3b)

DRAW THE DIAGRAM

Using Pythagoras theory

5² = 3² + x²

x² = 5² – 3²

X²= 25 – 9

X² = √16

X= 4cm

CosX = adjacent/hyp

= 4/5

Tan X = opp/adj. = 3/4

5cos x – 4tan x

5(4/5)- 4(3/4)

20/5 – 12/4

4-3= 1

4ai)

sum of angle in a D =180degree

xdegree + 90degree + 180degree – (3x+15)=180degree

xdegree + 90degree + 180degree – 3x+15=180degree

-2x=180degree – 255

+2x/2=+75/2

x=37.5

4aii)

<RsQ =180 – (3x+15)

<RsQ =180-(3*37.5+15)

=180-(112.5 + 15)

=180 – 127.5

<RsQ= 52.5degree

4b)

2N4seven =15Nnine

2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree

9*49+N*7+4*1=1*81+5*9+N*1

98+7N+4=81+45+N

7N+102=126+N

7N-N=126-102

6N/6 =24/6

N=4

6a) Draw the diagram

6b) Number that passed = 60% * 240 = 144

number that failed = 240 – 144 = 96

therefore, 28 + 2x + x + 14 + 6 + 6 -x + 8 = 96

2x + 62 = 96

2x = 96 – 62

2x = 34

x = 34/2

x = 17

6bi) faulty brakes cars= 8+6+x+6-x

= 8 + 6 + 6

= 20

bii) Only one fault = 28 + x + 2x

= 28 + 3x

= 28 + 3(17)

= 28 + 51

= 79

NO9) Using cosine rule,

|TQ|ˆ² = 4ˆ² + 6 ˆ² – 2(4)(6) cos30°

|TQ|ˆ² = 16 + 36 – 48(0.8660)

|TQ|ˆ² = 52 – 41.568

|TQ|ˆ² = 10.432

TQ = √10.432

TQ = 3.23CM

From similar triangles;

|PT|/|TQ| = |PS|/|SR|

4/3.23 = 10/|SR|

4|SR| = 32.3

|SR| = 32.3/4

|SR| = 8CM (nearest whole number)

9b) Now: |PT|/|PQ| = |PS|/|PR|

4/6 = 10/|PR|

4|PR| = 60

|PR| = 60/4

= 15CM

Therefore, |QR| = 15 – 6 = 9cm

Area of TQRS = Area of ΔPRS – Area of ΔPQT

=1/2(PS)(PR)sin30° – 1/2(PT)(PQ)sin30°

= 1/2sin30°(|PS| (PR) – |PT| |PQ|)

=1/2 * 1/2((10)(15) – (4)(6))

=1/4(150 – 24)

=1/4 * 126

=31.5 apx 32cm²

10a) Using Pythagoras theorem from SPQ

|SQ|^2 = 12^2 + 5^2

= 144+25

=169

SQ= sqroot of 169

= 13cm

Sin tita= 5/13 = 0.3846

Tita= Sin^-1(0.3846)

= 22.6degrees

From PRQ

Sin tita= |PR|/12

Sin 22.6 = PR/12

Sin 22.6= PR/12

PR= 12xsin 22.6

PR= 12×0.3843

PR= 4.61cm

(10bii)Let the height at which m touches the wall= y

Cos x^degrees= 8/10= 0.8

x^degrees= Cos^-1(0.8)

= 36.87degrees

Sin x^degrees = y/12

Sin 36.87= y/12

y= 12xsin36.87

y= 12×0.60000

y= 7.2m

5a) m+n+s+p+q/5 = 12

m+n+s+p+q =60

Now; (m+4) + (n-3) + (s+6) + (p-2) + (q+8)

= (m+n+s+p+q) + (4-3+6-2+8)/5

= 60+13/5

= 73/5 = 14.6

5b) 75% of 500 = 375 people

Number of people above 65 yrs = 500 – 375 =125

25% of 500 = 125

Number of people below 15 yrs = 125

Number between 15yrs and 65yrs

= 500 – (125 + 125)

= 500 – 250

= 250 people

(7a)

(y-y1)/(x-x1)=(y2-y1)/(x2-x1)

(y-5)/(x-2)=(-7-5)/(-4-2)

(y-5)/(x-2)=-12/-6

(y-5)/(x-2)=2

Cross multiply

y-5=2(x-2)

y-5=2x-4

2x-y-4+5=0

2x-y+1=0

(7bi)

DRAW THE DIAGRAM

(7bii)

(I)

p^2=q+r^2-2qrcosP

p^2=8^2+5^2-2*8*5*cos90

p^2=64+25-0

p^2=89

p=sqroot(89)

p=9.4339km

therefore |QR|=9.43km(3 sf)

(II)

q/sinQ=p/sinP

8/sinQ=9.4339/sin90

sinQ=(8*sin90/9.4339

sinq=(8*1)/9.4339 =0.8480

Q=sin^1(0.8480)=57.99 degrees

but Q=30+ A

A=Q-30

=57.99-30

A=27.99 degrees

The bearing of R from Q

=180-A

180-27.99

=155.01

=>152 degrees