SPM Additional Mathematics 2019, Paper 2 (Question 13)

Question 13:
Solution by scale drawing is not accepted.
Diagram 6 shows a quadrilateral ABCD such that AC and BD are straight lines.

Diagram 6

It is given that the area of ∆
ABC = 6 cm2 and ABC is obtuse.
(a) Find
(i) ABC,
(ii) the length, in cm, of AC,
(iii) BAC            [7 marks]
(b) Given BD = 7.3 cm and BCD = 90°, calculate the area in cm2, of ∆ACD. [3 marks]

Solution:
(a)(i)

(a)(ii)

(a)(iii)
$\begin{array}{l}\frac{\mathrm{sin}\angle BAC}{3.5}=\frac{\mathrm{sin}121}{6.532}\\ \mathrm{sin}\angle BAC=\frac{3.5×\mathrm{sin}121}{6.532}\\ \angle BAC={27}^{o}20‘\end{array}$

(b)