MATH SOLVE

7 months ago

Q:
# Consider circle H with a 6 centimeter radius. If the length of minor arc ST is 11/12 π, what is the measure of ∠RST? Assume RS ≅ TS.A) 15° B) 25° C) 30° D) 45°

Accepted Solution

A:

check the picture below.

since the arcs of ST and RS are twins, and the circle has a total radians of 2π, then the arcRT is just 2π - arcST - arcRS.

As you can see, arcRT is the "intercepted arc" by the "inscribed angle" RST, thus

[tex]\bf \widehat{RT}=2\pi -\widehat{ST}-\widehat{RS}\implies \widehat{RT}=2\pi -\cfrac{11\pi }{12}-\cfrac{11\pi }{12} \\\\\\ \widehat{RT}=2\pi -\cfrac{22\pi }{12}\implies \widehat{RT}=\cfrac{2\pi }{12}\implies \widehat{RT}=\cfrac{\pi }{6}\\\\ -------------------------------\\\\ \measuredangle RST=\cfrac{\pi }{6}\cdot \cfrac{1}{2}\implies \measuredangle RST=\cfrac{\pi }{12}[/tex]

since the arcs of ST and RS are twins, and the circle has a total radians of 2π, then the arcRT is just 2π - arcST - arcRS.

As you can see, arcRT is the "intercepted arc" by the "inscribed angle" RST, thus

[tex]\bf \widehat{RT}=2\pi -\widehat{ST}-\widehat{RS}\implies \widehat{RT}=2\pi -\cfrac{11\pi }{12}-\cfrac{11\pi }{12} \\\\\\ \widehat{RT}=2\pi -\cfrac{22\pi }{12}\implies \widehat{RT}=\cfrac{2\pi }{12}\implies \widehat{RT}=\cfrac{\pi }{6}\\\\ -------------------------------\\\\ \measuredangle RST=\cfrac{\pi }{6}\cdot \cfrac{1}{2}\implies \measuredangle RST=\cfrac{\pi }{12}[/tex]