Question

Calculate ADB.

Answer 1

We are given a circle with center O and points A, B, C, D on the circumference. We are given two central angles and asked to find an inscribed angle, ADB.

The key geometric theorem is: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

The angle we want to find, ADB, is the angle subtended by the arc AB at point D on the circumference.
The angle subtended by the same arc AB at the center is AOB.
We are given AOB = 50^.
According to the theorem, the inscribed angle is half the central angle.
ADB = (1)/(2) × AOB ADB = (1)/(2) × 50^ ADB = 25^ The measure of ADB is 25^.

Answer 2

We need to find all three interior angles of the triangle ABC inscribed in the circle.

We will use the same theorem as in the previous part for angles subtended by arcs. The sum of angles in a triangle is 180^.

1. Calculate ACB:
This angle is subtended by the arc AB. The central angle for this arc is AOB = 50^.
ACB = (1)/(2) × AOB = (1)/(2) × 50^ = 25^ 2. Calculate BAC:
This angle is subtended by the arc BC. The central angle for this arc is BOC = 80^.
BAC = (1)/(2) × BOC = (1)/(2) × 80^ = 40^ 3. Calculate ABC:
We can use the fact that the sum of angles in ABC is 180^.
ABC = 180^ - ( BAC + ACB) ABC = 180^ - (40^ + 25^) ABC = 180^ - 65^ ABC = 115^ Alternative method for ABC:
The angle ABC subtends the major arc AC. The central angle for the minor arc AC is AOC = AOB + BOC = 50^ + 80^ = 130^.
The central angle for the major arc AC is the reflex angle AOC = 360^ - 130^ = 230^.
ABC = (1)/(2) × (reflex AOC) = (1)/(2) × 230^ = 115^ Both methods give the same result.

The angles of triangle ABC are: BAC = 40^, ACB = 25^, and ABC = 115^.