If o is the center of circle of radius r and ab is a chord of the circle at a distance r/2 from o,then bao is
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oa=ob=r
ab=r/2
draw a perpendicular odfrom center to ab
in triangle oad we have
sinθ=ad/oa
=1/4=90°/4
hence<aob=45°
in triangle aob we have
<aob+<oab+<oba=180°
45°+2<oab=180°
<oab=135°/2=67.5
ab=r/2
draw a perpendicular odfrom center to ab
in triangle oad we have
sinθ=ad/oa
=1/4=90°/4
hence<aob=45°
in triangle aob we have
<aob+<oab+<oba=180°
45°+2<oab=180°
<oab=135°/2=67.5
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