+0  
 
0
13
1
avatar+297 

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?

 May 4, 2024
 #1
avatar+1459 
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∘​.

 May 4, 2024

2 Online Users