Question 33691: A regular Pentagon ABCDE is incribed in circle O. Chords EC and DB intersect at F (not diameter). Chord DB is Extended to G (outside circle), and Tangent GA is drawn.
find measure of angle BDE
find measure of angle BFC
find measure of angle AGD
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! A regular Pentagon ABCDE is incribed in circle O. Chords EC and DB intersect at F (not diameter). Chord DB is Extended to G (outside circle), and Tangent GA is drawn.
find measure of angle BDE
find measure of angle BFC
find measure of angle AGD
HOPE YOU HAVE A FIGURE OF THE ABOVE..ASSUMING THAT I AM GIVING THE SOLUTION.
SUM OF ALL INTERIOR ANGLES IN A PENTAGON =(2*5-4)*90 =540
SINCE IT IS REGULAR PENTAGON EACH ANGLE =540/5=108
AS PER GIVEN DATA,A,B,E,D LIE ON A CIRCLE...SO ABED IS A CYCLIC QUADRILATERAL..HENCE SUM OF OPPOSITE ANGLES=180=ANGLE EAB+ANGLE BDE
BUT ANGLE EAB=108 AS PROVED ABOVE
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HENCE ANGLE BDE=180-108=72
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SIMILARLY FROM CYCLIC QUADRILATERAL ECBA,WE GET ANGLE ECB=72
NOW IN A REGULAR PENTAGON EACH SIDE SPANS 1/5 OF PERIMETER OF CIRCUM CIRCLE WITH O AS CENTRE.HENCE EACH SIDE SUBTENDS AN ANGLE OF 360/5=72 AT CENTRE AND HALF OF THAT THAT IS 36 AT THE CIRCUMFERENCE..
HENCE ANGLE ADB=ANGLE DBC =36
NOW IN TRIANGLE CFB...WE HAVE...
ANGLE FCB =ANGLE ECB =72..PROVED ABOVE.
ANGLE FBC=ANGLE DBC=36......PROVED ABOVE
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HENCE ANGLE BFC=180-72-36=72
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ABDE IS A CYCLIC QUADRILATERAL...AB IS A CHORD...AG IS A TANGENT TO CIRCUM CIRCLE..HENCE ANGLE BAG = ANGLE IN THE OPPOSITE SEGMENT =ANGLE ADB=36..PROVED ABOVE
HENCE ANGLE EAG=ANGLE EAB + ANGLE BAG=108+36=144
IN QUADRILATERALDEAG...WE HAVE
ANGLE AGD=360-ANGLE EDB-ANGLE EAG-ANGLE DEA
=360-72-144-108=36
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HENCE ANGLE AGD=36
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