In ` A B C ,(a+b+c)(b+c-a)=k b c` if `k<0` (b) `k >0` `=4`

In A B C ,= if (a+b+c)(a-b+c)=3a c , then find /_Bdot

If a ,b ,a n dc are in H.P., then th value of ((a c+a b-b c)(a b+b c-a c))/((a b c)^2) is ((a+c)(3a-c))/(4a^2c^2) b. 2/(b c)-1/(b^2) c. 2/(b c)-1/(a^2) d. ((a-c)(3a+c))/(4a^2c^2)

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 B C , if (sinA)/(csinB)+(sinB)/c+(sinC)/b=c/(a b)+b/(a c)+a/(b c), then the value of angle A is 120^0 (b) 90^0 (c) 60^0 (d) 30^0

The straight line a x+b y+c=0 , where a b c!=0, will pass through the first quadrant if (a) a c >0,b c >0 (b) ac >0 or b c 0 or a c >0 (d) a c<0 or b c<0

If in a triangle ABC, (1 + cos A)/(a) + (1 + cos B)/(b) + (1+ cos C)/(c) = (k^(2) (1 + cos A) (1 + cos B) (1 + cos C))/(abc) , then k is equal to

Prove that =|1 1 1 a b c b c+a^2a c+b^2a b+c^2|=2(a-b)(b-c)(c-a)

If in A B C , side a , b , c are in A.P. then (a) B > 60^0 (b) B<60^0 (c) Blt=60^0 (d) B=|A-C|

If in the expansion of (1+x)^n ,a ,b ,c are three consecutive coefficients, then n= (a c+a b+b c)/(b^2+a c) b. (2a c+a b+b c)/(b^2-a c) c. (a b+a c)/(b^2-a c) d. none of these

The value of the determinant |k a k^2+a^2 1k b k^2+b^2 1k c k^2+c^2 1| is k(a+b)(b+c)(c+a) k a b c(a^2+b^(f2)+c^2) k(a-b)(b-c)(c-a) k(a+b-c)(b+c-a)(c+a-b)