Question

in the given figure lines CA and CB touch the circle with centre D in A and B if angle ADB is 125 degree then
1) find the measure of angle ACB
2)find the type of quadrilateral ABCD.​

Answer 1

Answer:

1. 45 degree

2. ADBC

I have given the answer

If it is wrong tell me

Answer 2

AC and BC are 2 tangents to the circle with centre D.

We know that;

Tangent drawn from an external point to a circle are equal in lengths.

So, AC = AB

Also,

By Tangent Perpendicularity theorem,

Radius is perpendicular to the tangent at the point of contact.

So, ∠CAD = ∠CBD = 90°

In Quadrilateral ABCD,

By Angle Sum Property,

∠ACB + ∠CBD + ∠DAC + ∠ADB = 360°

➥ ∠ACB + 90° + ∠90° + ∠ADB = 360°

➝ ∠ACB + ∠ADB + 180° = 360°

➝ ∠ACB + ∠ADB = 360° - 180°

➝ ∠ACB + ∠ADB = 180°

Here, it is already given that ∠ADB = 125°

∠ACB + ∠ADB = 180°

➝ ∠ACB + 125° = 180°

➝ ∠ACB = 180° - 125°

∴ ∠ACB = 55°

Since angles are of different measures,

It is an irregular polygon or irregular quadrilateral