The sides a, b, c of a triangle satisfy the r | vedclass

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(B) Given relations for the sides of the triangle are:$c^2 = 2ab$  $(i)$$a^2 + c^2 = 3b^2$  $(ii)$Substituting $(i)$ into $(ii)$:$a^2 + 2ab = 3b^2$$a^

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$45$

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(B) Given relations for the sides of the triangle are:$c^2 = 2ab$  $(i)$$a^2 + c^2 = 3b^2$  $(ii)$Substituting $(i)$ into $(ii)$:$a^2 + 2ab = 3b^2$$a^2 + 2ab - 3b^2 = 0$$(a + 3b)(a - b) = 0$Since $a$ and $b$ are side lengths,$a, b > 0$,so $a = b$.Substituting $a = b$ into $(i)$:$c^2 = 2(b)(b) = 2b^2$$c = \sqrt{2}b$Now,we have sides $a = b$,$b = b$,and $c = \sqrt{2}b$.Since $a^2 + b^2 = b^2 + b^2 = 2b^2 = c^2$,the triangle is a right-angled triangle with the right angle at $C$ (i.e.,$\angle C = 90^{\circ}$).Since $a = b$,the triangle is an isosceles right-angled triangle.Therefore,$\angle A = \angle B = 45^{\circ}$.The measure of $\angle BAC$ is $45^{\circ}$.

The sides $a, b, c$ of a triangle satisfy the relations $c^2=2ab$ and $a^2+c^2=3b^2$. Then the measure of $\angle BAC$,in degrees,is

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