(Long Questions) – Question 6

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


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