Question

solve this please its urgent​

Answer 1

Answer:

Given that O is the centre of smaller circle and ∠APB=75

∘

.

We know, angle subtended by an arc at centre of circle is doubled the angle subtended at any point of circle .

So, ∠AOB=2∠APB

⟹ ∠AOB=2×75

∘

=150

∘

.

Now, since ACOB is a cyclic quadrilateral,

∠ACB+∠AOB=180

∘

...[Opposite angles of cyclic quadrilateral are supplementary]

⇒∠ACB+150

∘

=180

∘

⇒∠ACB=180

∘

−150

∘

⇒∠ACB=30

∘

Also, since ABOD is a cyclic quadrilateral,

∠ADB+∠AOB=180

∘

...[Opposite angles of cyclic quadrilateral are supplementary]

⇒∠ADB+150

∘

=180

∘

⇒∠ADB=180

∘

−150

∘

⇒∠ADB=30

∘

.

Hence, ∠AOB=150

∘

, ∠ACB=30

∘

and ∠ADB=30

∘

.

Step-by-step explanation:

thanku

Mark in brainliest

Answer 2

Answer:

angle ADB = 128°

Step-by-step explanation:

CENTRAL ANGLE AND THE ANGKE BETWEEN TWO TANGENTS ARE ALWAYS SUPPLEMENTARY.

THEREFORE,

ANGLE ACB + ANGLE ADB = 180°

I.E. ANGLE ADB = 180-52 = 128°