In $△ A B C$ , if $cos A = sin B - cos C,$ then triangle is

Right angled triangle

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Equilateral triangle

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Isosceles triangle

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None of these

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Solution

The correct option is A Right angled triangle

$c o s A = s i n B - c o s C c o s A + c o s C = s i n B 2 c o s ((\frac{C + A}{2})) c o s ((\frac{A - C}{2})) = s i n B a s w e k n o w A + B + C = π ∴ A + C = π - B o r 2 c o s ((\frac{π}{2}) - (\frac{B}{2})) c o s ((\frac{A - C}{2})) = s i n B 2 s i n (\frac{B}{2}) c o s (\frac{A - C}{2}) = 2 s i n (\frac{B}{2}) c o s (\frac{B}{2}) ∴ (\frac{A - C}{2}) = (\frac{B}{2}) o r A = B + C n o w A + B + C = π o r A + A = π ∴ A = (\frac{π}{2}) H e n c e R i g h t a n g l e d t r i a n g l e$

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