Question

If $co{s}^{2}A+co{s}^{2}C=si{n}^{2}B,$ then $△$ ABC is
[MP PET 1991]

• A. Equilateral
• B. Right angled
• C. Isosceles
• D. None of these

Answer 1

The correct option is B

Right angled

Trick: Obviously it is not an equilateral triangle because A=B=C=60^o does not satisfy the given condition. But B = 90^o then $si{n}^2$ B = 1 and

$co{s}^{2}A+co{s}^{2}C=co{s}^{2}A+co{s}^2(\frac{\pi }{2}-A)$
$=co{s}^{2}A+si{n}^{2}A=1$
Hence this satisfy the condition, so it is a right angle triangle but not necessarily isosceles.