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# need some help

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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

Mar 24, 2022

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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.

Mar 24, 2022
edited by proyaop  Mar 24, 2022
edited by proyaop  Mar 24, 2022
edited by proyaop  Mar 24, 2022
edited by proyaop  Mar 24, 2022