Can there exist triangles ABC satisfying the following relations? Wrire yes or no giving reasons:
$(a + b)^{2} = c^{2} + a b$ and $\sqrt{2} (s i n A + c o s A) = \sqrt{3} .$

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Solution

Yes.
$a^{2} + b^{2} - c^{2} + a b = 0 \Rightarrow 2 a b c o s C + a b = 0$
$c o s C = - \frac{1}{2} \Rightarrow= C = 120^{0}$
Also from 2nd relation
We have $s i n (A + \frac{π}{4} = \frac{\sqrt{3}}{2} = s i n \frac{p i}{3} ∴ A = 15^{0}$
and hence $B = 45^{0}$ . Such a $△$ is possible.

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