See complete Problem 182 at:
www.gogeometry.com/problem/p182_overlapping_circles_angle.htm
Overlapping Circles, Find an angle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, September 22, 2008
Elearn Geometry Problem 182
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x=(180°-mDAG)/2=(180°-mDAB-mBAE)/2=(180°-60°-45°)/2=37.5°
ReplyDeleteConnect AB, AD, and BD, revealing Equilateral Triangle ABD; (AB=AD=BD=r, Angle BAD=60°) and AB is perpendicular to EF and =EB, (any line, ray, or segment crossing circle center is always perpendicular to the tangent point where it meets at,) Angle BAE=45°. Thus Angle DAG=Angle BAD+Angle BAE=(60+45)°=105°.
ReplyDeleteSince DAG is isoceles, (AD=AD=r,)
Therefore, Angle AGD=((180-105)/2)°=Angle x=37.5°.
AB=BE=r entonces <BAE=45=<BAG=2<GDB
ReplyDeleteAD=AG=r entonces <AGD=<ADG
ABD equilatero entonces <GDB=60-x
Teniamos que <GAB=45=2<GDB Y GDB=60-x entonces x=37,5
1) AG=AC=AD ==> G is incenter of tr. CDE. 2) angle <CBE=30 degs, or <CDE=15 degs, thus 2x= <CED+<CDE=37.5 degs.
ReplyDeleteBest regards
Tr. ABD is equilateral, Tr. ABE is right isocelese
ReplyDeleteHence < DAG = 105 and so x = (180 - 105)/2 = 37.5
Sumith Peiris
Moratuwa
Sri Lanka
Triangle ABD is an equalateral triangle
ReplyDelete<ABD=60
<ACD=30 (< at centre is twice < at circumference)
x=<ACD=30 (< in same seg.)
Can anyone point out what's wrong for this solution? I cannot find any mistakes on this but the answer is the same with all you guys.