50°

25°

40°, 40°

\(\angle\)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°

2, 2, 10, 10 (in cm)

\(\angle\)FEH =\(\angle\)AED = 180⁰-\(\left( \frac { 1 }{ 2 } \angle A+\frac { 1 }{ 2 } \angle B \right) \)

\(\angle\)FGH =\(\angle\)CGB
= 180⁰-\(\left( \frac { 1 }{ 2 } \angle A+\frac { 1 }{ 2 } \angle B \right) \)
Adding,
\(\angle\)FEH +\(\angle\)FGH = 180⁰
\(\therefore\)  EFGH is a cyclic quadrilateral.

\(\angle\)ACD = 90°, \(\angle\)AED = 90°   (angle in semi-circle)
\(\angle\)ADE = 180° - (60° + 90°)
= 30° (angle sum property)
\(\angle\)ABC = 180° - \(\angle\)CDA
= 180°- 70° = 110°  (ADCB is a cyclic quadrilateral)
\(\angle\)BCA = 180 - (110° + 30°) = 40°  (Angle sum property)
\(\angle\)BCD = \(\angle\)BCA + \(\angle\)ACD
Therefore
\(\angle\)BCD = 40° + 90° = 130°.

Given: O is the centre of a circle.
To Prove: x + y = z
Proof: \(\angle 3=\angle 4\)
| Angles in the same segment of a circle are equal
\(\angle z=2\angle 3\)
⇒ \(\angle z=\angle 3+\angle 3\)
⇒ \(\angle z=\angle 3+\angle 4\)      .....(1)
Now ​​​​​ \(\angle y=\angle 3+\angle 1\)  .....(2)
   | An exterior angle of a triangle is equal to the sum of its two interior opposite angles
(1) - (2) gives
 \(\angle z-\angle y=\angle 4-\angle 1\)
As \(4=\angle x+\angle 1\Rightarrow \angle 4-\angle 1=\angle x\)
 | An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒ \(\angle 4-\angle 1=\angle x\)         ....(4)
From (3) and (4),
\(\angle z-\angle y=\angle x\)
⇒ \(\angle x+\angle y=\angle z\)
⇒ x+y=z

\(\angle c=\angle BAE\)
| An exterior angle of a cyclic quadrilateral is equal to its interior opposite angle
⇒ ​​​​​\(\angle c=62°\)    ....(1)
In \(\Delta AEC\), \(\angle ACE+\angle CAE+\angle d=180°\)
| Angle sum property of a triangle
⇒\(43°+62°+\angle d=180°\)
⇒  \(\angle d=75°\)  ..........(2)
\(\angle a+\angle d=180°\)
 | Opposite angles of a cyclic quadrilateral are supplementary
⇒ ​​​​​​ \(\angle a+75°=180°\)
⇒ ​​​​​​​\(\angle a=105°\)  ..........(3)
In \(\Delta FDE\),
​ \(\angle c+(180°-\angle d)+\angle b=180°\)
 | Angle sum property of a triangle
⇒ \(62°+(180°-75°)+\angle b=180°\)
⇒ ​​​​​​​ \(\angle b=13°\) 

(i) x = 2 x 35° = 70°
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(ii) \(x=\frac { 1 }{ 2 } \times 110°=55°\)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iii) \(x=\frac { 1 }{ 2 } \times 70°=35°\)
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iv) x = 180° - (90° + 55°) I ∵ Angle in a semi-circle is 90°
= 180° - 145° = 35°
(v) \(x=\frac { 1 }{ 2 } \times \left( 180°-120° \right) \)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle \(=\frac { 1 }{ 2 } \left( 60° \right) =30°\). 

If two chords of a circle are equidistant from the centre, then they are equal.

Theorem

\(AQ=QB=\frac { 1 }{ 2 } AB=4\quad cm\) 
\(OA^{ 2 }=OQ^{ 2 }+AQ^{ 2 }\) 
\(⇒\ OQ=3cm\) 
\(CP=PD=\frac { 1 }{ 2 } CD=3\ cm\) 
\(OC^{ 2 }=OP^{ 2 }+CP^{ 2 }x\) 
\( ⇒\ OP=4\ cm\) 
\(PQ=OP+OQ=7\ cm\)

\(QM=MR=\frac { 1 }{ 2 } QR=\frac { 15 }{ 2 } cm\) 
\(PM=MS=\frac { 1 }{ 2 } PS=\frac { 20 }{ 2 } cm=10cm\) 
\(\ PQ=PM-QM\) 

\(AC=\sqrt { OA^{ 2 }+OC^{ 2 } } \) 
\(\quad =\sqrt { 13^{ 2 }-5^{ 2 } } =12\quad cm\)

\(AC=CB=\frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } \times 16=8\quad cm\) 
\(OA=\sqrt { OC^{ 2 }+AC^{ 2 } }\) 
\(=\sqrt { 6^{ 2 }+8^{ 2 } } =10\quad cm\)

\(AC=BC=\frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } (24)=12\quad cm\)

\(OA=\sqrt { OC^{ 2 }+AC^{ 2 } } \)
\(=\sqrt { 5^{ 2 }+12^{ 2 } } =13\quad cm\) 

\(BM=MC=\frac { 1 }{ 2 } BC=\frac { 1 }{ 2 } (12)=6 \ cm\) 
\(AB=\sqrt { AM^{ 2 }+BM^{ 2 } } =\sqrt { 8^{ 2 }+6^{ 2 } } =10\ cm\)