Points $T$ and $U$ lie on a circle centered at $O$, and point $P$ is outside the circle such that $\overline{PT}$ and $\overline{PU}$ are tangent to the circle. If $\angle TOP = 45^{\circ}$, then what is the measure of minor arc $TU$, in degrees?

rtsdylifdts May 4, 2024

#1**0 **

Since PT and PU are tangent to the circle centered at O, then ∠PTO and ∠PUO are right angles. Additionally, since ∠TOP=45∘, then △PTO is a 45-45-90 triangle.

In a 45-45-90 triangle, each of the acute angles measures 45∘, and the hypotenuse (opposite the right angle) has a length 2 times the length of either leg. Therefore, TO=OP⋅2.

Because TU is a minor arc and ∠TOP=45∘, then arc TU intercepts a central angle ∠TOU that also measures 45∘.

Since central angles and their corresponding minor arcs have the same measure in degrees, then minor arc TU has a measure of 45∘.

blackpanther May 4, 2024