(Long Questions) – Question 6

Posted on April 22, 2020 by user

Question 6:
Diagram below shows trapezium ABCD.

(a) Calculate
(i) ∠BAC.
(ii) the length, in cm, of AD.
(b) The straight line AB is extended to B’ such that BC = B’C.
(i) Sketch the trapezium AB’CD.
(ii) Calculate the area, in cm², of ∆BB’C.

Solution:
(a)(i)

\begin{array}{l} 5^{2} = 4^{2} + 7^{2} - 2 (4) (7) \cos ∠ B A C \\ 25 = 16 + 49 - 56 \cos ∠ B A C \\ 56 \cos ∠ B A C = 40 \\ \cos ∠ B A C = \frac{40}{56} \\ ∠ B A C = \cos^{- 1} \frac{40}{56} \\ = 44^{o} 25' \end{array}

(a)(ii)

\begin{array}{l} \frac{A D}{\sin ∠ D C A} = \frac{7}{\sin 115^{o}} \\ \frac{A D}{\sin 44^{o} 25'} = \frac{7}{\sin 115^{o}} \leftarrow (∠ D C A = ∠ B A C) \\ A D = \frac{7}{\sin 115^{o}} \times \sin 44^{o} 25' \\ A D = 5.406 cm \end{array}

(b)(i)

(b)(ii)

\begin{array}{l} \frac{\sin ∠ A B C}{7} = \frac{\sin 44^{o} 25'}{5} \\ \sin ∠ A B C = \frac{\sin 44^{o} 25'}{5} \times 7 \\ = 78^{o} 28' \\ ∠ A B C = 180^{o} - 78^{o} 28' \\ ∠ A B C = 101^{o} 32' (obtuse angle) \\ ∠ C B B' = 180^{o} - 101^{o} 32' = 78^{o} 28' \\ ∠ B C B' = 180^{o} - 78^{o} 28' - 78^{o} 28' = 23^{o} 4' \\ Area of △ B B' C = \frac{1}{2} \times 5 \times 5 \times 23^{o} 4' \\ = 4.898 {cm}^{2} \end{array}

(Long Questions) – Question 5

Posted on April 22, 2020 by user

Question 5:

The diagram below shows a triangle ABC.

(a) Calculate the length, in cm, of AC.

(b) A quadrilateral ABCD is now formed so that AC is a diagonal, ∠ACD = 45° and AD = 14 cm. Calculate the two possible values of ∠ADC.

(c) By using the acute ∠ADC from (b), calculate

(i) the length, in cm, of CD,

(ii) the area, in cm², of the quadrilateral ABCD

Solution:
(a)

Using cosine rule,

AC² = AB² + BC² – 2 (AB)(BC) ∠ABC

AC² = 16² + 12² – 2 (16)(12) cos 70^o

AC² = 400 – 131.33

AC² = 268.67

AC = 16.39 cm

(b)

\begin{array}{l} Using sine rule, \\ \frac{\sin ∠ A D C}{16.39} = \frac{\sin 45^{\circ}}{14} \\ \sin ∠ A D C = \frac{16.39 \times \sin 45^{\circ}}{14} \\ \sin ∠ A D C = 0.8278 \\ ∠ A D C = {55.87}^{\circ} or (180^{\circ} - {55.87}^{\circ}) \\ ∠ A D C = {55.87}^{\circ} or 124 {.13}^{\circ} \end{array}

(c)(i)

\begin{array}{l} Acute angle of ∠ A D C = {55.87}^{\circ} \\ ∠ C A D = 180^{\circ} - 45^{\circ} - {55.87}^{\circ} = {79.13}^{\circ} \\ \frac{C D}{\sin {79.13}^{\circ}} = \frac{14}{\sin 45^{\circ}} \\ C D = \frac{14 \times \sin {79.13}^{\circ}}{\sin 45^{\circ}} = 19.44 cm \end{array}

(c)(ii)

\begin{array}{l} Area of quadrilateral A B C D \\ = Area of Δ A B C + Area of Δ A C D \\ = \frac{1}{2} (16) (12) \sin 70^{\circ} + \frac{1}{2} (16.39) (14) \sin {79.13}^{\circ} \\ = 90.21 + 112.67 \\ = 202.88 {cm}^{2} \end{array}

(Long Questions) – Question 4

Posted on April 22, 2020 by user

Question 4:
Diagram below shows a quadrilateral PQRS.

(a) Find
(i) the length, in cm, of QS.
(ii) ∠QRS.
(iii) the area, in cm², of the quadrilateral PQRS.
(b)(i) Sketch a triangle S’Q’R’ which has a different shape from triangle SQR such that S’R’ = SR, S’Q’ = SQ and ∠S’Q’R’ = ∠SQR.
(ii) Hence, state ∠S’R’Q’.

