Question

Which of the following statements is/are true

$I$: lf $(a+b+c)(a+b-c)=ab$ then $\angle C={60}^0$

$II$: lf ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$ in a $\Delta ABC$ then it is right angled triangle.

• A. only $I$
• B. only $II$
• C. both $I$ and $II$
• D. neither $I$ nor $II$

Answer 1

The correct option is B only $II$

$(a+b{)}^2-c^2=ab$

$a^2+b^2+2ab-c^2=ab$

$a^2+b^2-c^2=-ab$

$\frac{a^2+b^2-c^2}{ab}=-1$

$cosc=\frac{a^2+b^2-c^2}{2ab}$

$cosc=\frac{-1}{2}$

$\angle c=120$

Now,

$1-si{n}^{2}A+1-si{n}^{2}B+si{n}^{2}C=2$

$si{n}^{2}C=si{n}^{2}A+si{n}^{2}B$

$si{n}^{2}C-si{n}^{2}A=si{n}^{2}B$

$\frac{1-cos2C}{2}-\frac{(1-cos2A)}{2}=si{n}^{2}B$

$\frac{cos2A-cos2C}{2}=si{n}^{2}B$

$\frac{2sin(A+C)\cdot sin(C-A)}{2}=si{n}^{2}B$

$sinB\cdot sin(C-A)=si{n}^{2}B$

$sin(C-A)=sinB$

$C-A=B$

$C=A+B$

$\therefore \angle C=\frac{\pi }{2}$

$\therefore$ only II is true.