Matematik Tambahan SPM 2017, Kertas 2 (Soalan 15)

Soalan 15 (10 markah):
Penyelesaian secara lukisan berskala tidak diterima.
Rajah 8 menunjukkan sisi empat PQRS pada suatu satah mengufuk.
VQSP ialah sebuah piramid dengan keadaan PQ = 12 m dan V adalah 5 m tegak di atas P.
Cari
(a) ∠QSR,
(b) panjang, dalam m, bagi QS,
(c) luas, dalam m², bagi satah condong QVS.

 

Penyelesaian: 
(a)

$\begin{array}{l}\frac{\sin \angle QSR}{20.5}=\frac{\sin {64}^o}{22}\\ \sin \angle QSR=\frac{\sin {64}^o}{22}\times 20.5\\ \sin \angle QSR=0.8375\\ \angle QSR={56}^{o}52‘\end{array}$

(b)

$\begin{array}{l}\angle QRS={180}^o-{64}^o-{56}^{o}52‘\\ ={59}^{o}8‘\\ \frac{QS}{\sin {59}^{o}8‘}=\frac{22}{\sin {64}^o}\\ QS=\frac{22}{\sin {64}^o}\times \sin {59}^{o}8‘\\ QS=21.01\text{ m}\end{array}$

(c)

$\begin{array}{l}Q{V}^2=P{Q}^2+V{P}^2\\ QV=\sqrt{{12}^2+5^2}\\ QV=13\text{ m}\\ \\ S{V}^2=P{S}^2+V{P}^2\\ SV=\sqrt{{10}^2+5^2}\\ SV=\sqrt{125}\text{ m}\\ \\ QS=21.01\text{ m}\\ \\ {21.01}^2={13}^2+{\left(\sqrt{125}\right)}^2-2\left(13\right)\left(\sqrt{125}\right)kos\theta \\ 26\left(\sqrt{125}\right)kos\theta =169+125-441.42\\ kos\theta =\frac{169+125-441.42}{26\left(\sqrt{125}\right)}\\ kos\theta =-0.5071\\ \theta ={120}^{o}28‘\\ \\ \text{Luas }\Delta QVS\\ =\frac{1}{2}\times 13\times \sqrt{125}\times \sin {120}^{o}28‘\\ =62.64{\text{ m}}^2\end{array}$