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