Question

10. In a triangle ABC, A is equal to 62º and B and C are bisected by BO and CO. Calculate the measure of BOC.​

Answer 1

Answer:

In ∆ABC , A=62°

=A+B+C=180° {angle sum property}

=62°+ B+C=180°

=B+ C=(180-62)°

=B+C=118°

=B/2+C/2=59° {divided by 2 both sides}

=OBC+OCB=59°........(i)

NOW in ∆BOC,

=OBC+OCB +BOC =180°

=59°+ BOC = 180° { From (i)}

=BOC= (180-59)°

=BOC= 121°

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Answer 2

Answer:

In ∆ABC , A=62°

=A+B+C=180° {angle sum property}

=62°+ B+C=180°

=B+ C=(180-62)°

=B+C=118°

=B/2+C/2=59° {divided by 2 both sides}

=OBC+OCB=59°........(i)

NOW in ∆BOC,

=OBC+OCB +BOC =180°

=59°+ BOC = 180° { From (i)}

=BOC= (180-59)°

=BOC= 121°