Glven the following diagram, IfmXCOF = 150°, then mZBOC =
ADIS
B
D

∠ BOC = 30°

Step-by-step explanation:

Given ∠ COF = 150°

∠ BOC + ∠ COF = 180° ( BF is a straight angle ) , that is

∠ BOC + 150° = 180° ( subtract 150° from both sides )

∠ BOC = 30°

2/3

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)