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

We have,
cos A=`(sinB)/(sinC)`
`rArr(b^(2)+c^(2)-a^(2))/(2bc)=b/c(because(sinA)/a=(sinB)/b=(sinC)/c=k)`
`rArrb^(2)+c^(2)-a^(2)=2b^(2)`
`rArrc^(2)-a^(2)=b^(2)`
`rArrc^(2)=a^(2)+b^(2)`
`rArrDelta` ABC right angled triangle at `angle`C.