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Find the exact length of the radius of the circle in which a central angle [tex]\theta = \frac{\pi}{18}[/tex] intercepts an arc of length [tex]s = \frac{5\pi}{6}[/tex] meters.

[tex]r = \, \square \, \text{m}[/tex]

Sagot :

To solve for the radius [tex]\( r \)[/tex] of the circle, we can use the relationship between the arc length [tex]\( s \)[/tex], the radius [tex]\( r \)[/tex], and the central angle [tex]\( \theta \)[/tex] (in radians). The formula that relates these quantities is:

[tex]\[ s = r \theta \][/tex]

We are given:
[tex]\[ \theta = \frac{\pi}{18} \quad \text{and} \quad s = \frac{5\pi}{6} \][/tex]

Our task is to find the radius [tex]\( r \)[/tex]. We can rearrange the formula to solve for [tex]\( r \)[/tex]:

[tex]\[ r = \frac{s}{\theta} \][/tex]

Substitute the given values into the equation:

[tex]\[ r = \frac{\frac{5\pi}{6}}{\frac{\pi}{18}} \][/tex]

To simplify the division of fractions, we multiply by the reciprocal:

[tex]\[ r = \frac{5\pi}{6} \cdot \frac{18}{\pi} \][/tex]

Notice that the [tex]\( \pi \)[/tex] terms in the numerator and denominator cancel each other out:

[tex]\[ r = \frac{5 \cdot 18}{6} \][/tex]

Further simplify the expression by performing the multiplication and division:

[tex]\[ r = \frac{90}{6} \][/tex]

[tex]\[ r = 15 \][/tex]

Therefore, the exact length of the radius is:

[tex]\[ r = 15 \, \text{m} \][/tex]