Question

If ${\cos }^{2}A+{\cos }^{2}C={\sin }^{2}B$, then $△ABC$ is

• A. equilateral
• B. right angled
• C. isosceles
• D. None of the above

Answer 1

Answer: (B)

Given, ${\cos }^{2}A+{\cos }^{2}C={\sin }^{2}B$
Obviously it is not an equilateral triangle because $A=B=C=60^{\circ }$ does not satisfy the given condition.
But $B=90^{\circ }$, then ${\sin }^{2}B=1$ and ${\cos }^{2}A+{\cos }^{2}C={\cos }^{2}A+{\cos }^2\left(\frac{\pi }{2}-A\right)$
Hence, this satisfies the condition, so it is a right angled triangle but not necessarily isosceles triangle.