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°