Find Arc Height

Question

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Answer 1

 

In the picture AIFHB is the semicircle, and the arc IFHB stands on the chord BI which is 25º inclined at B to the semicircle diameter AB. The radius of the semicircle is 5 units long. AC=5 is the same length as radii EF=AE=EI=EH=EB. The radius EH=5 bisects the chord BI perpendicularly. The height of the arc is GH.

Triangle BEI is isosceles, with radial arms EI=BE=5. EG/BE=sin(25).

Therefore EG=5sin(25). EG+GH=EH, 5sin(25)+GH=5, so the height of the arc, GH=5(1-sin(25))=2.8869 approx.