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find the area of the triangle
1. C=110 degrees, a=6, b=10
2. B=130 degrees, a=92, c=30


Sagot :

[tex]1.\ \ \ |\angle \ ACB|=110^0\\\\.\ \ \ \ \ a=|BC|=6\ [u]\ \ \ and\ \ \ b=|AC|=10\ [u]\\\\Area= \frac{1}{2} \cdot |AC|\cdot |BC|\cdot sin(|\angle \ ACB|)\\\\Area= \frac{1}{2} \cdot 6\cdot 10\cdot sin110^0=30\cdot sin (180^0-70^0)=\\\\.\ \ \ \ \ \ =30\cdot sin70^0\approx30\cdot 0.9397=28.191\ [u^2]\\\\[/tex]

[tex]2.\ \ \ |\angle \ ABC|=130^0\\\\.\ \ \ \ \ a=|BC|=92\ [u]\ \ \ and\ \ \ c=|AB|=30\ [u]\\\\Area= \frac{1}{2} \cdot |AB|\cdot |BC|\cdot sin(|\angle \ ABC|)\\\\Area= \frac{1}{2} \cdot 30\cdot 92\cdot sin130^0=1380\cdot sin (180^0-50^0)=\\\\.\ \ \ \ \ \ =1380\cdot sin50^0\approx1380\cdot 0.7660=1057.08\ [u^2]\\\\[/tex]