Question

In fig, Triangle ODC is congruent to Triangle OBA. Angle BOC=125, Angle CDO =70. Find Angle DOC, DCO, OAB ​

Answer 1

Answer:

So,

DOC = 60°

DCO = 50°

OAB = 60°

Answer 2

Given:

• ∠BOC=125
• ∠CDO =70

Solution

DOB is a straight line.

So, ∠DOC + ∠ COB = 180°

$\leadsto$ ∠DOC = 180° – 125°

$\leadsto$ 55°

In ΔDOC, Sum of the measures of the angles of a triangle is 180º

$\circ \: \: \: {\boxed{\tt\red{ Sum \ of \ angles_{( \Delta )} = 180° }}}$

So, ∠DCO + ∠ CDO + ∠ DOC = 180°

Putting values of ∠ CDO and ∠ DOC ;

$\leadsto$ ∠DCO + 70º + 55º = 180°

$\leadsto$ ∠DCO = 55°

We Know that,

ΔODC ∝ ¼ ΔOBA,

So, ΔODC ~ ΔOBA.

We also Know that,

Corresponding angles are equal in similar triangles.

$\leadsto$ ∠OAB = ∠OCD

$\leadsto$ ∠ OAB = 55°

$\leadsto$ ∠OAB = ∠OCD

$\leadsto$ ∠OAB = 55°

Hence,

• ∠DOC = 55°
• ∠DCO = 55°
• ∠OAB = 55°