Try this beautiful problem from Geometry from AMC-8, 2000, Problem-24, based on angles of Star

## Angles of Star | AMC-8, 2000 | Problem 24

If \(\angle A = 20^\circ\) and \(\angle AFG =\angle AGF\), then \(\angle B+\angle D =\)

- \(90\)
- \(70\)
- \(80\)

**Key Concepts**

Geometry

Star

Triangle

## Check the Answer

Answer:\(80\)

AMC-8, 2000 problem 24

Pre College Mathematics

## Try with Hints

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 \(\triangle BFD\),\((\angle BFD +\angle B +\angle D\))=\(180^\circ\)

\(\Rightarrow \angle B +\angle D=180^\circ -100^\circ\)

\(\Rightarrow \angle B +\angle D=80^\circ\)

## Other useful links

- https://www.cheenta.com/probability-amc-8-2004-problem-no-22/
- https://www.youtube.com/watch?v=0yiR_y8YXc0