Question

In the given figure, ∆ODC ~ ∆OBA, ∠ BOC = 125° and ∠ CDO = 70 find, (i)∠DOC, (ii) ∠ DCO, (iii) ∠ OAB.

Plz give me answer ( for math experts) ​

Answer 1

Given

• ∆ODC ~ ∆OBA
• ∠ BOC = 125° and ∠ CDO = 70

To find out:

(i)∠DOC

(ii) ∠ DCO

(iii) ∠ OAB.

Solution:

Since, BD is a line and OC is a ray on it ,

( i ) ∠DOC + ∠BOC = 180°

∠DOC + 125° = 180°

∠DOC = 180° - 125°

∠DOC = 55°

( ii ) In ∆ CDO ,We have

∠CDO + ∠DOC + ∠DCO = 180°

⇒ 70° + 55° + ∠DCO = 180°

⇒ 125° + ∠DCO = 180°

⇒ ∠DCO = 180° - 125°

⇒ ∠DCO = 55°

( iii ) It is given that ∆ODC ~ ∆OBA

Therefore,

∠ODC = ∠OBA , ∠OCD = ∠OAB

∠OBA = 70° and ∠OAB = 55°

Answer 2

$\large\underline\mathrm{Given:-}$

• ∆ODC~∆OBA

• ∆BOC = 125° And ∆CDO = 70°

$\large\underline\mathrm{To \: find}$

$\implies$ (1) ∆DOC

$\implies$ (2) ∆DCO

$\implies$ (3) ∆OAB

$\large\underline\mathrm{Solution}$

• BD is line and OC is a ray0n it.

$\implies$ (1) ∆DOC + ∆BOC = 180°

$\implies$ ∆DOC + 125° = 180°

$\implies$ ∆DOC = 180° - 125°

$\implies$ ∆DOC = 55°

• ∆CDO

$\implies$ ∆CDO + ∆DOC + ∆DCO = 180°

$\implies$ 70° + 55° + ∆DOC = 180°

$\implies$ 125° + ∆DCO = 180°

$\implies$ ∆DCO = 180° - 125°

$\implies$ ∆DCO = 55°

$\implies$ (3) ∆ODC~∆OBA

$\implies$ ∆ODC = ∆OBA

$\implies$ ∆OCD = ∆OAB

$\implies$ ∆OBA = 70° and ∆OAB = 55°