Question

In the adjoining figure of $\Delta ABC,\angle BCA={120}^{\circ }$ and $AB=c$,$BC=a$,$AC=b$ then:

• A. $c^2=a^2+b^2+ba$
• B. $c^2=a^2+b^2-ba$
• C. $c^2=a^2+b^2-2ba$
• D. $c^2=a^2+b^2+2ba$

Answer 1

The correct option is A $c^2=a^2+b^2+ba$

Given In $△ABC,\angle BCA={120}^{\circ }$

As we know that

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$⟹\cos {120}^{\circ }=\frac{a^2+b^2-c^2}{2ab}$

$⟹-\frac{1}{2}=\frac{a^2+b^2-c^2}{2ab}$

$⟹-ab=a^2+b^2-c^2$

$⟹c^2=a^2+b^2+ab$