Question

touch the circle with centre D in A and
In the given figure, lines CA and CB,
9
B. If < ADB = 125° then
i. Find the measure of Z ACB
ii. Find the type of the quadrilateral
O ABCD​

Answer 1

Answer:

In given figure ABCD is a cyclic quadrilateral.

We know that, sum of the opposite angles in a cyclic quadrilateral is 180∘

So, ∠BCD + ∠BAD = 180∘

⇒  125∘ + ∠BAD = 180∘

⇒  ∠BAD = 180∘ - 125∘

∴  ∠BAD = 55∘   ----   (1)

Since, ∠ABD is an angle in semi-circle.

∴ ∠ABD = 90∘      --- (2)

In △ABD,

∠ADB + ∠ABD + ∠BAD = 180∘

∠ADB + 90∘ + 55∘ = 180∘    (From 1 and 2)

∴  ∠ADB = 35∘