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:

$/_ABO=/_BAO$

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