9.4.2 Solution of Triangles Long Questions (Question 3 & 4)

Question 3:


Solution:
(a)
$\begin{array}{l}\frac{QS}{\sin P}=\frac{PS}{\sin Q}\\ \frac{QS}{\sin {85}^{\circ }}=\frac{13.1}{\sin {28}^{\circ }}\\ QS=\frac{13.1\times \sin {85}^{\circ }}{\sin {28}^{\circ }}\\ QS=27.8\text{ cm}\end{array}$

(c)
$\begin{array}{l}\text{Using cosine rule,}\\ Q{S}^2=Q{R}^2+R{S}^2-2\left(QR\right)\left(RS\right)\cos \angle QRS\\ {27.8}^2={6.4}^2+{25.98}^2-2\left(6.4\right)\left(25.98\right)\cos \angle QRS\\ 772.84=715.92-332.54\cos \angle QRS\\ \cos \angle QRS=\frac{715.92-772.84}{332.54}\\ \cos \angle QRS=-0.1712\\ \angle QRS={99.86}^{\circ }\end{array}$

Question 4:
Diagram below shows a quadrilateral PQRS.



(a) Find
(i) the length, in cm, of QS.
(ii) ∠QRS.
(iii) the area, in cm², of the quadrilateral PQRS.
(b)(i) Sketch a triangle S’Q’R’ which has a different shape from triangle SQR such that S’R’ = SR, S’Q’ = SQ and ∠S’Q’R’ = ∠SQR.
(ii) Hence, state ∠S’R’Q’.



Solution:
(a)(i)
$\begin{array}{l}\angle P=180-76-34=70\\ \frac{QS}{\sin 70}=\frac{8}{\sin 34}\\ QS=\frac{8\times \sin 70}{\sin 34}\\ =13.44\text{ cm}\end{array}$

(a)(ii)
$\begin{array}{l}{13.44}^2=6^2+9^2-2\left(6\right)\left(9\right)\cos \angle QRS\\ 108\cos \angle QRS=6^2+9^2-{13.44}^2\\ \cos \angle QRS=\frac{6^2+9^2-{13.44}^2}{108}\\ \angle QRS={\cos }^{-1}\left(-0.5892\right)\\ ={126}^{o}6‘\end{array}$

(a)(iii)
$\begin{array}{l}\text{Area of }PQRS\\ =\text{Area of }PQS+\text{Area of }QRS\\ =\left(\frac{1}{2}\times 8\times 13.44\times \sin 76\right)+\left(\frac{1}{2}\times 6\times 9\times \sin {126}^{o}6‘\right)\\ =52.16+21.82\\ =73.98{\text{ cm}}^2\end{array}$

(b)(i)



(b)(ii)
$\begin{array}{l}\angle S‘R‘Q‘=\angle S‘RR‘\\ =180-{126}^{o}6‘\\ ={53}^{o}54‘\end{array}$