12. In the given figure, AODC-AOBA
BOC = 1250, CDO = 70°
Find JOAB
D C
70
OX125
A
B
Answers
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5
Answer:
∠DOC+125
o
=180
o
(linear pair)
⇒ ∠DOC=180
o
−125
o
=55
o
In △DOC
∠DCO+∠CDO+∠DOC=180
o
(sum of three angles of △ODC)
⇒ ∠DCO+70
o
+55
o
=180
o
⇒ ∠DCO+125
o
=180
o
⇒ ∠DCO=180
o
−125
o
=55
o
Now we are given that △ODC∼△OBA
⇒ ∠OCD=∠OAB (Corresponding angles of similar triangles)
⇒ ∠OAB=∠OCD=∠DCO=55
o
i.e., ∠OAB=55
o
Hence we have,
∠DOC=55
o
;∠DCO=55
o
;∠OAB=55
o
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