Bab 10 Penyelesaian Segitiga

Posted on September 19, 2016 by user

Soalan 4:

Dalam rajah di bawah, ABC ialah sebuah segi tiga. AGJB, AHC dan BKC ialah garis lurus. Garis lurus JK adalah berserenjang kepada BC.

Diberi bahawa BG = 40cm, GA = 33 cm, AH = 30 cm, GAH = 85^o dan JBK = 45^o.
(a) Hitung panjang, dalam cm, bagi

i. GH

ii. HC

(b) Luas segi tiga GAH adalah dua kali luas segi tiga JBK. Hitung panjang, dalam cm, bagi BK.

(c) Lakar segi tiga A’B’C’ yang mempunyai bentuk yang berlainan daripada segi tiga ABC dengan keadaan A’B’ = AB, A’C’ = AC dan ∠ A’B’C’ = ∠ ABC.

Penyelesaian:

(a)(i)

Guna petua kosinus,

GH² = AG² + AH² – 2 (AG)(AH) kos ∠ GAH

GH²= 33²+ 30² – 2 (33)(30) kos 85^o

GH² = 1089 + 900 – 172.57

GH² = 1816.43

GH = 42.62 cm

(a)(ii)

∠ ACD = 180^o – 45^o– 85^o= 50^o

Guna petua sinus,

\begin{array}{l} \frac{A C}{\sin 45^{\circ}} = \frac{73}{\sin 50^{\circ}} \\ A C = \frac{73 \times \sin 45^{\circ}}{\sin 50^{\circ}} \end{array}

AC = 67.38 cm

Oleh itu, HC = 67.38 – 30 = 37.38 cm

(b)

Area of ∆ GAH = ½ (33)(30) sin 85^o = 493.12 cm²

Katakan panjang BK = JK = x

2 × Area of ∆ JBK = Area of ∆ GAH

2 × [½ (x)(x)] = 493.12

x² = 493.12

x = 22.21 cm
BK = 22.21 cm

(c)

Bab 10 Penyelesaian Segitiga

Posted on September 19, 2016 by user

Soalan 3:

Rajah di bawah menunjukkan sebuah segi tiga ABC.

(a) Hitungkan panjang, dalam cm, bagi AC.

(b) Suatu sisi empat ABCD dibentuk dengan keadaan AC ialah pepenjuru, ∠ACD = 45° dan AD = 14 cm.

Hitung dua nilai yang mungkin bagi ∠ADC.

(c) Dengan menggunakan ∠ADC yang tirus dari (b), hitungkan
i. panjang, dalam cm, bagi CD,
ii. luas, dalam cm², sisi empat ABCD itu

Penyelesaian:

(a)

Guna petua kosinus,

AC² = AB² + BC² – 2 (AB)(BC) kos ∠ABC

AC² = 16² + 12² – 2 (16)(12) kos 70^o

AC² = 400 – 131.33

AC² = 268.67

AC = 16.39 cm

(b)

\begin{array}{l} Guna petua sinus, \\ \frac{\sin ∠ A D C}{16.39} = \frac{\sin 45^{\circ}}{14} \\ \sin ∠ A D C = \frac{16.39 \times \sin 45^{\circ}}{14} \end{array}

sin ∠ ADC = 0.8278

∠ ADC = 55.87^o atau (180^o – 55.87^o)

∠ ADC = 55.87^o atau 124.13^o

(c)(i)

sudut tirus ADC = 55.87^o

∠ CAD = 180^o – 45^o – 55.87^o= 79.13^o

\begin{array}{l} \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)

Luas sisi empat ABCD

= Luas ∆ ABC + Luas ∆ ACD

= ½ (16)(12) sin 70^o+ ½ (16.39)(14) sin79.13^o

= 90.21 + 112.67

= 202.88 cm²

3.7.1 The Role of Human Nervous System (Structured Question 1 & 2)

Posted on September 2, 2016 by user

Question 1:
Diagram below shows a cross-section of part of the nervous system.

(a)(i) Name structure P.

(a)(ii) State the function of P.

(b)(i) Why is Q swollen at the dorsal root?

(b)(ii) Complete the Diagram with the neurones involved in a reflex action. Mark the direction of the impulse movement on the neurones.

(c) Compare two structures of a sensory neurone and a motor neurone.

(d) If the spinal nerve is cut off at R, what is the effect on the organ which is connected to it?
Explain your answer.

(e) Azmin’s finger accidentally touches a flame.
Explain briefly how his reflex action functions to avoid the injury.

Answer:
(a)(i)
P : Spinal cord

(a)(ii)
Function of P : Controls reflex actions.

(b)(i)
To place the cell bodies of the afferent neurones.

(b)(ii)

(c)

(d)
The organ is unable to respond. Impulses cannot flow to the effector.

(e)
- The receptor detects heat and triggers an impulse.
- The impulse is sent to the spinal cord through the afferent neurone.
- The impulse flows through the afferent neurone which synapses with the interneurone and then synapses with the efferent neurone.
- The efferent neurone sends an impulse to the effector.
- The hand is pulled away from flame.