Solution:
(a)(i)

\begin{array}{l} ∠ P = 180 - 76 - 34 = 70 \\ \frac{Q S}{\sin 70} = \frac{8}{\sin 34} \\ Q S = \frac{8 \times \sin 70}{\sin 34} \\ = 13.44 cm \end{array}

(a)(ii)

\begin{array}{l} {13.44}^{2} = 6^{2} + 9^{2} - 2 (6) (9) \cos ∠ Q R S \\ 108 \cos ∠ Q R S = 6^{2} + 9^{2} - {13.44}^{2} \\ \cos ∠ Q R S = \frac{6^{2} + 9^{2} - {13.44}^{2}}{108} \\ ∠ Q R S = \cos^{- 1} (- 0.5892) \\ = 126^{o} 6' \end{array}

(a)(iii)

\begin{array}{l} Area of P Q R S \\ = Area of P Q S + Area of Q R S \\ = (\frac{1}{2} \times 8 \times 13.44 \times \sin 76) + (\frac{1}{2} \times 6 \times 9 \times \sin 126^{o} 6') \\ = 52.16 + 21.82 \\ = 73.98 {cm}^{2} \end{array}

(b)(i)

(b)(ii)

\begin{array}{l} ∠ S' R' Q' = ∠ S' R R' \\ = 180 - 126^{o} 6' \\ = 53^{o} 54' \end{array}

(Long Questions) – Question 3

Posted on April 22, 2020 by user

Question 3:

The diagram shows a trapezium PQRS. PS is parallel to QR and QRS is obtuse. Find

(a) the length, in cm, of QS,

(b) the length, in cm, of RS,

(c) ∠QRS,

(d) the area, in cm², of triangle QRS.

Solution:
(a)

\begin{array}{l} \frac{Q S}{\sin P} = \frac{P S}{\sin Q} \\ \frac{Q S}{\sin 85^{\circ}} = \frac{13.1}{\sin 28^{\circ}} \\ Q S = \frac{13.1 \times \sin 85^{\circ}}{\sin 28^{\circ}} \\ Q S = 27.8 cm \end{array}

(b)
∠RQS = 180^o – 85^o – 28^o

∠RQS = 67^o

Using cosine rule,

RS² = QR² + QS² – 2 (QR)(QS) ∠RQS

RS² = 6.4² + 27.8² – 2 (6.4)(27.8) cos 67^o

RS² = 813.8 – 139.04

RS² = 674.76

RS = 25.98 cm

(c)

\begin{array}{l} Using cosine rule, \\ Q S^{2} = Q R^{2} + R S^{2} - 2 (Q R) (R S) \cos ∠ Q R S \\ {27.8}^{2} = {6.4}^{2} + {25.98}^{2} - 2 (6.4) (25.98) \cos ∠ Q R S \\ 772.84 = 715.92 - 332.54 \cos ∠ Q R S \\ \cos ∠ Q R S = \frac{715.92 - 772.84}{332.54} \\ \cos ∠ Q R S = - 0.1712 \\ ∠ Q R S = {99.86}^{\circ} \end{array}

(d)
Area of triangle QRS

= ½ (QR)(RS) sin R

= ½ (6.4) (25.98) sin 99.86^o

= 81.91 cm²

(Long Questions) – Question 2

Posted on April 22, 2020 by user

Question 2:

In the diagram below, ABC is a triangle. AGJB, AHC and BKC are straight lines. The straight line JK is perpendicular to BC.

It is given that BG= 40cm, GA = 33 cm, AH = 30 cm, GAH = 85^o and JBK= 45^o.

(a) Calculate the length, in cm of

i. GH

ii. HC

(b) The area of triangle GAH is twice the area of triangle JBK. Calculate the length, in cm,

of BK.

(c) Sketch triangle which has a different shape from triangle ABC such that, A’B’ = AB,

A’C’ = AC and ∠A’B’C’ = ∠ABC.

Solution:
(a)(i)

Using cosine rule,

GH² = AG² + AH² – 2 (AG)(AH) ∠GAH

GH²= 33²+ 30² – 2 (33)(30) cos 85^o

GH² = 1089 + 900 – 172.57

GH² = 1816.43

GH = 42.62 cm

(a)(ii)

\begin{array}{l} ∠ A C D = 180^{\circ} - 45^{\circ} - 85^{\circ} = 50^{\circ} \\ Using sine rule, \\ \frac{A C}{\sin 45^{\circ}} = \frac{73}{\sin 50^{\circ}} \\ A C = \frac{73 \times \sin 45^{\circ}}{\sin 50^{\circ}} \\ A C = 67.38 cm \\ ∴ H C = 67.38 - 30 = 37.38 cm \end{array}

