Can someone help me solve for angle BOC and angle BOA?

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Answer 1 · 2018-06-04T10:16:06

#color(blue)(/_BOC=100^@)#

#color(blue)(/_BAO=30^@)#

From diagram:

#Delta BOC # is isosceles. ( the two equal sides are the radii of the circle )

Therefore:

#/_CBO=/_BCO=40^@#

#/_BOC=180^@-(/_CBO+/_BCO)=>180^@-80^@=100^@#

Sum of angles in a triangle:

#/_BOA +/_AOD=180^@# ( angles on the straight line BD )

Therefore:

#/_BOA=180^@-/_AOD=>180^@-60^@=120^@#

#Delta BOA# is also isosceles:

Therefore:

#/_ABO+/_BAO+/_BOA=180^@#

#/_ABO+/_BAO=180^@-/_BOA=60^@#

Since:

#/_ABO=/_BAO#

#/_BAO=60^@/2=30^@#