Q. MQ 41. MQ is the sine of angle P. PQ 4. cos Q MQ PQ 43. tan B AC BC 44. sin B cos A x x x 9 1 x 9 15 m x x x m

19 47. a. b. AB BC AD DE AB AB 50 5 AB AC BC AE DE AC AC 70 5 AC CE AE AC ft mile 130 ft; 4 1 mile ft mile 1980 ft 8 1 mile x x x 3,960, ft tan 57 x 37 x 37 tan feet 50. sin 75 x 4 x 4 sin meters 51. tan 0 x 5 x 5 tan meters height of sculpture: meters 5. cos 5 60 x x cos 5 60 x feet cos northern component: sin 7 n 60 n 60 sin 7 57 knots eastern component: cos 7 e 60 e 60 cos knots 54. a. tan A 7 15 ( ) 7 A tan 1 5 south of west 15 b km 34

TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions

TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii)