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With usual notations, in triangle A B C ...

With usual notations, in triangle `A B C ,acos(B-C)+bcos(C-A)+c"cos"(A-B)` is equal to `a b c//R^2` (b) `(a b c)/(4R^2)` `(4a b c)/(R^2)` (d) `(a b c)/(2R^2)`

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With usual notations, in triangle A B C ,acos(B-C)+bcos(C-A)+c"cos"(A-B) is equal to(a) (a b c)/R^2 (b) (a b c)/(4R^2) (c) (4a b c)/(R^2) (d) (a b c)/(2R^2)

In a triangle A B C ,(r_1+r_2)/(1+cosC) is equal to (2a b)/(c"Delta") (b) ((a+b))/(c"Delta") (a b c)/(2"Delta") (d) (a b c)/("Delta"^2)

In any triangle A B C , prove that: acos((B-C)/2)=(b+c)sin(A/2)

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In a triangle A B C , right angled at C ,t a n A+t a n B is equal to a. a+b b. (c^2)/(a b) c. (a^2)/(b c) d. (b^2)/(a c)

In a A B C(b csin^2A)/(cos A+cosB cos C) is equal to b^2+C^2 b. b c c. a^2 d. a^2+b c

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For any triangle ABC, prove that (b+c)cos((B+C)/2)=acos((B-C)/2)

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