8.3 Common Tangents (Sample Questions)

Posted on April 23, 2020 by user

Example 1:

In the diagram, ABG and CDG are two common tangents to the circle with centres O and F respectively. Find the values of

(a) x, (b) y, (c) z.

Solution:

(a)

In ∆ BFG,

angle BFG = ½ × 56^o = 28^o

angle FBG = 90^o ← (tangent is 90^o to radius)

x^o+ 28^o + 90^o = 180^o

x^o= 180^o – 118^o

x^o= 62^o

x = 62

(b)

angle AOF = x^o = 62^o← (AO// BF)

y^o= 2 × 62^o

y^o= 124^o

y = 124

(c)

Exterior angle of AOC = 360^o – 124^o = 236^o

Angle EOC = ½ × 236^o = 118^o

z^o= (180^o – 118^o) × ½ ← (∆ EOC is isosceles triangle, angle OEC = angle OCE)

z^o= 31^o

z = 31

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 4:

In figure above, ABCD is a tangent to the circle CEF at point C. EGC is a straight line. The value of y is

Solution:

\begin{array}{l} ∠ C E F = ∠ D C F = 70^{\circ} \\ ∠ A E G + 70^{\circ} + 210^{\circ} = 360^{\circ} \\ ∠ A E G = 80^{\circ} \\ In cyclic quadrilateral A B G E, \\ ∠ A B G + ∠ A E G = 180^{\circ} \\ y^{\circ} + 80^{\circ} = 180^{\circ} \\ y = 100 \end{array}

Question 5:

In figure above, PAQ is a tangent to the circle at point A. AEC and BED are straight lines. The value of y is

Solution:

∠ABD = ∠ACD = 40^o

∠ACB = ∠PAB = 60^o

y= 180^o– ∠ACB – ∠CBD – ∠ABD

y= 180^o– 60^o – 25^o– 40^o = 55^o

Question 6:

In figure above, KPL is a tangent to the circle PQRS at point P. The value of x is

Solution:

∠PQS = ∠SPL= 55^o

∠SPQ = 180^o– 30^o – 55^o= 95^o

In cyclic quadrilateral,

∠SPQ + ∠SRQ = 180^o

95^o+ x^o = 180^o

x = 85^o

8.3 Common Tangents (Part 1)

Posted on April 23, 2020 by user

8.3 Common Tangents

A common tangent to two circles is a straight line that touches each of the circles at only one point.

1. Intersect at two points

(a) Circles of the same size

Number of common tangents	Properties of common tangents
Two common tangents: AB and CD	AC = BD AB = CD AB parallel to // OR parallel to // CD

(b) Circles of different sizes

Number of common tangents	Properties of common tangents
Two common tangents: ABE and CDE	AB = CD BE = DE OA // RB OC // RD

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 7:

In figure above, APB is a tangent to the circle PQR at point P. QRB is a straight line. The value of x is

Solution:

∠PQR = ∠RPB = 45^o

∠QPR = (180^o– 45^o) ÷ 2 = 67.5^o

∠PQR + ∠BPQ + x^o = 180^o

45^o+ (67.5^o + 45^o) + x^o = 180^o

x = 22.5^o

Question 8:

The figure above shows two circles with respective centres O and V. AB is a common tangent to the circles. OPRV is a straight line. The length, in cm, of PR is

Solution:

\begin{array}{l} \cos 86^{o} = \frac{O M}{O V} \\ 0.070 = \frac{1}{O V} \\ O V = \frac{1}{0.070} \\ O V = 14.29 c m \\ ∴ P R = 14.29 - 5 - 4 \\ = 5.29 c m \end{array}

8.1 Tangents to a Circle

Posted on April 23, 2020 by user

8.1 Tangents to a Circle

1. A tangent to a circle is a straight line that touches the circle at only one point and the point is called the point of contact.

2. If a straight line cuts a circle at two distinct points, it is called a secant. The chord is part of the secant in a circle.

3. Tangent to a circle is perpendicular to the radius of the circle that passes through the point of contact.

If ABC is the tangent to the circle at B, then

∠

ABO =

∠

CBO = 90^o.

Properties of Two Tangents to a Circle from an External Point

In the diagram, BA and BC are two tangents from an external point B. The properties of the tangents are as follows.

(a) BA = BC

(b)

∠

ABO =

∠

CBO = x^o

(c)

∠

AOB =

∠

COB = y^o

(d) ∠ OAB =

∠

OCB = 90^o

(e)

∠

AOC +

∠

ABC = 180^o

(f) ∆ AOB and ∆ COB are congruent

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 1:

In figure above, FAD is a tangent to the circle with centre O. AEB and OECD are straight lines. The value of y is

Solution:

∠OAD = 90^o

∠AOD= 180^o– 90^o – 34^o= 56^o

y = 56^o÷ 2 = 28^o

Question 2:

In figure above, PQR is a tangent to the circle QSTU at Q and TUPV is a straight line. The value of y is

Solution:

∠QTS = ∠RQS = 40^o

∠SQT= ∠QTS = 40^o (isosceles triangle)

∠PQT= 180^o– 40^o – 40^o= 100^o

∠TPQ= 180^o– 115^o = 65^o

y = 180^o– 100^o – 65^o= 15^o

Question 3:

In figure above, ABC is a tangent to the circle BDE with centre O, at B.

Find the value of y.

