Question

Can someone help me solve (a) and (c) pls.Thanks ​

Answer 1

Answer:

Area of ABCD = 959.93 units²

Step-by-step explanation:

a). By applying Sine rule in the ΔABD,

  $\frac{\text{SinA}}{46}=\frac{\text{Sin}\angle{DBA}}{35}$

  $\frac{\text{Sin110}}{46}=\frac{\text{Sin}\angle{DBA}}{35}$

 Sin∠DBA = $\frac{35\times \text{Sin}(110)}{46}$

 m∠DBA = $\text{Sin}^{-1}(0.714983)$

 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($\frac{24.36}{2}$)

                           = 35×46Sin(12.18)

                           = 339.68 units²

   Area of ΔBCD = BD×BC×Sin($\frac{59.92}{2}$)°

                            = 46×27×(0.4994)

                            = 620.25 units²

   Area of ABCD = 339.68 + 620.25

                            = 959.93 units²