Question

please help me please don't scam​

Answer 1

$\large\underline{\sf{Solution-}}$

As AOB is diameter of a circle.

We know, Angle in semi-circle is right angle.

$\rm \implies\:\boxed{ \tt{ \: \angle ACB = 90 \degree \: }} - - (1)$

Also, it is given that

$\rm :\longmapsto\:\angle PCA = 120\degree$

$\rm :\longmapsto\:\angle PCB + \angle ACB = 120\degree$

$\rm :\longmapsto\:\angle PCB + 90\degree = 120\degree$

$\rm :\longmapsto\:\angle PCB = 120\degree - 90\degree$

$\rm \implies\:\boxed{ \tt{ \: \angle PCB = 30\degree \: }} - - - (2)$

We know,

↝ Angle in alternate segment are equal.

$\rm \implies\:\angle PCB = \angle CAB$

$\rm \implies\:\boxed{ \tt{ \: \angle CAB = 30\degree \: }} - - - (3)$

$\red{\bigg \{ \because \: of \: equation \: (2) \: \bigg \}}$

Now,

We know that,

Angle subtended at the centre by an arc is double the angle subtended on the circumference by the same arc.

$\rm \implies\:\angle COB = 2 \: \angle CAB$

$\rm \implies\:\angle COB = 2 \: \times 30\degree$

$\rm \implies\:\boxed{ \tt{ \: \angle COB = 60\degree \: }}$

Additional Information :-

1. Angle in same segments are equal.

2. Sum of the opposite angles of a cyclic quadrilateral is supplementary.

3. Exterior angle of a cyclic quadrilateral is equals to interior opposite angle.

4. Equal chords subtends equal angles at the centre.

5. Equal chords are equidistant from the centre.

6. The perpendicular drawn from centre bisects the chord.