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find measure of arc SE
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Note that chord HG is congruent to chord SE because they are equidistant from the center. Does that help?
and how do i go on from there
how would i find the radius of this
would it be 100
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\[\tan 40=\frac{6}{\frac{1}{2}HG}\]
but i dont know the measure of HG
That is exactly the point. Solve that equation for HG
i got measure of arc se is 100 is that correct
\[\frac{1}{2}HG \tan 40=6\] \[HG \tan 40=12\] \[HG=\frac{12}{\tan 40}\]
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\[HG=\frac{12}{.839}=14.3\]
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