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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°

Consider circle H with a 6 centimeter radius If the length of minor arc ST is 1112 π what is the measure of RST Assume RS TS A 15 B 25 C 30 D 45 class=

Respuesta :

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]
Ver imagen jdoe0001
isyllus

Answer:

The measure of ∠RST=15°

A is correct.

Step-by-step explanation:

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

Length of minor arc [tex]=\dfrac{11}{12}\pi[/tex]

RS ≅ TS   (Given)

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

Major arc RT [tex]=\dfrac{11}{12}\pi+\dfrac{11}{12}\pi=\dfrac{11}{6}\pi[/tex]

Minor arc RT [tex]=2\pi \times - \dfrac{11}{6}\pi=\dfrac{\pi}{6}[/tex]

Central angle of minor arc = ∠RHT

Therefore, [tex]\angle RHT=\dfrac{\pi}{6}[/tex]

[tex]\angle RST=\dfrac{1}{2}\angle RHT[/tex]

[tex]\angle RST=\dfrac{1}{2}\times \dfrac{\pi}{6}=\dfrac{\pi}{12}[/tex]

Now we will change radian to degree

[tex]\angle RST=\dfrac{\pi}{12}\times \dfrac{180}{\pi}=15^\circ[/tex]

Hence, The measure of ∠RST=15°

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