Mathematics and Mathematical Literacy - Thutong

Mathematics(NSC)/Grade 11/ P2 87 ExemplarMEMORANDUM4.2PQ =P'Q ' ==22( 4 −1) + ( 4 − 2) = 132( 3 −12) + ( 6 −12)9 × 13 = 313Area PQRS = 13 × 13 = 132=117Area P´Q´R´S´ = 3 13 × 3 13 = 9× 13 = 117The length of the sides of PQRS increase bya factor of 3 to give the length of the sides ofP´Q´R´S´. The area of PQRS increased by a factor of 9to give the area of P´Q´R´S´ . This is 3² i.ethe square of the increase of the length of thesides. 65.1.1 −tanx.cosx tan x−− sin x − sin xx cos x sin x= sin . + .cos x sin x cos x sin x1 cos x + 1= 1+ or cos x cos x65.1.2− cos 60tan 451−= 2 11= −245.2.1 cos x (2 cos x – 1) 15.2.2 cos x = 0 ∴ x = 90° + 360°n or 270° + 360°n n ε Z(add on the period of the cos graph i.e.360°nto get general solution)ORcos x = 1 2 ∴ x = 60° + 360°n or 300° + 360°n,n ε Z 55.3.1 sin (180°+58°) = - sin 58° = - k 25.3.2 sin 2 58° + cos 2 58° = 1 ∴ cos 2 58° = 1 – k 22∴cos58= 1−k 36.10,5 or 1 2 16.2 Sipho, Ray and Vishnu get - 0,17Lorraine gets 0,23 46.32sin θ1−2cos θ 2sin θ1+2cos θ2 2cos θ − sin θ= 2 2cos θ + sin θ722= cos θ − sin θ 2 22= 1−sin θ − sin θ = 1−2sin θ 222or cos θ − ( 1−cos θ ) = 2cos θ −17.1 39,69 cm 17.23sin18x= − 5Reference angle is 36,87° ∴ 18 x = 216,87 + 360 n ∴ x = 12 + 20 n OR ∴ 18 x = 323,13 + 360 n ∴ x = 18 + 20 n ∴ x = 12 , 18, 32 or 38 68.1 y - x 28.2 In ∆PAB:PB 5= sin( 90 + x) sin( y − x)5cos x∴ PB = sin( y − x)38.3 In ∆PBT:PTsin y = PB5cos xsiny∴ PT =sin( y − x)29.1 1bcsin x 219.2 DAKˆ = 360 − 90 − 90 − x= 180 − x 1∴ΔDAK= bcsin(180− x)21= bcsin x 2= ΔABC310.1 Sum of lengths is 42,4Mean length is 4,24 210.2Length(cm)x i − x ( x x) 2i −3,2 -1,04 1,08163,6 -0,64 0,40965 0,76 0,57764,1 -0,14 0,01964,3 0,06 0,00364,7 0,46 0,21163,4 -0,84 0,70565,2 0,96 0,92164,6 0,36 0,12964,3 0,06 0,00364,0644,064Standard deviation = = 0, 67 96Copyright reserved