Question Number 164310 by ZiYangLee last updated on 16/Jan/22
Answered by Rasheed.Sindhi last updated on 16/Jan/22
$${Let}\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}}\:={c} \\ $$$${Obviously}\:{c}\geqslant{a}\:\&\:{c}\geqslant{b} \\ $$$${So}\:{the}\:{greater}\:{angle}\:{is}\:{opposite}\:{to}\:{c} \\ $$$${Now},\:{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\left(\mathrm{120}°\right) \\ $$$$\Rightarrow{Opposite}\:{angle}\:{to}\:{c}\:{is}\:\mathrm{120}° \\ $$$${i}-{e}\:{the}\:{greatest}\:{angle}\:{is}\:\mathrm{120}° \\ $$$$\begin{array}{|c|}{\:\:\:\underset{\:\:\:\left(\gamma\:{is}\:{opposite}\:{angle}\:{to}\:{c}\right)} {\overset{\boldsymbol{{Cosine}}\:\boldsymbol{{law}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\gamma}}\:}\\\hline\end{array}\: \\ $$
Commented by mr W last updated on 16/Jan/22
$${nicely}\:{combined}! \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jan/22
$$\mathcal{T}{hanks}\:\boldsymbol{{sir}}! \\ $$