NYGH 2015-S3EOY-IM1 ans

1 A −15 17 8 A − 90o 2015 S3 EOY IM1 Solutions No. 1 (i) 17 8sin =A (ii) AA tan)180(tan −=−° 15 8= (iii) AA sin)90(cos =°− 17 8= 2 Y X m c xb x ay −= xy (given) x − b a xaby += 2 )2lg( −y )2ln( −y x blg bln alg aln abxxy xb x ay +−= −= abxy aby aby x x lglg)2lg( lg)2lg( 2 +=− =− += 3 (a) 2 3 16 38sin ==∠AEC 2 3sin =∠CED 3 (b) °=−°=∠ − 120)2 3(sin180 1CED (take obtuse ∠) °=∠⇒ 30ECD (∠ sum of ∆) OR °==∠ − 60)2 3(sin 1CEA °=∠⇒ 30ECD (ext. ∠ = sum of int. opp. ∠) OR °==∠ − 60) 38 12(sin 1BCA °=∠⇒ 30ECD (adj. ∠s on a str. line) Since °=∠=∠ 30CDEECD , ∆CED is isosceles 3 (c) Using the property rt ∠ in a semicircle, AD is the diameter of circle through A, B and D AD 1230sin =° OR ( ) 83816 22 =−== ECED cm 24=⇒ AD cm ⇒ radius = 12 cm

2 4 (a) (i) 4 (a) (ii) 4 (b) 1+= bxay 5 (i) 213 15grad =− −= )1(21 −=− XY 12 −= XY 122 −=− yxy OR 2 )1(442 xy −−±= xy xyy =− =+− 2 2 )1( 12 xxy ±=±= 12 42 1+= xy (rej 1+− x ) 5 (ii) sub 9=x , 4=y 6 (i) Let l be the perpendicular bisector of PQ 1 of grad1)5(7 012 of grad =⇒−=−− −−= lPQ midpt of PQ )6,1(2 120,2 75 −=   −+−= 1)6( −=−− xy eqn of l is 7−= xy 6 (ii) At pt C, 627133 =⇒−=− xxx 3=x , 4−=y C (3, −4) 6 (iii) 80))4(0()35( 22 =−−+−−=CP units eqn of circle is 80)4()3( 22 =++− yx 7 (a) (i) °=∠ 36AOE (∠ at centre = 2∠ at ce) (ii) °=∠ 18XAE (∠ in alt. seg) y x 0 y x 0 x = 0 y = 0

3 7 (b) (i) °+°=∠ 1868ABC ( ext. ∠ of cyclic quad) °= 86 °=°+°=∠+∠ 1809486BCDABC By interior ∠s, // lines, BA is parallel to CX OR °−°=∠ 94180BAD (∠s in opp. segment) °= 86 °=∠=∠ 86ADXBAD By alternate ∠s, // lines, BA is parallel to CX 7 (b) (ii) °=∠ 90OAX (tan ⊥ rad) °−°−°=∠ 5290180AXD (interior ∠s, BA // CX) °= 38 OR °−°=∠ 5290BAY (tan ⊥ rad) °= 38 °=∠=∠ 38BAYAXD (corr. ∠s, BA // CX) 7 (c) CX is not a tangent ⇒ °≠∠ 90ODX (tan ⊥ rad) ⇒ °≠∠+∠ 180ODXOAX Since we can’t use the reasoning ∠s in opp. segment, AODX is not a cyclic quadrilateral 8 (a) (b) grad at (2, 0.5) = 1.0 to 1.4 (actual 1.25) (c) 02 3 2 9 =+− xx 4352 =+− xx ⇒ add in line 4=y =x 0.3 – 0.4, 4.1 – 4.2 (d) xxx 2 34352 −=+− 0392 7 =+− xx 06187 2 =+− xx (e) (i) Draw xy −= 3 (ii) 1=c 9 (a) (i) °−+= 42cos)120)(123(2120123 222AB 1.87...128.87 ==AB m (3 s.f.) x xy 3 52 +−= xy −= 3 xy −= 1 4=y

4 9 (a) (ii) 128.87 42sin 123 sin °=∠BAC    °=∠ − 128.87 42sin123sin 1BAC °= 843.70 (3 d.p.) Bearing of C from A °= 8.070 (1 d.p.) 9 (b) let the angle of depression be θ 123 65tan =θ 123 65tan 1−=θ °=°= 9.27...854.27 (1 d.p.) 9 (c) Let the shortest distance of B from AC be d 12342sin d=° d××1202 1 °= 42sin123d °= 42sin)120)(123(2 1 ...303.82= = 82.3 m (3 s.f.) 9 (d) °−°−°=∠ 843.7042180ABC (∠ sum of ∆) °= 157.67 (3 d.p.) °−°=∠ 42157.67EAC (ext. ∠ = sum of int. opp. ∠) °= 157.25 (3 d.p.) area of °=∆ 157.25sin)128.87)(120(2 1AEC 2220= m2 (3 s.f.) 10 (i) S(−4, 0), R(2, 0) x-coord of T 12 24 −=+−= sub 9,1 −=−= yx T (−1, −9) 10 (ii) 10)4)(2( +−=+− xxx 010822 =−+−+ xxx 01832 =−+ xx 0)3)(6( =−+ xx 3or6 =−= xx 7or16 == yy P (−6, 16) Q (3, 7) 10 (iii) Area of quadrilateral PQRS 001670 24632 2 1 −−= [ ])6442()4814(2 1 −−−+= 84= sq units

5 10 (iv) x-coord of X = −1 sub 11,1 =−= yx X (−1, 11) 10 (v) By similar triangles, QX : XP = 4 : 5 OR 2444 22 =+=QX units 2555 22 =+=XP units and ΔRXPRQX∆ share the same height 5 4 ofarea ofarea =∆ ∆ RXP RQX 11 The circle is tangent to the lines 102 += xy and 102 −= xy at A and B respectively Eqn of diameter is xy 2 1−= xx 2 1102 −=+ OR xx 2 1102 −=− 4−=x OR 4=x )2,4(−A , )2,4( −B radius 2024 22 =+= units Area of circle ( )220π= π20= sq units k = 20 y x 102 += xy 102 −= xy 5−=x 5−=y 5=y 5=x A B