Question-164310

Question Number 164310 by ZiYangLee last updated on 16/Jan/22

Answered by Rasheed.Sindhi last updated on 16/Jan/22

L e t \sqrt{a^{2} + b^{2} + a b} = c

O b v i o u s l y c ⩾ a & c ⩾ b

S o t h e g r e a t e r a n g l e i s o p p o s i t e t o c

N o w, c^{2} = a^{2} + b^{2} + a b = a^{2} + b^{2} - 2 a b (- \frac{1}{2})

c^{2} = a^{2} + b^{2} - 2 a b \cos (120 °)

\Rightarrow O p p o s i t e a n g l e t o c i s 120 °

i - e t h e g r e a t e s t a n g l e i s 120 °

\begin{matrix} \underset{(γ i s o p p o s i t e a n g l e t o c)}{\overset{C o s i n e l a w}{c^{2} = a^{2} + b^{2} - 2 a b \cos γ}} \end{matrix}

Commented by mr W last updated on 16/Jan/22

n i c e l y c o m b i n e d!

Commented by Rasheed.Sindhi last updated on 16/Jan/22

T h a n k s s i r!