Answer:
The area of triangle is \(110.10cm^{2}\)
Step-by-step explanation:
It is given that AB=19cm, BC=12cm and ∠B=105°
Now, Area of the triangle ABC=\(\frac{1}{2}acsinB\)
=\(\frac{1}{2}{\times}12{\times}19{\times}sin105^{\circ}\)
=\(6{\times}19{\times}(0.965)\)
=\(110.10cm^{2}\)
Thus, the area of triangle is \(110.10cm^{2}\)
