If in ` A B C ,A=pi/7,B=(2pi)/7,C=(4pi)/7` then `a^2+b^2+c^2` must be `R^2` (b) `3R^2` (c) `4R^2` (d) `7R^2`

In trgABC , a=7, b=2, B= pi /4 then, c= ?

(-4/7 a^2 b)xx(-2/3 b^2c)xx(-7/6 c^2a)

If ' O ' is the circum centre of the A B C and R_1,\ R_2,\ R_3 are the radii of the circumcircle of triangle O B C ,\ O C A\ a n d\ O A B respectively then a/(R_1)+b/(R_2)+c/(R_3) has the value equal to (a b c)/(2R^3) (b) (R^3)/(a b c) (c) (4)/(R^2) (d) (a b c)/(R^3)

In triangle ABC, if r_(1)=2r_(2)=3r_(2), then a:b is equal to (5)/(4) (b) (4)/(5)(c)(7)/(4)(d)(4)/(7)

In !ABC,angleA=pi/2b=4,c=3 ,then the value of R/r is equal to

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

In a triangle ABC,a=4,b=3,/_A=60^(@) then c is root of the equation c^(2)-3c-7=0 (b) c^(2)+3c+7=0(c)c^(2)-3c+7=0(d)c^(2)+3c-7=0

If (3-tan^(2)((pi)/(7)))/(1-tan^(2)((pi)/(7)))=k cos((pi)/(7)) then the value 1-tan^(2)((pi)/(7)) of k is (a) (b)2(c)3(d)4