In `DeltaABC`, with usual notations, if `cosA=(sinB)/(sinC)`, then the triangle is

Options

A. Acute angled triangle
B. Equilateral triangle
C. Obtuse angled triangle
D. Right angled triangle

Answer

D

Explanation

Key idea Use sine rule,
`(sinA)/(a)=(sinB)/(b)=(sinC)/(c)`
We have,
`cosA=(sinB)/(sinC)`
`implies(b^(2)+c^(2)-a^(2))/(2bc)=(b)/(c)" "(because(sinA)/(a)=(sinB)/(b)=(sinC)/(c)=k)`
`impliesb^(2)+c^(2)-a^(2)=2b^(2)`
`impliesc^(2)-a^(2)=b^(2)`
`impliesc^(2)=a^(2)+b^(2)`
`impliesDeltaABC` right angled triangle at `angleC`.