(b)

\begin{array}{l} Area of Δ G A H = \frac{1}{2} (33) (30) \sin 85^{\circ} = 493.12 {cm}^{2} \\ Let length of B K = J K = x \\ 2 \times Area of Δ J B K = Area of Δ G A H \\ 2 \times [\frac{1}{2} (x) (x)] = 493.12 \\ x^{2} = 493.12 \\ x = 22.21 cm \\ B K = 22.21 cm \end{array}

(c)

(Long Questions) – Question 1

Posted on April 22, 2020 by user

Question 1:

The diagram shows a quadrilateral ABCD. The area of triangle BCD is 12 cm² and ∠BCD is acute. Calculate

(a) ∠BCD,

(b) the length, in cm, of BD,

(c) ∠ABD,

(d) the area, in cm², quadrilateral ABCD.

Solution:

(a) Given area of triangle BCD = 12 cm²

½ (BC)(CD) sin C = 12

½ (7) (4) sin C = 12

14 sin C = 12

sin C = 12/14 = 0.8571

C = 59^o

∠BCD = 59^o

(b)
Using cosine rule,

BD² = BC² + CD² – 2 (7)(4) cos 59^o

BD² = 7²+ 4² – 2 (7)(4) cos 59^o

BD² = 65 – 28.84

BD² = 36.16

BD = 36.16
BD = 6.013 cm

(c)
Using sine rule,

\begin{array}{l} \frac{A B}{\sin 35^{\circ}} = \frac{6.013}{\sin A} \\ \frac{10}{\sin 35^{\circ}} = \frac{6.013}{\sin A} \\ \sin A = \frac{6.013 \times \sin 35^{\circ}}{10} \\ \sin A = 0.3449 \\ A = {20.18}^{\circ} \\ ∠ A B D = 180^{\circ} - 35^{\circ} - {20.18}^{\circ} \\ ∠ A B D = {124.82}^{\circ} \end{array}

(d)
Area of quadrilateral ABCD

= Area of triangle ABD + Area of triangle BCD

= ½ (AB)(BD) sin B + 12 cm

= ½ (10) (6.013) sin 124.82 + 12

= 24.68 + 12

= 36.68 cm²

10.3 Area of Triangle

Posted on April 22, 2020 by user

10.3 Area of Triangle

Example:

Calculate the area of the triangle above.

Solution:

\begin{array}{l} A r e a = \frac{1}{2} a b \sin C \\ A r e a = \frac{1}{2} (7) (4) \sin 40^{o} \\ A r e a = 9 {cm}^{2} \end{array}

Example:

Diagram above shows a triangle ABC, where AC = 8 cm and ∠C = 32^o. Point D lies on straight line BC where BD = 10 cm and ∠ADB = 70^o . Calculate
(a) the length, in cm, of CD,
(b) the area, in cm² of ∆ ADC,
(c) the area, in cm² of ∆ ABC,
(d) the length, in cm, of AB.

Solution:
(a)

\begin{array}{l} ∠ A D C + 70^{o} = 180^{o} \\ ∠ A D C = 110^{o} \\ ∠ C A D = 180^{o} - 110^{o} - 32^{o} \\ ∠ C A D = 38^{o} \\ Using Sine Rule, \\ \frac{8}{\sin 110^{o}} = \frac{C D}{\sin 38^{o}} \\ C D \sin 110^{o} = 8 \sin 38^{o} \\ C D (0.940) = 8 (0.616) \\ C D = 5.243 \end{array}

(b)

\begin{array}{l} Area of △ A D C \\ = \frac{1}{2} (8) (5.243) sin 32^{o} \\ = 20.972 (0.530) \\ = 11.12 {cm}^{2} \end{array}

(c)

\begin{array}{l} Area of △ A B C \\ = \frac{1}{2} (8) (10 + 5.243) sin 32^{o} \\ = 60.972 (0.530) \\ = 32.315 {cm}^{2} \end{array}

(d)

\begin{array}{l} Use Cosine Rule for △ A B C, \\ A B^{2} = {15.243}^{2} + 8^{2} - 2 (15.243) (8) \cos 32^{o} \\ A B^{2} = 232.35 + 64 - 206.83 \\ A B^{2} = 89.52 \\ A B = 9.462 cm \end{array}

10.2 The Cosine Rule

Posted on April 22, 2020 by user

10.2 The Cosine Rule

The cosine rule can be used when

(i) two sides and the included angle, or

(ii) three sides of a triangle are given.

(A) If you know 2 sides and 1 angle between them [included angle] ⇒ Cosine rule

Example:

Calculate the length of AC, x, in cm for the triangle above.

