1 of 2

D

A

B

C

O

Soln.

∠DOC

+

∠BOC

=

180°

∠DOC

+

125

=

180

∠DOC

125

180

=

–

∠DOC

55°

=

[Angles in a linear pair]

…(i)

∴

∴

125°

∠BOC

+

∠CDO

=

∠DCO

[Exterior angle is equal to sum of its

two interior opposite angles]

?

55°

ΔODC ~ ΔOBA, ∠BOC = 125°, ∠CDO = 70°

Find ∠DOC, ∠DCO and ∠OAB

?

∠DCO

+

125

=

70

∠DCO

70

125

=

–

∴

∴

∠DCO

55°

=

∴

70°

…(ii)

55°

∴

EX.6.3 (Q.2)

2 of 2

Soln.

ΔODC ~ ΔOBA

∠DCO

∠OAB

=

∠OAB

55°

=

[Given]

[corresponding angles of similar triangles]

∴

?

55°

D

A

B

C

O

125°

55°

70°

55°

∠DCO

55°

=

…(ii)

Q. Given : ΔODC ~ ΔOBA, ∠BOC = 125°, ∠CDO = 70°

Find : ∠DOC, ∠DCO and ∠OAB