ln `Delta ABC`, `tan A=3`,`tan B=(2)/(3)` `c=sqrt(10)` then circum radius of `Delta` is
(1) `11sqrt((13)/(10))`
(2) `(10sqrt(13))/(11)`
(3) `(sqrt(130))/(11)`
(4) `(5sqrt(13))/(11)`

Simplify (sqrt(13)-sqrt(11))/(sqrt(13)+sqrt(11)) + (sqrt(13)+sqrt(11))/(sqrt(13)-sqrt(11))

Slove (3-sqrt(11)) (3+ sqrt(11))

If combined equation of two sides AB and AC of a triangle is x^(2)-xy-2y^(2)+x+y=0 and (1,1) is orthocenter then circumradius of Delta ABC is 1) (5sqrt(2))/(3) 2) (10sqrt(2))/(3) 3) (3sqrt(2))/(5) 4) (3)/(10sqrt(2))

(1)/(sqrt(11-2sqrt(30)))-(3)/(sqrt(7-2sqrt(10)))-(4)/(sqrt(8+4sqrt(3)))

The sum to n terms of the series (1)/(sqrt(7)+sqrt(10))+(1)/(sqrt(10)+sqrt(13))+(1)/(sqrt(13)+sqrt(16))+... is

If x = (5-sqrt(3))/(5+sqrt(3)) and y = (5+sqrt(3))/(5-sqrt(3)) , show that x -y = - (10sqrt(3))/(11)

if cos^(-1)((2)/(3x))+cos^(-1)((3)/(4x))=(pi)/2, x>3/4 then x= (a) (sqrt(145))/(11) (b) (sqrt(145))/(12) (c) (sqrt(146))/(10) (d) (sqrt(146))/(11)

If G is the centroid of triangle ABC and BC=3,CA=4 AB=5 then BG= (sqrt(13))/(3) (2sqrt(13))/(3) sqrt(13) (4sqrt(13))/(3)

Multiply 5 sqrt(11) by 3sqrt(11)

the value of |((3-i sqrt(2))^(2))/(1+i2)| is equal to (i) (11)/(sqrt(5)) (ii) (18)/(sqrt(7)) (iii) (5)/(sqrt(11)) (iv) (13)/(sqrt(3))