Question

Given below there are certain matching type questions, where two columns are given.Each item of column$I$ has to be matched with the items of column $II$, by encircling the correct matches:

Answer 1

A.)$a^4-2(b^2+c^2)a^2+b^4+b^2{c}^2+c^4=0$
$\Rightarrow {(b^2+c^2-a^2)}^2-{(b^2+c^2)}^2+b^4+b^2{c}^2+c^4=0$
$\Rightarrow {(b^2+c^2-a^2)}^2=b^2{c}^2$
$\Rightarrow {\left(\frac{b^2+c^2-a^2}{2bc}\right)}^2=\frac{1}{4}$
$\Rightarrow {\cos }^{2}A=\frac{1}{4}$
$\therefore \cos A=\pm \frac{1}{2}$
$\therefore A={60}^0$ or ${120}^0$
B.)$\because \cos C=\frac{a^2+b^2-c^2}{2ab}$
or $a^2+b^2-c^2=2ab\cos C$
Squaring both sides, we get
$a^4+b^4+c^4+2a^2{b}^2-2b^2{c}^2-2c^2{a}^2=4a^2{b}^2{\cos }^{2}C$
$\Rightarrow 3a^2{b}^2=4a^2{b}^2{\cos }^{2}C$
$\Rightarrow {\cos }^{2}C=\frac{3}{4}$
or $\cos C=\pm \frac{\sqrt{3}}{2}$
$\therefore C={30}^0$ when $\cos C=\frac{\sqrt{3}}{2}$
and $C=180-30={150}^0$ when $\cos C=-\frac{\sqrt{3}}{2}$
C)$\because a^4+b^4+c^4+2a^2{c}^2=2a^2{b}^2+2b^2{c}^2$
$\Rightarrow b^4-2(a^2+c^2)b^2+{(a^2+c^2)}^2=0$
$\Rightarrow {[b^2-(a^2+c^2)]}^2=0$
$\Rightarrow b^2-a^2-c^2=0$
$\therefore b^2=a^2+c^2$
$\Rightarrow \cos B=0$
$\therefore B={90}^0$