Question

In the adjoining figure, O is the centre of the circle. If ∠ABP = 350



∠BAQ = 650

, Find [1] ∠PAB [2] ∠QBA​

Answer 1

${\huge{\bold{\underline{Answer:-}}}}$

We know that BD is the diameter of the circle

Angle in a semicircle is a right angle

∠BAD = 90o

Consider △ BAD

Using the angle sum property

∠ADB + ∠BAD + ∠ABD = 180o

By substituting the values

∠ADB + 90o + 35o = 180o

On further calculation

∠ADB = 180o – 90o – 35o

By subtraction

∠ADB = 180o – 125o

So we get

∠ADB = 55o

We know that the angles in the same segment of a circle are equal

∠ACB = ∠ADB = 55o

So we get

∠ACB = 55o

Therefore, ∠ACB = 55o.

Answer 2

$\huge \tt \red {answer}$

We know that BD is the diameter of the circle

Angle in a semicircle is a right angle

∠BAD = 90o

Consider △ BAD

Using the angle sum property

∠ADB + ∠BAD + ∠ABD = 180o

By substituting the values

∠ADB + 90o + 35o = 180o

On further calculation

∠ADB = 180o – 90o – 35o

By subtraction

∠ADB = 180o – 125o

So we get

∠ADB = 55o

We know that the angles in the same segment of a circle are equal

∠ACB = ∠ADB = 55o

So we get

∠ACB = 55o

Therefore, ∠ACB = 55o.