Question

Assertion :STATEMENT- 1 : In a triangle, $a^2+b^2+c^2=ca+ab\sqrt{3}$ . Then the triangle is right angled with $A={90}^o,B={60}^o,C={30}^o$. Reason: STATEMENT- 2 : If ${\alpha }^2+{\beta }^2=0,\alpha ,\beta \in R$, then $\alpha =\beta =0$

• A. Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation for Statement -1
• B. Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement -1 is True, Statement -2 is False
• D. Statement -1 is False, Statement -2 is True

Answer 1

The correct option is A Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation for Statement -1

STATEMENT- 1 $4(a^2+b^2+c^2)=4ca+4ab\sqrt{3}$
(multiply by 4)
$(a^2+4c^2-4ac)+(4b^2+3a^2-4\sqrt{3}ab)=0$
$\Rightarrow (2c-a{)}^2+(2b-\sqrt{3}a{)}^2=0$
$\Rightarrow a=2c$ & $b=\frac{\sqrt{3}a}{2}$
So, $cosA=\frac{\frac{3}{4}{a}^2+\frac{a^2}{4}-a^2}{2bc}=0$
So, $\angle A={90}^0,\angle B={60}^0,\angle C={30}^0$
STATEMENT- 2 : It is a standard result.