Categories
AMC 8 Geometry Math Olympiad

Angles of Star | AMC 8, 2000 | Problem 24

Try this beautiful problem from GeometryAMC-8, 2000 ,Problem-24, based triangle. You may use sequential hints to solve the problem.

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 \(\angle B+\angle D =\)

Angles of Star
  • \(90\)
  • \(70\)
  • \(80\)

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……..

Angles of Star

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

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

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

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *