Answer:
Area of ABCD = 959.93 units²
Step-by-step explanation:
a). By applying Sine rule in the ΔABD,
[tex]\frac{\text{SinA}}{46}=\frac{\text{Sin}\angle{DBA}}{35}[/tex]
[tex]\frac{\text{Sin110}}{46}=\frac{\text{Sin}\angle{DBA}}{35}[/tex]
Sin∠DBA = [tex]\frac{35\times \text{Sin}(110)}{46}[/tex]
m∠DBA = [tex]\text{Sin}^{-1}(0.714983)[/tex]
m∠DBA = 45.64°
Therefore, m∠ADB = 180° - (110° + 45.64°) = 24.36°
m∠ADB = 24.36°
c). Area of ABCD = Area of ΔABD + Area of ΔBCD
Area of ΔABD = AD×BD×Sin([tex]\frac{24.36}{2}[/tex])
= 35×46Sin(12.18)
= 339.68 units²
Area of ΔBCD = BD×BC×Sin([tex]\frac{59.92}{2}[/tex])°
= 46×27×(0.4994)
= 620.25 units²
Area of ABCD = 339.68 + 620.25
= 959.93 units²
