Given:
Sides of triangle a = 1200, b =859, c = 956.
Required:
Angles of a triangle.
Explanation:
We know that
[tex]\begin{gathered} CosA=\frac{b^2+c^2-a^2}{2bc} \\ CosB=\frac{a^2+c^2-b^2}{2ac} \\ CosC=\frac{a^2+b^2-c^2}{2ab} \end{gathered}[/tex]
Now,
[tex]\begin{gathered} CosA=\frac{859^2+956^2-1200^2}{2\times859\times956} \\ CosA=\frac{737881+913936-1440000}{1642408} \\ A=Cos^{-1}(0.13) \end{gathered}[/tex]
And
[tex]\begin{gathered} CosB=\frac{1200^2+956^2-859^2}{2\times1200\times956} \\ CosB=\frac{1440000+913936-737881}{2294400} \\ A=Cos^{-1}(0.14) \end{gathered}[/tex]
And
[tex]\begin{gathered} CosC=\frac{1440000+737881-913936}{2\times1200\times859} \\ C=Cos^{-1}(0.61) \end{gathered}[/tex]
Answer:T
These are three angles of triangle.
