9.4.3 Solution of Triangles Long Questions (Question 5 & 6)

Question 5:


Solution:
(a)


(b)

$\begin{array}{l}\text{Using sine rule,}\\ \frac{\sin \angle ADC}{16.39}=\frac{\sin {45}^{\circ }}{14}\\ \sin \angle ADC=\frac{16.39\times \sin {45}^{\circ }}{14}\\ \sin \angle ADC=0.8278\\ \angle ADC={55.87}^{\circ }\text{ or }\left({180}^{\circ }-{55.87}^{\circ }\right)\\ \angle ADC={55.87}^{\circ }\text{ or 124}{\text{.13}}^{\circ }\end{array}$

(c)(i)
$\begin{array}{l}\text{Acute angle of }\angle ADC={55.87}^{\circ }\\ \angle CAD={180}^{\circ }-{45}^{\circ }-{55.87}^{\circ }={79.13}^{\circ }\\ \frac{CD}{\sin {79.13}^{\circ }}=\frac{14}{\sin {45}^{\circ }}\\ CD=\frac{14\times \sin {79.13}^{\circ }}{\sin {45}^{\circ }}=19.44\text{ cm}\end{array}$

(c)(ii)
$\begin{array}{l}\text{Area of quadrilateral }ABCD\\ \text{= Area of }\Delta ABC+\text{Area of }\Delta ACD\\ =\frac{1}{2}\left(16\right)\left(12\right)\sin {70}^{\circ }+\frac{1}{2}\left(16.39\right)\left(14\right)\sin {79.13}^{\circ }\\ =90.21+112.67\\ =202.88{\text{ cm}}^2\end{array}$

Question 6:
Diagram below shows trapezium ABCD.
(a) Calculate
(i) ∠BAC.
(ii) the length, in cm, of AD.
(b) The straight line AB is extended to B’ such that BC = B’C.
(i) Sketch the trapezium AB’CD.
(ii) Calculate the area, in cm², of ∆BB’C.  



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


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


(b)(i)




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