Geometry / Mr. Hansen 4/5/2005

Name: _______________________

Mr. Hansen’s Angle-Arc Review Puzzle and Solutions

	Given: ¤C with points of tangency at T and E; arc DT = 70° Find: All angles and arcs.
	Solutions (don’t peek until you’ve tried!)
1.	By the rule of Angle Half SAD, ÐDTF = 35°.
2.	Since segment CT is a radius to a point of tangency, ÐCTG = 90°. Since ÐFTC = 90°, ÐDTC = 55°.
3.	ÐDCT is a central angle, hence 70° because of the given arc.
4.	ÐBCE = 70° by vert. Ðs.
5.	ÐDCB = ÐECT = 110° by supp. Ðs.
6.	Arcs BD and TE are both 110° by central Ðs found in step 5; arc BE = 70° by central Ð found in step 4.
7.	ÐA = 20° by the rule of Angle Half SAD, since ½(110 – 70) = 20.
8.	ÐAEC = 90° (radius to a point of tangency).