Let $T$ be a point inside square $EFGH$ such that $TE = \sqrt{6}$, $TF = 2 \sqrt{3}$, and $TG = 3 \sqrt{2}$. Find $\angle ETF,$ in degrees.

H              G

     T

E              F

Let s  =side of square

Let  angle TFE =  A

Let  angle TFG  = 90 - A   so that  cos (90 - A) =  sin A

Using the Law of Cosines twice

TG^2  =  s^2  + TF^2  -2 *s * sqrt (12) * sinA

TE^2  = s^2  + TF^2  - 2*s * sqrt (12) * cos A

18 = s^2 + 12  - 2*s*sqrt(12)* sin (A)

6  =  s^2 + 12  - 2*s*sqrt(12)*cos (A)              simplify this system

6 =   s^2  - s * 2sqrt (12) * sin (A)

-6  = s^2 - s * 2sqrt (12) * cos (A)

sin A  =  (6 - s^2) / (s*2sqrt (12))

cos A = (-6 -s^2) /(s*2sqrt (12))

Using the identity

sin^2 A + cos^2A = 1

(6 -s^2)^2 + (-6 -s^2)^2 =  (s*2sqrt(12))^2

36 - 12s^2 + s^4  +  36 +12s^2 + s^4  =  48s^2

2s^4 - 48s^2 + 72   = 0

s^4 - 24s^2 + 36 =  0

s^4  -24s^2  =   -36

s^4 -24s^2  + 144  =  -36+ 144

(s^2 - 12)^2  =  108

s^2 - 12 = sqrt (108)

s^2  =  sqrt (108) + 12

s^2  =  6sqrt (3) + 12

Using the Law of Cosines once more

s^2  =  TE^2 + TF^2  - 2 * sqrt (6) * sqrt (12) cos ( ETF)

6sqrt (3) + 12  =  6 + 12  - 2 * sqrt (72) * cos (ETF)       ...     {2 sqrt (72)  =  12sqrt(2) }

6sqrt (3) - 6  = - 12sqrt (2) * cos (ETF)

sqrt  (3) - 3  =  - 2sqrt (2) * cos (ETF)

cos (ETF)  =   ( sqrt(3) - 3)  / ( -2sqrt (12))         

arccos  [ 3 -sqrt (3) ] / [ sqrt (48) ]   =  ETF  ≈  79.45°