In Fig- - ODC - OBA- BOC - 125 - and - CDO - 70 - Find - DOC- DCO and - OAB

By linear angle property, ∠ COD+∠ COB=180^o ⇒∠ COD=180 125=55^o In Δ COD, 55+70+∠ DCO=180^o ⇒∠ DCO=55^o As Δ ODC∼Δ OBA ⇒∠ OCD=∠ OAB ...

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Standard IX

Mathematics

Lines Parallel to the Same Line

In Fig., Δ O...

Question

In Fig., $Δ O D C \sim Δ O B A, ∠ B O C = 125^{\circ}$ and $∠ C D O = 70^{\circ}$ . Find $∠ D O C, ∠ D C O$ and $∠ O A B$

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Solution

By linear angle property,
$∠ C O D + ∠ C O B = 180^{o}$
$\Rightarrow ∠ C O D = 180 - 125 = 55^{o}$
In $Δ C O D, 55 + 70 + ∠ D C O = 180^{o}$
$\Rightarrow ∠ D C O = 55^{o}$
As $Δ O D C \sim Δ O B A$
$\Rightarrow ∠ O C D = ∠ O A B = 55^{o}$ .

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In Fig.,ΔODC∼ΔOBA,∠BOC=125∘ and ∠CDO=70∘. Find ∠DOC,∠DCO and ∠OAB

By linear angle property,∠COD+∠COB=180o⇒∠COD=180−125=55oIn ΔCOD,55+70+∠DCO=180o⇒∠DCO=55oAs ΔODC∼ΔOBA⇒∠OCD=∠OAB=55o.

In Fig., $Δ O D C \sim Δ O B A, ∠ B O C = 125^{\circ}$ and $∠ C D O = 70^{\circ}$ . Find $∠ D O C, ∠ D C O$ and $∠ O A B$

By linear angle property,
$∠ C O D + ∠ C O B = 180^{o}$
$\Rightarrow ∠ C O D = 180 - 125 = 55^{o}$
In $Δ C O D, 55 + 70 + ∠ D C O = 180^{o}$
$\Rightarrow ∠ D C O = 55^{o}$
As $Δ O D C \sim Δ O B A$
$\Rightarrow ∠ O C D = ∠ O A B = 55^{o}$ .