Answer: 66.5
Step-by-step explanation:
Let θ represent ∠MLK
[tex]\text{Then }cos\ \theta =\dfrac{9}{15}\quad \rightarrow \quad \theta = cos^{-1}\bigg(\dfrac{9}{15}\bigg)\quad \rightarrow \quad \theta = 53^o[/tex]
Arc length (s) = radius (r) · θ (theta must be in radians)
[tex]\stackrel\frown{JM}\ =\ \stackrel\frown{KM}\text{ so }\stackrel\frown{JK}\ =\ 2\stackrel\frown{KM}\\\\.\qquad \rightarrow \quad \stackrel\frown{JK}\ =\ 2\bigg(15\cdot 53^o\cdot \dfrac{\pi}{180}\bigg)\\\\.\qquad \rightarrow \quad \stackrel\frown{JK}\ =\ 2\bigg(\dfrac{53\pi}{12}\bigg)\\\\\.\qquad \rightarrow \quad \stackrel\frown{JK}\ =\ \dfrac{53\pi}{6}[/tex]
[tex]\stackrel\frown{JK}+ \stackrel\frown{JPK}=\text{Circumference of the circle}\\\\\dfrac{53\pi}{6}+\stackrel\frown{JPK}=2\pi\cdot 15\\\\.\qquad \stackrel\frown{JPK}=30\pi-\dfrac{53\pi}{6}\\\\\\.\qquad \stackrel\frown{JPK}=\dfrac{180\pi}{6}-\dfrac{53\pi}{6}\\\\\\.\qquad \stackrel\frown{JPK}=\dfrac{127\pi}{6}\\\\\\.\qquad \stackrel\frown{JPK}=66.5[/tex]
