Question

plz ans it correctly
..​

Answer 1

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

Consider, triangle OAB

OA = OB [ Radius of circle ]

⟹ ∠BAO = ∠ABO = x (say) --------(1)

[ Angle opposite to equal sides are equal ]

Now, in triangle OAB

We know, sum of all angles of a triangle is supplementary.

$\rm \: \angle \: BOA + x + x = 180 \degree \:$

$\rm \: \angle \: BOA + 2x = 180 \degree \:$

$\rm\implies \:\angle \:BOA = 180\degree - 2x - - - - (2)$

Let assume that ∠CAD = y

Now, In triangle CAD

• ∠CDA = 90° (Given)

• ∠CAD = y

Using angle sum property, we get

$\rm \: 90\degree + y + \angle \:ACD = 180\degree$

$\rm\implies \:\angle \:ACD = 90\degree - y$

So,

$\rm\implies \:\angle \:ACB = 90\degree - y - - - (2)$

Now, ∠BOA is at the centre subtended by an arc AB and ∠ACB on the circumference of circle.

As, we know angle subtended at the centre of circle by an arc is double the angle subtended on the circumference by the same arc.

$\rm\implies \:\angle \:BOA = 2\angle \:ACB$

$\rm \: 180\degree - 2x = 2(90\degree - y)$

$\rm \: 180\degree - 2x = 180\degree - 2y$

$\rm \: - 2x = - 2y$

$\rm \: x = y$

$\rm\implies \:\boxed{ \rm{ \:\angle \:BAO \: = \: \angle \:CAD \: }}$

$\rule{190pt}{2pt}$

Additional information :-

1. Angle in same segments are equal.

2. Angle in semi-circle is right angle.

3. Equal chords are equidistant from the centre.

4. Perpendicular drawn from centre bisects the chord.

Answer 2

Answer:

Consider, triangle OAB

OA = OB [ Radius of circle ]

⟹ ∠BAO = ∠ABO = x (say) --------(1)

[ Angle opposite to equal sides are equal ]

Now, in triangle OAB

We know, sum of all angles of a triangle is supplementary.

Let assume that ∠CAD = y

Now, In triangle CAD

∠CDA = 90° (Given)

∠CAD = y

Using angle sum property, we get

So,

Now, ∠BOA is at the centre subtended by an arc AB and ∠ACB on the circumference of circle.

As, we know angle subtended at the centre of circle by an arc is double the angle subtended on the circumference by the same arc.