ㄥACB= 70^o
ㄥADB=ㄥACB
⇒ ㄥADB=70^o (Angles in the same segment of a circle)
In AB,
ㄥDAB+ㄥADB+ㄥDBA=180^o (Angle sum property of triangles)
⇒ 60^o+70^o+ㄥDBA=180^o
⇒ㄥDBA=50^o

Draw OP perpendicular to xy from the centre to a chord bisecting it.
OP \(\bot\) to chord BC
\(\Rightarrow\) BP = PC      ...(i)
Similarly, AP = PD       ..(ii)
Subtracting eqn. (i) from eqn. (ii), we get
AP- BP = PD-PC
or AB = CD

\(\angle\)AOB = 80°
\(\Rightarrow\) \(\angle\)ADB = 40° (\(\because\) \(\angle\)AOB = 2 \(\angle\)ADB)
\(\angle\)ACB = \(\angle\)ADB
= 40°  (angles in the same segment)

Since sum of the opposite pairs of angles in a cyclic quadrilateral is 180°.
Hence ㄥB+ㄥD=180°
ㄥB=180°-70°
=110°
Again, AB II CD and AD is its transversal, so
ㄥA+ㄥD=180°
ㄥA=180°-70°
=110°
and ㄥA+ㄥC=180°
⇒ 110°+ㄥC=180°
ㄥC=180°-110°
=70°

Given AB and CD are the chords of a circle with centre at O such that AB = CD

To Prove: \(\angle\)AOB = \(\angle\)COD
Proof: In \(\Delta\)AOB and \(\Delta\)COD
AO = CO        (radii of same circle)
AB = CD         (given)
BO = DO        (radii of same circle)
\(\Delta\)AOB\(\cong \) \(\Delta\)COD   (SSS)
\(\angle\)AOB = \(\angle\)COD  (c.p.c.t.) 2 Hence Proved.

Given, AB = 10 cm
ON = 12cm

Also, ON \(\bot\) AB
and AN = BN (\(\therefore\) Perpendicular drawn from the centre of the circle bisects the chord)
In\(\Delta\)ONB,
OB² = ON² + NB²         (By Pythagoras theorem)
\(\therefore\) OB² = 12² + 5²        (\(\because\) BN = 5 cm)
= 144+25=169
\(\therefore\) OB = 13 cm
Hence, the radius of the circle is 13 cm.

ㄥBAC =ㄥBDC=45° (angles in the same segment)
In ΔDBC,
ㄥDBC+ㄥBCD+ㄥCDB=180° [Angle sum property]
ㄥBCD = 80°

20°

\(\angle\)AOD + \(\angle\)BOD = 180° (linear pair)
\(\angle\)BOD = 180°- \(\angle\)AOD
\(\angle\)BOD = 180°- 75° = 105°
\(\angle\)CED =90° (angle in semi-circle)
\(\angle\)CDE =90° - \(\angle\)OCE \(\Rightarrow\) 90° - 40° = 50°
\(\angle\)OBE = \(\angle\)OBD
\(\angle\)OBD = 180°- (105° + 50°)
(In \(\Delta\)DBO, Angle sum property of \(\Delta\))
\(\angle\)OBE = \(\angle\)OBD = 25°

36, 60

\(\angle\)ACB=1/2 * \(\angle\)AOB
=1/2*90^o
=45^o
In \(\Delta\)ACB, \(\angle\)CAB=180^o-(30^o+45^o)
=105^o
\(\angle\)OAB=\(\angle\)OBA
=45^o
(Angles opp. to equal sides of triangle are equal as OA = OB radius of same circle)
\(\angle\)CAO = 105°- \(\angle\)OAB
= 105°-45°
= 60°

70°

2, 2, 10, 10 (in cm)

6 cm