Geometry Problem 1528. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.
Details: Click on the figure below.
More Details
To post a solution to this problem click Comment underneath the post, or click into the line that says, “Enter Comment.” Type what you want to say and press Publish to post your solution.
To post a solution to this problem click Comment underneath the post, or click into the line that says, “Enter Comment.” Type what you want to say and press Publish to post your solution.
ODQE is a square and hence < FQE = 45
ReplyDeleteBut QF = QE (radii of circle Q)
Therefore < QEF = <QFE = 67.5 degrees
Sumith Peiris
Moratuwa
Sri Lanka
90=<DQE=2<DCE, CD=CE by symmetry, so <QFE=90-<DCE/2=67.5
ReplyDelete