10.4.8 Penyelesaian Segitiga, SPM Praktis (Kertas 2)

Soalan 8:
Rajah menunjukkan sisi empat kitaran PQRS.


(a) Hitung
(i) panjang, dalam cm, bagi PR,
(ii) ∠PRQ.
(b) Cari
(i) luas, dalam cm², bagi ∆ PRS.
(ii) jarak terdekat, dalam cm, dari titik S ke PR.


Penyelesaian:
(a)(i)
$\begin{array}{l}P{R}^2=7^2+8^2-2\left(7\right)\left(8\right)kos{80}^o\\ P{R}^2=113-19.4486\\ PR=\sqrt{93.5514}\\ PR=9.6722\text{ cm}\end{array}$


(a)(ii)
$\begin{array}{l}\text{Dalam sisi empat kitaran}\\ \angle PQR+\angle PSR=180\\ \angle PQR+80=180\\ \angle PQR={100}^o\\ \\ \frac{\sin \angle QPR}{3}=\frac{\sin 100}{9.6722}\\ \sin \angle QPR=0.3055\\ \angle QPR={17}^{o}47\text{'}\\ \\ \angle PRQ={180}^o-{100}^o-{17}^{o}47\text{'}\\ ={62}^{o}13\text{'}\end{array}$


(b)(i)
$\begin{array}{l}\text{Luas }△PRS\\ =\frac{1}{2}\times 7\times 8\sin {80}^o\\ =27.5746{\text{ cm}}^2\end{array}$


(b)(ii)

$\begin{array}{l}\text{Luas }△PRS=27.5746\\ \frac{1}{2}\times 9.6722\times h=27.5746\\ h=\frac{27.5746\times 2}{9.6722}\\ =5.7018\text{ cm}\\ \text{Jarak terdekat}=5.7018\text{ cm}\end{array}$