In the figure 6.35 triangles ODC and OBA are similar. The values of angles DOC , DCO and OAB are respectively
3 points
55, 55, 55
45, 45, 45
55, 45, 55
45, 55, 45
Answers
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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Step-by-step explanation:
angle BOC = angle AOD = 125°
so,
angle ODC + angle OCD = 125° ( By exterior angle property)
70° + angle OCD = 125°
angle OCD = 125°- 70°
= 55°
so,
angle OCD = angle OAB ( interior alternate angles)
angle DOC + angle OCD + angle ODC = 180°
DOC + 55° + 70°= 180°
DOC = 180° - 125°
= 55°
55,55,55