Solution:

∠BOD= 2 × ∠BED

= 2 × 35^o = 70^o

∠ODB = ∠OBD

= (180^o– 70^o) ÷ 2 = 55^o

∠ EDB = ∠ EBA = 75^o

y^o+ ∠ ODB = 75^o

y^o+ 55^o = 75^o

y= 20^o

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 4:

John has a collection of stamps from Thailand, Indonesia and Singapore. He picks one stamp at random. The probability of picking a Thailand stamp is

\frac{1}{5}

and the probability of picking an Indonesia stamp is

\frac{8}{15} .

John has 20 Singapore stamps. Calculate the total number of stamps in his collection.

Solution:

\begin{array}{l} Probability of picking a Singapore stamp \\ = 1 - \frac{1}{5} - \frac{8}{15} \\ = \frac{4}{15} \\ Given that John has 20 Singapore stamps, \\ ∴ Total number of stamp in John's collection \\ = \frac{15}{4} \times 20 \\ = 75 \end{array}

Question 5:
A box contains 36 green erasers and a number of red erasers.
If an eraser is picked randomly from the box, the probability of picking a red eraser is

\frac{5}{8}

. Find the number of red erasers.

Solution:
Probability of picking a green eraser = 1 –

\frac{5}{8}

\frac{3}{8}

Total number of erasers in the box = 36 ×

\frac{8}{3}

= 96

Hence, number of red erasers in the box = 96 – 36 = 60

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 1:

A box contains 5 pink marbles and 21 yellow marbles. Sharon puts another 4 pink marbles and 1 yellow marble inside the box. A marble is chosen at random from the box.

What is the probability that a pink marble is chosen?

Solution:
Total number of pink marbles = 5 + 4 = 9
Total number of yellow marbles = 21 + 1 = 22
Total number of marbles = 9 + 22 = 31

∴ P (pink marble) = \frac{9}{31}

Question 2:
In a group of 80 prefects, 25 are girls. Then 10 boys leave the group.
If a prefect is chosen at random from the group, state the probability that the prefect chosen is a boy.

Solution:
Number of boys = 80 – 25 = 55

After 10 boys left,
Number of boys = 55 – 10 = 45

Total number of prefects = 25 + 45 = 70

∴ P (b o y) = \frac{45}{70} = \frac{9}{14}

Question 3:
60 people have taken part in a singing contest. If a person is chosen at random from all the contestants, the probability of getting a male contestant is

\frac{8}{15}

. If there are 12 males and 3 females who did not qualify to the second round, find the probability that a male is chosen from the second round contestants.

Solution:
Number of male contestants = 60 ×

\frac{8}{15}

= 32

After 12 males and 3 females left,
Total number of contestants left = 60 – 15 = 45

Number of male contestant = 32 – 12 = 20

∴ P (male contestant) = \frac{20}{45} = \frac{4}{9}

7.3 Probability of an Event

Posted on April 23, 2020 by user

(A) Probability of an Event

1. The probability of an event A, P(A) is given by

2. If P(A) = 0, then the event A will certainly not occur.

3. If P(A) = 1, then the event A will certainly to occur.

Example 1:

Table below shows the distribution of a group of 80 pupils playing a game.

	Form Four	Form Five
Girls	28	16
Boys	12	24

A pupil is chosen at random from the group to start the game.

What is the probability that a boy from Form Five will be chosen?

Solution:

Let

A = Event that a boy from Form Five

S = Sample space

n(S) = 28 + 12 + 16+ 24 = 80

n(A) = 24

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ = \frac{24}{80} = \frac{3}{10} \end{array}

(B) Expected Number of Times an Event will Occur

If the probability of an event A and the number of trials are given, then the number of times event A occurs

= P(A) × Number of trials

Example 2:

In a football training session, the probability that Ahmad scores a goal in a trial is ⅝. In 40 trials are chosen randomly, how many times is Ahmad expected to score a goal?

Solution:

Number of times Ahmad is expected to score a goal

= ⅝ × 40

= 25

(C) Solving Problems

Example 3:

Kelvin has 30 white, blue and red handkerchiefs. If a handkerchief is picked at random, the probability of picking a white handkerchief is

\frac{2}{5} .

Calculate

(a) the number of white handkerchiefs.

(b) the probability of picking a blue handkerchief if 8 of the handkerchiefs are red in colour.

Solution:

Let

W = Event that a white handkerchief is picked.

B = Event that a blue handkerchief is picked.

R = Event that a red handkerchief is picked.

S = Sample space

(a)
n(S) = 30

\begin{array}{l} n (W) = P (W) \times n (S) \\ = \frac{2}{5} \times 30 = 12 \end{array}

(b)
Given n(R) = 8

n(B) = 30 – 12 – 8 = 10

\begin{array}{l} P (B) = \frac{n (B)}{n (S)} \\ = \frac{10}{30} = \frac{1}{3} \end{array}

7.1 Sample Space

Posted on April 23, 2020 by user

7.1 Sample Space

(A) Experiment

1. An experiment is a process or an action in making an observation to obtain the required results.

2. The result of the experiment is called the outcome.

(B) Listing all the possible outcomes of an experiment

We can list all the possible outcomes of an experiment by carrying out the experiment or by reasoning.

Example 1:

Six cards as shown in the diagram above are placed in a box. A card is drawn at random from the box. List all the possible outcomes.

Solution:

All the possible outcomes are O, R, A, N, G, E.

(C) Sample space of an experiment

1. A sample space is the set of all the possible outcomes of an experiment.

2. The letter S is used to represent the sample space and all the possible outcomes are written in brackets, { }.

Example 2:

A letter is chosen from the word ‘GARDEN’.

(a) List all the possible outcomes.

(b) Write the sample space, S, using set notation.

Solution:

(a) The possible outcomes are G, A, R, D, E and N

(b) Sample space, S = {G, A, R, D, E, N }