Bab 9 Trigonometri II

Posted on June 11, 2016 by user

9.3.2 Trigonometri II, SPM Praktis (Kertas 1)

Soalan 4:

Dalam rajah di atas, WZY ialah satu garis lurus. ∠XYZ = 90^o, ∠XWZ = 30^odan WZ = XZ = 30cm. Cari panjang XY.

Penyelesaian:

∠WXZ = ∠XWZ = 30^o

Maka ∠XZY = 30^o+ 30^o= 60^o

$\begin{array}{l} \sin ∠ X Z Y = \frac{X Y}{X Z} \\ \sin 60^{o} = \frac{X Y}{30} \end{array}$

XY= sin 60^o × 30

XY = 25.98cm

Soalan 5:

Dalam rajah di atas, PQS ialah satu segitiga bersudut tegak. Diberi bahawa SR = 6cm, PQ = 12 cm dan 5SR = 2PS. Cari nilai cos α dan tan β.

Penyelesaian:

$\begin{array}{l} 5 S R = 2 P S \\ P S = \frac{5}{2} S R \\ P S = \frac{5}{2} (6) \\ P S = 15 c m \\ k o s α = \frac{P Q}{P S} \\ k o s α = \frac{12}{15} = \frac{4}{5} \end{array}$

Dalam ∆ PQS, guna Teori Pythagoras,

$\begin{array}{l} Q S = \sqrt{P S^{2} - P Q^{2}} \\ Q S = \sqrt{15^{2} - 12^{2}} = 9 c m \\ \tan β = - \tan ∠ P S Q \leftarrow \begin{array}{l} 90^{\circ} < β < 180^{\circ} \\ (d a l a m sukuan II), \\ tan β is negatif \end{array} \\ \tan β = - \frac{P Q}{Q S} \\ \tan β = - \frac{12}{9} = - \frac{4}{3} \end{array}$

Soalan 6:

Dalam rajah di atas, ADC ialah satu garis lurus,

$\sin q = \frac{3}{5} dan \tan p = \frac{1}{2}$ . Hitung jarak AC.

Penyelesaian:

$\begin{array}{l} Diberi \sin q = \frac{B D}{A B} = \frac{3}{5} \\ \frac{B D}{30} = \frac{3}{5} \\ B D = \frac{3}{5} \times 30 \\ B D = 18 c m \end{array}$

Dalam ∆ ABD, guna Teori Pythagoras,

$\begin{array}{l} A D = \sqrt{A B^{2} - B D^{2}} \\ A D = \sqrt{30^{2} - 18^{2}} = 24 c m \\ Given tan p = \frac{B D}{D C} = \frac{1}{2} \\ \frac{18}{D C} = \frac{1}{2} \\ D C = 36 c m \end{array}$

Oleh itu, jarak AC = 24 + 36 = 60cm