[tex]\huge\bold{Given :}[/tex]
Angle AOB = 100°
Angle OBC = 68°
Angle BCO = [tex]x°[/tex]
[tex]\huge\bold{To\:find :}[/tex]
The value of [tex]x°[/tex].
[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf {C.\:x°\:=\:32°}}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}[/tex]
An exterior angle of a triangle is equal to sum of two opposite interior angles.
And so we have,
➪ ∠ AOB = ∠ OBC + ∠ BCO
➪ 100° = 68° + [tex]x°[/tex]
➪ [tex]x°[/tex] = 100° - 68°
➪ [tex]x°[/tex] = 32°
Therefore, the value of [tex]x°[/tex] is 32°.
[tex]\large\mathfrak{{\pmb{\underline{\blue{To\:verify}}{\blue{:}}}}}[/tex]
∠ AOB = ∠ OBC + ∠ BCO
✒ 100° = 68° + 32°
✒ 180° = 100°
✒ L. H. S. = R. H. S.
[tex]\boxed{Hence\:verified.}[/tex]
(Note: Kindly refer to the attached file.)
[tex]\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{ヅ}}}}}[/tex]