Solution:

\begin{array}{l} b^{2} = a^{2} + c^{2} - 2 a c \cos B \\ x^{2} = 4^{2} + 7^{2} - 2 (4) (7) \cos 50^{\circ} \\ x^{2} = 16 + 49 - 56 (0.6428) \\ x^{2} = 65 - 35.997 \\ x^{2} = 29.003 \\ x = 5.385 cm \end{array}

(B) If you know 3 sides ⇒ Cosine rule

Example:

Calculate ÐBAC for the triangle above.

Solution:

\begin{array}{l} \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \\ \cos ∠ B A C = \frac{7^{2} + 6^{2} - 8^{2}}{2 (7) (6)} \\ \cos ∠ B A C = 0.25 \\ ∠ B A C = \cos^{- 1} 0.25 \\ ∠ B A C = {75.52}^{o} \end{array}

10.1 The Sine Rule

Posted on April 22, 2020 by user

10.1 The Sine Rule

In a triangle ABC in which the sides BC, CA and AB are denoted by a, b, and c as shown, and A, B, C are used to denote the angles at the vertices A, B, C respectively,

The sine rule can be used when

(i) two sides and one non-included angle or

(ii) two angles and one opposite side are given.

(A) If you know 2 angles and 1 side ⇒ Sine rule

Example:

Calculate the length, in cm, of AB.

Solution:

∠ACB = 180^o – (50^o + 70^o) = 60^o

\begin{array}{l} \frac{A B}{\sin 60^{o}} = \frac{4}{\sin 50^{o}} \\ A B = \frac{4 \times \sin 60^{o}}{\sin 50^{o}} \\ A B = 4.522 cm \end{array}

(B) If you know 2 sides and 1 angle (but not between them) ⇒ Sine rule

Example:

Calculate ∠ACB.

Solution:

\begin{array}{l} \frac{28}{\sin 54^{o}} = \frac{26}{\sin ∠ A C B} \\ \sin ∠ A C B = \frac{26 \times \sin 54^{o}}{28} \\ \sin ∠ A C B = 0.7512 \\ ∠ A C B = {48.7}^{o} \end{array}

(C) Case of ambiguity (2 possible triangles)

Example

Calculate ∠ACB, θ.

Solution:

Two possible triangle with these measurement

AB = 26cm BC = 28 cm Ð BAC = 54^o

\begin{array}{l} \frac{26}{\sin θ} = \frac{28}{\sin 54^{o}} \\ \sin θ = 0.7512 \\ θ = \sin^{- 1} 0.7512 \\ θ = {48.7}^{o}, 180^{o} - {48.7}^{o} \\ θ = {48.7}^{o} (Acute angle), {131.3}^{o} (Obtuse angle) \end{array}

Long Questions (Question 2 & 3)

Posted on April 22, 2020 by user

Question 2:
Given the equation of a curve is:
y = x² (x – 3) + 1
(a) Find the gradient of the curve at the point where x = –1.
(b) Find the coordinates of the turning points.

Solution:
(a)

\begin{array}{l} y = x^{2} (x - 3) + 1 \\ y = x^{3} - 3 x^{2} + 1 \\ \frac{d y}{d x} = 3 x^{2} - 6 x \\ When x = - 1 \\ \frac{d y}{d x} = 3 {(- 1)}^{2} - 6 (- 1) \\ = 9 \\ Gradient of the curve is 9 . \end{array}

(b)
At turning points,

\frac{d y}{d x} = 0

3x² – 6x = 0
x² – 2x = 0
x (x – 2) = 0
x = 0, 2

y = x² (x – 3) + 1
When x = 0, y = 1
When x = 2,
y = 2² (2 – 3) + 1
y = 4 (–1) + 1 = –3
Therefore, coordinates of the turning points are (0, 1) and (2, –3).

Question 3:
It is given the equation of the curve is y = 2x (1 – x)⁴ and the curve pass through T(2, 4).
Find
(a) the gradient of the curve at point T.
(b) the equation of the normal to the curve at point T.

Solution:
(a)

\begin{array}{l} y = 2 x {(1 - x)}^{4} \\ \frac{d y}{d x} = 2 x \times 4 {(1 - x)}^{3} (- 1) + {(1 - x)}^{4} \times 2 \\ = - 8 x {(1 - x)}^{3} + 2 {(1 - x)}^{4} \\ At T (2, 4), x = 2. \\ \frac{d y}{d x} = - 16 (- 1) + 2 (1) \\ = 16 + 2 \\ = 18 \end{array}

(b)

\begin{array}{l} Equation of normal: \\ y - y_{1} = - \frac{1}{\frac{d y}{d x}} (x - x_{1}) \\ y - 4 = - \frac{1}{18} (x - 2) \\ 18 y - 72 = - x + 2 \\ x + 18 y = 74 \end{array}