Since we know that triangle ECT is right, dropping a perpendicular from point I (not labeled...it's where EC intersects the circle) to ET, creates similar triangle EIT'
We can find the length of IT' using trigonometry.
IT′=16sin(17)≈4.68so the ratio 16/4.68 = (16+r)/r
Solving this for r, we get16r≈4.86(16)+4.68r11.32r≈74.88r≈6.61
Using this in the above formula gets uss=πrθ/180≈π(6.61)(146)/180≈16.9
The given answer of 17 is really close.
