**In the figure below, O is the centre of the circle and AE is parallel to BC. DF = DE, Angle OAB = 58 degrees and Angle FED = 50 degrees.**

**a) Find Angle GBC**

b) Find Angle DCB

Solution

b) Find Angle DCB

Solution

**a)**

Angle ABO = 58 deg (isosceles triangle)

Angle BOG = (58 + 58) deg = 116 deg (exterior angles)

Angle OGB = [(180 - 116) divided by 2] = 32 deg

Angle GBC = 32 deg (alternate angles)

**Answer: 32 degrees**

**b)**

Angle FDE = (180 - 50 - 50) deg = 80 deg

Angle FDE = Angle GDC = 80 deg

Angle DCB = (180 - 80) deg = 100 deg

Answer: 100 degrees

Answer: 100 degrees

Printer Friendly Version

## No comments:

Post a Comment