10.4.5 Penyelesaian Segitiga, SPM Praktis (Kertas 2)

Soalan 5:
Rajah di bawah menunjukkan trapezium ABCD.

(a) Hitung
(i) ∠BAC.
(ii) panjang, dalam cm, bagi AD.
(b) Garis lurus AB dipanjangkan ke B’ dengan keadaan BC = B’C.
(i) Lakar trapezium AB’CD.
(ii) Hitung luas, dalam, cm², bagi ∆BB’C.  


Penyelesaian:
(a)(i)
$\begin{array}{l}{5}^2=4^2+7^2-2\left(4\right)\left(7\right)\text{kos}\angle BAC\\ 25=16+49-56\text{kos}\angle BAC\\ 56\text{kos}\angle BAC=40\\ \text{kos}\angle BAC=\frac{40}{56}\\ \angle BAC={\text{kos}}^{-1}\frac{40}{56}\\ ={44}^{o}25\text{'}\end{array}$


(a)(ii)
$\begin{array}{l}\frac{AD}{\sin \angle DCA}=\frac{7}{\sin {115}^o}\\ \frac{AD}{\sin {44}^{o}25\text{'}}=\frac{7}{\sin {115}^o}\leftarrow \left(\angle DCA=\angle BAC\right)\\ AD=\frac{7}{\sin {115}^o}\times \sin {44}^{o}25\text{'}\\ AD=5.406\text{ cm}\end{array}$


(b)(i)




(b)(ii)
$\begin{array}{l}\frac{\sin \angle ABC}{7}=\frac{\sin {44}^{o}25\text{'}}{5}\\ \sin \angle ABC=\frac{\sin {44}^{o}25\text{'}}{5}\times 7\\ ={78}^{o}28\text{'}\\ \angle ABC={180}^o-{78}^{o}28\text{'}\\ \angle ABC={101}^{o}32\text{'}\left(\text{sukuan kedua}\right)\\ \\ \angle CBB\text{'}={180}^o-{101}^{o}32\text{'}={78}^{o}28\text{'}\\ \angle BCB\text{'}={180}^o-{78}^{o}28\text{'}-{78}^{o}28\text{'}={23}^{o}4\text{'}\\ \text{Luas bagi }△BB\text{'}C=\frac{1}{2}\times 5\times 5\times {23}^{o}4\text{'}\\ =4.898{\text{ cm}}^2\end{array}$