Question

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

Answer 1

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

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

Explanation:

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^@`