Question

Find mJPK ................

Answer 1

Answer: 66.5

Explanation:

Let θ represent ∠MLK

`\text{Then }cos\ \theta =(9)/(15)\quad \rightarrow \quad \theta = cos^(-1)\bigg((9)/(15)\bigg)\quad \rightarrow \quad \theta = 53^o`

Arc length (s) = radius (r) · θ (theta must be in radians)

`\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 (\pi)/(180)\bigg)\\\\.\qquad \rightarrow \quad \stackrel\frown{JK}\ =\ 2\bigg((53\pi)/(12)\bigg)\\\\\.\qquad \rightarrow \quad \stackrel\frown{JK}\ =\ (53\pi)/(6)`

`\stackrel\frown{JK}+ \stackrel\frown{JPK}=\text{Circumference of the circle}\\\\(53\pi)/(6)+\stackrel\frown{JPK}=2\pi\cdot 15\\\\.\qquad \stackrel\frown{JPK}=30\pi-(53\pi)/(6)\\\\\\.\qquad \stackrel\frown{JPK}=(180\pi)/(6)-(53\pi)/(6)\\\\\\.\qquad \stackrel\frown{JPK}=(127\pi)/(6)\\\\\\.\qquad \stackrel\frown{JPK}=66.5`