If in a `DeltaABC,`
`2(cosA)/(a)+(cosB)/(b)+(2cosC)/(c)=(a)/(bc)+(b)/(ca)`
Then, the value of the `angleA` is ...... Degree

Answer

`90^(@)`

Explanation

Given , `(2 cosA)/(a)+(cosB)/(b)+(2cosC)/(c)=(a)/(bc)+(b)/(ca)" ".....(i)`
We know that `cos A=(b^(2)+c^(2)-a^(2))/(2bc)`
`cos B=(c^(2)+a^(2)-b^(2))/(2ac)`
`and cos C=(a^(2)+b^(2)-c^(2))/(2ab)`
On putting these values in Eq. (i) we get
`(2(b^(2)+c^(2)-a^(2))
)/(2abc)=(c^(2)+b^(2)-b^(2))/(2abc)+2(a^(2)+b^(2)-c^(2))/(2abc)=(a)/(bc)+(b)/(ca)`
`=(a^(2)+b^(2))/(abc)`
`rArr 3b^(2)+c^(2)+a^(2)=2a^(2)+2b^(2)`
`rArr b^(2)+c^(2)=a^(2)`
Hene, the angle A is `90^(@)`