1.2 SPM Practice (Short Questions)


Question 11:
Express each of the following as a number in base two.
(a) 26 + 24 + 1
(b) 25 + 23 + 2 + 20

Solution:
(a) 2 6 + 2 4 +1 = 1 _ × 2 6 + 0 _ × 2 5 + 1 _ × 2 4 + 0 _ × 2 3 + 0 _ × 2 2 + 0 _ × 2 1 + 1 _ × 2 0 = 1010001 2

(b) 2 5 + 2 3 +2+ 2 0 = 1 _ × 2 5 + 0 _ × 2 4 + 1 _ × 2 3 + 0 _ × 2 2 + 1 _ × 2 1 + 1 _ × 2 0 = 101011 2



Question 12:
State the value of the digit 2 in the number 324175 , in base ten.

Solution:


Value of the digit 2
= 2 × 53
= 250



Question 13:
101102 + 1112 =

Solution:


Alternatively, use a scientific calculator to get the answer directly.



Question 14:
1100102 – 1112 =

Solution:


Alternatively, use a scientific calculator to get the answer directly.

1.2 SPM Practice (Short Questions)


Question 5:
What is the value of the digit 3, in base ten, in the number 43155?
 
Solution:
Identify the place value of each digit in the number first.
 
4
3
1
5
Place Value
53
52
51
50
Value of the digit 3 
= 3 × 52
= 75



Question 6:
Express 5(52 + 2) as a number in base 5.
 
Solution:
Step 1: Expand 5(52 + 2) first.
Step 2: write 5(52 + 2) in expanded notation for base 5.
5(52 + 2)
= 53 + 2 × 5
= 1 × 53 + 0 × 52 + 2 × 51+ 0 × 50
= 10205

53
52
51
50
1
0
2
05


Question 7:
1101102 – 111012 =
 
Solution:

Alternatively, use a scientific calculator to get the answer directly.

10.1 Angle of Elevation and Angle of Depression (Part 2)


3. The angle of elevation or angle of depression is always measured from the horizontal line. 




Example 3:
Diagram below shows two vertical poles JK and NL on a horizontal plane. is a point on NL such that JK = ML.

The angle of depression of point J from point is
 
Solution:
The angle of depression of point J from N is the angle between line JN and the horizontal line through N.

Angle of depression of J from N
= angle of elevation of N from J
=∠NJM

SPM Practice (Short Questions)


8.4 Circles, SPM Practice (Short Questions)
Question 9:

In figure above, ABC is a tangent to the circle centre O at B. AED is a straight line.
Find the value of y.

Solution:
ABO = 90o
BOE = 2 × 40o = 80o
In quadrilateral AEOB,
AEO = 360– ∠ABO  – ∠BOE – 35o
= 360– 90o – 80o – 35= 155o
yo+ ∠AEO = 180
yo+ 155o = 180o
y= 180o  – 155o
y o = 25



Question 10:
 

In figure above, ABC is a tangent to the circle centre O, at point B.
The value of is

Solution:
OBC = 90
BOD = 2 × 50o = 100o
In quadrilateral BODC,
x= 360– ∠BOD – ∠OBC – 120o
= 360– 100o – 90o – 120o
= 50

8.3 Common Tangents (Part 3)


3. Do not intersect
(a) Circles of the same size



Number of common tangents
Properties of common tangents
Four common tangents:
AB, CD, PS and RQ
AB = CD = OV
PS = RQ
AB // OV // CD


(b) Circles of different sizes

 

Number of common tangents
Properties of common tangents
Four common tangents:
AB, CD, PS and RQ
AB = CD
BT = DT
PS = RQ
OA // VB
OC // VD

8.3 Common Tangents (Part 2)


2. Intersect at one point
(a) Circles of the same size


 
Number of common tangents
Properties of common tangents
Three common tangents:
AB, CD and PQ
AC = PQ = BD
AB = OR = CD
AB // OR // CD
AC // PQ // BD
PQ perpendicular to OR
 

(b) Circles of different sizes
(i) External contact
 


Number of common tangents
Properties of common tangents
Three common tangents:
ABE, CDE and PQ
AB = CD
BE = DE
OA // RB
OC // RD
PQ perpendicular to ORE


(ii) Internal contact
 

Number of common tangents
Properties of common tangents
One common tangent: PQR
ONQ perpendicular to PR

8.2 Angle between Tangent and Chord (Example 3 & 4)


Example 3:
In the diagram, PQR is a tangent to the circle QSTU at Q.


Find the values of
(a) x   (b) y
 
Solution:
(a)
UTS + ∠UQS = 180o ←(opposite angle in cyclic quadrilateral QSTU)
105o + ∠ UQS = 180o
UQS = 75o
x+ 75o + 20o = 180o(the sum of angles on a straight line PQR = 180o)
x+ 95o = 180o
= 85o
 
(b) 
PQU = ∠ QSU  ← (angle in alternate segment)
85o = 35o + y
y = 50o



Example 4:

In the diagram, ABC is a tangent to the circle BDE with centre O, at B.
Find the value of x.
 
Solution:


B E D = C B D = 54 B D E = 180 54 2 = 63 Isosceles triangle E B D = E D B A B E = B D E = 63 In A B E , x + 45 + 63 = 180 x + 108 = 180 x = 72

8.2 Angle between Tangent and Chord (Example 1 & 2)


Example 1:

In the diagram, ABC is a tangent to the circle BDE at B.
The length of arc BD is equal to the length of arc DE.
Find the value of p.
 
Solution:
Angle BED = 82o  ← (angle in alternate segment)
Angle DBE = 82o  ← (Arc BD = Arc DE, BDE is an isosceles triangle)
Therefore p= 180o – 82o – 82= 16o



Example 2:
In the diagram, PQR is a tangent to the circle QSTU at Q.

Find the value of y.
 
Solution:
Angle QUT 
= 180o– 98o(opposite angle in cyclic quadrilateral QSTU )
= 82o
Angle QTU = 75o  ← (angle in alternate segment)
Therefore y= 180o – (82o + 75o)  ← (Sum of interior angles in ∆ QTU)
= 23o