Question

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°

Answer 1

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


`\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}`

Answer 2

The measure of ∠RST=15°

A is correct.

Explanation:

Given: Circle with centre H. Radius of circle is 6 cm.

Length of minor arc
`=(11)/(12)\pi`

RS ≅ TS (Given)

arc RS = arc ST ( equal choed subtended equal arc.

Major arc RT
`=(11)/(12)\pi+(11)/(12)\pi=(11)/(6)\pi`

Minor arc RT
`=2\pi * - (11)/(6)\pi=(\pi)/(6)`

Central angle of minor arc = ∠RHT

Therefore,
`\angle RHT=(\pi)/(6)`

`\angle RST=(1)/(2)\angle RHT`

`\angle RST=(1)/(2)* (\pi)/(6)=(\pi)/(12)`

Now we will change radian to degree

`\angle RST=(\pi)/(12)* (180)/(\pi)=15^\circ`

Hence, The measure of ∠RST=15°