Try this beautiful problem from Geometry from AMC-8, 2000, Problem-24, based on angles of Star
If \(\angle A = 20^\circ\) and \(\angle AFG =\angle AGF\), then \(\angle B+\angle D =\)
Geometry
Star
Triangle
Answer:\(80\)
AMC-8, 2000 problem 24
Pre College Mathematics
Find the \(\angle AFG\)
Can you now finish the problem ..........
sum of the angles of a Triangle is \(180^\circ\)
can you finish the problem........
we know that the sum of the angles of a Triangle is \(180^\circ\)
In the \(\triangle AGF\) we have,\((\angle A +\angle AGF +\angle AFG) =180^\circ \)
\(\Rightarrow 20^\circ +2\angle AFG=180^\circ\)(as \(\angle A =20^\circ\) & \(\angle AFG=\angle AGF\))
\(\Rightarrow \angle AFG=80^\circ\) i.e \(\angle EFD=\angle 80^\circ\)
So the \(\angle BFD=\frac{360^\circ -80^\circ-80^\circ}{2}=100^\circ\)
Now in the \(\triangle BFD\),\((\angle BFD +\angle B +\angle D\))=\(180^\circ\)
\(\Rightarrow \angle B +\angle D=180^\circ -100^\circ\)
\(\Rightarrow \angle B +\angle D=80^\circ\)
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