In `DeltaABC, a=2, b=sqrt(6)` and `c=sqrt(3)+1`, find `angleA`.

1.	In `DeltaABC, a=2, b=sqrt(6)` and `c=sqrt(3)+1`, find `angleA`.
Answer» `cosA=(b^(2)+c^(2)-a^(2))/(2bc)` `=(sqrt(6))^(6)+((sqrt(3)+1)^(2)-2^(2))/(2sqrt(6)(sqrt(3)+1))` `=(2sqrt(3)(sqrt(3)+1))/(2sqrt(6)(sqrt(3)+1))` `=1/sqrt(2)=cos45^(@)` `A=45^(@)`.

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