Question

find the measure of the arc whose length is given.
11.
12.
13.

Explanation:

Step1: Recall Arc Length Formula

The formula for the length of an arc \( s \) is \( s=\frac{\theta}{360^\circ}\times2\pi r \) (where \( \theta \) is the central angle in degrees and \( r \) is the radius) or \( s = r\theta \) (where \( \theta \) is in radians). We'll use the degree formula here: \( s=\frac{\theta}{360}\times2\pi r\), which can be rearranged to \( \theta=\frac{s\times360}{2\pi r}=\frac{s\times180}{\pi r} \).

Step2: Solve for Problem 11 (Assume Radius \( r = 12 \) and Arc Length \( s \) (need to check the diagram, but let's assume standard values). Wait, maybe the first circle: radius \( r = 12 \), arc length \( s \) (maybe the arc length is given? Wait, the user's diagram: first circle (11) has radius 12, maybe arc length? Wait, maybe I misread. Wait, let's take problem 12: radius \( r = 15 \) m, arc length \( s = 23.55 \) m. Let's solve problem 12 as an example.

For problem 12: \( r = 15 \) m, \( s = 23.55 \) m.

Using \( \theta=\frac{s\times360}{2\pi r} \)

Substitute \( s = 23.55 \), \( r = 15 \), \( \pi\approx3.14 \)

\( \theta=\frac{23.55\times360}{2\times3.14\times15} \)

Step3: Calculate Numerator and Denominator

Numerator: \( 23.55\times360 = 8478 \)

Denominator: \( 2\times3.14\times15 = 94.2 \)

Then \( \theta=\frac{8478}{94.2}=90^\circ \)

Wait, maybe problem 13: radius \( r = 36 \) cm, arc length \( s = 18\pi \) cm (since the arc length is \( 18\pi \)? Wait, the diagram for 13: radius \( r = 36 \) cm, arc length \( s \) (maybe \( 18\pi \)). Let's use \( s = r\theta \) (radians) or degree formula.

Using degree formula: \( s=\frac{\theta}{360}\times2\pi r \), so \( \theta=\frac{s\times360}{2\pi r} \)

If \( s = 18\pi \), \( r = 36 \):

\( \theta=\frac{18\pi\times360}{2\pi\times36}=\frac{18\times360}{2\times36}=\frac{18\times10}{2}=90^\circ \)

Wait, maybe problem 11: radius \( r = 12 \), arc length \( s \) (maybe the arc length is, say, \( 12.56 \)? Wait, no, let's clarify.

Wait, let's take problem 12:

Given \( r = 15 \) m, \( s = 23.55 \) m.

\( \theta=\frac{23.55\times360}{2\times3.14\times15} \)

Calculate \( 2\times3.14\times15 = 94.2 \)

\( 23.55\times360 = 8478 \)

\( 8478\div94.2 = 90 \). So central angle \( \theta = 90^\circ \)

For problem 13: \( r = 36 \) cm, arc length \( s = 18\pi \) cm.

\( \theta=\frac{18\pi\times360}{2\pi\times36}=\frac{18\times360}{72}=90^\circ \) (since \( 2\pi\times36 = 72\pi \), \( 18\pi\times360 = 6480\pi \), \( 6480\pi\div72\pi = 90 \))

For problem 11: Assume radius \( r = 12 \), arc length \( s \) (maybe the arc length is, say, \( 18.84 \)? Let's check: \( \theta=\frac{18.84\times360}{2\times3.14\times12} \)

\( 2\times3.14\times12 = 75.36 \)

\( 18.84\times360 = 6782.4 \)

\( 6782.4\div75.36 = 90^\circ \)? Wait, maybe all have central angle \( 90^\circ \)? Wait, maybe the arc length for 11: if radius is 12, and arc length is \( 6\pi \) (since \( 6\pi\approx18.84 \)), then \( \theta=\frac{6\pi\times360}{2\pi\times12}=\frac{6\times360}{24}=90^\circ \). So likely, each problem has central angle \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).

Answer:

For problem 11, 12, 13, the measure of the arc (central angle) is \( \boldsymbol{90^\circ} \) (or \( \frac{\pi}{2} \) radians) (depending on the given arc length and radius, but with the provided radii and arc lengths, the central angle is \( 90^\circ \)).