Question

In the given figure, <BAC = 32°, <BCE = 161º. Find <ACB, <ABC and <DBC.
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Answer 1

Step-by-step explanation:

<ACB = 180°-<BCA

=180°-161

= 19°

<ABC=180°-(32°+19°)

=129°

<DBC=180°-129°

=51°

Answer 2

Answer:

<ABC=129

<ACB=19

<DBC=51

Step-by-step explanation:

<ABC+<BAC=<BCE. [exterior angle property]

<ABC=161-32

<ABC=129

<ABC+<BAC+<ACB=180 [Angle sum property]

<ACB=180-161

<ACB=19

<BAC+<ACB=<DBC [exterior angle property]

<DBC=32+19

<DBC=51

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