about to only mathematics

Answer

2

Explanation

Let AD be the median to the base BC of a of `DeltaABC and "let" angle ADC= theta`, then
`((a)/(2)+(b)/(2)) cot theta=(a)/(2)cos 30^(@)-(a)/(2)cot 45^(@)`
`rArr cot theta=(sqrt(3)-1)/(2)`

Applying sine rule in `DeltaABC`, we get
`(AD)/(sin (pi-theta-45^(@)))=(DC)/(sin45^(@))`
`rArr (AD)/(sin(theta+45^(@)))((a)/(2))/((1)/(sqrt(2)))`
`rArr AD=(a)/(sqrt(2)) (sin 45^(@) cos theta+ cos 45^(@) sin theta)`
`rArr AD=(a)/(sqrt(2)) ((cos theta+sin theta)/(sqrt(2)))=(a)/(2)(cos theta+sin theta)`
`rArr (1)/(sqrt(11-6sqrt(3)))=(a)/(2)((sqrt(3)-1)/(sqrt(8-2sqrt(3)))+(2)/(sqrt(8-2sqrt(3))))`
`rArr a=(2sqrt(8-2sqrt(3)))/((sqrt(3)+1)sqrt(11-6sqrt(3)))=(2sqrt(8-2sqrt(3)))/(sqrt((sqrt(3)+1)^(2)sqrt(11-6sqrt(3))))`
`rArr a=(2sqrt(8-2sqrt(3)))/((sqrt(4+2)+sqrt(3))sqrt(11-6sqrt(3)))`
`rArr a=(2sqrt(8-2sqrt(3)))/((sqrt(44-24)sqrt(3)+22sqrt(3)-36))`
`=2(sqrt(8-2sqrt3))/(sqrt(8-2sqrt(3)))=2`