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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

But try the problem first…

Answer:$80$

Source

AMC-8, 2000 problem 24

Pre College Mathematics

## Try with Hints

First hint

Find the $\angle AFG$

Can you now finish the problem ……….

Second Hint

sum of the angles of a Triangle is $180^\circ$

can you finish the problem……..

Final Step

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

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

$\Rightarrow \angle B +\angle D=80^\circ$