sLet \(T\) 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.

Picture here: https://rb.gy/xzupss

Guest Mar 24, 2022

#1**+2 **

I would like to give the problem a shot:

First I would like to draw an altitute from T to FG and from T to EF. The points of intersection are A and B respectively.

Intuitively, you can see that triangle FAT is a 30-60-90 triangle, and TAG is a 45-45-90 triangle, and you could find that **angle ETF = 105 degrees**.

To give a serious go, we can use the fact that putting a rectangle with 2 sides on the sides of a 45-45-90 triangle will create 2 similar triangles that are 45-45-90 and a rectangle:

Since BT = FA because it is a rectangle, and TG = 3sqrt(2), then AG = 3, and FA = sqrt(3).

Tan TFA = 3 / sqrt(3) = sqrt(3), thus angle TFA = 60 degrees. Angle FAT is also 90 degrees. Angle FTB is congruent to angle TFA, so angle BTF = 60 degrees.

Since angle ETB is 45 degrees, and angle ETB + angle BTF = angle ETF, then 60 + 45 = angle ETF.

Thus, **angle ETF = 105 degrees**.

proyaop Mar 24, 2022