Question

Assertion :Statement 1:If in a triangle ${\sin }^{2}A+{\sin }^{2}B+si{n}^{2}C=2$, then one of the angles must be ${90}^{\circ }.$ Reason: Statement 2: For any triangle,${\sin }^{2}A+{\sin }^{2}B+si{n}^{2}C=2+2\cos A\cos B\cos C$

• A. Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
• B. Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Answer 1

The correct option is A Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}B+{\sin }^{2}C=2$------------1

In a triangle, $A+B+C=\pi$

${\sin }^A,{\sin }^{2}B,{\sin }^{2}C\le 1$

Any one of ${\sin }^2,{\sin }^{2}B,{\sin }^{2}C$ must be $=1$

So, any of the three angle must be ${90}^{\circ }$ , another two angles are acute angle and complementary.

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}B+{\sin }^{2}C=2+\cos A\cos B\cos C$

From this equation, if equation has to verify then ,

(1) $\cos A\cos B\cos C$ must be $0$

So, any of the angle must be ${90}^{\circ }$