Question

AB is tangent to the circle k(O) at B, and

AD is a secant, which goes through O. Point O is between A and D∈k(O). Find m∠BAD and m∠ADB, if measure of arc BD is 110°20'.

Answer 1

m∠BAD is 20° 20'

m∠ADB is 34° 50'

Explanation:

• When a tangent and a secant intersect outside a circle then the measure of the angle formed between them is one-half the positive difference of the measures of the intercepted arcs

• The measure of an inscribed angle is one-half the measure of its subtended arc

Look to the attached figure

In circle O

∵ AB is a tangent to circle O at B

∵ AD is a secant intersects circle O at C and D

∵ O ∈ AD

- That means AD divides the circle into two equal arc

∵ The measure of the circle is 360°

∴ m of ARC CBD =
`(1)/(2)` × 360° = 180°

To find m∠BAD use the first rule above

∵ m∠BAD =
`(1)/(2)` (m arc BD - m arc BC)

∵ m arc BD = 110° 20'

- Subtract the m of arc BD from 180° to find m of arc BC

∴ m arc BC = 179° 60' - 110° 20' = 69° 40'

- Substitute the measures of arcs BD and BC in the formula

of m∠BAD

∴ m∠BAD =
`(1)/(2)` (110° 20' - 69° 40') =
`(1)/(2)` (109° 80' - 69° 40')

∴ m∠BAD =
`(1)/(2)` (40° 40')

∴ m∠BAD = 20° 20'

∵ ∠ADB is an inscribed angle subtended by arc BC

- By using the 2nd rule above

∴ m∠ADB =
`(1)/(2)` m arc BC

∵ m arc BC = 69° 40'

∴ m∠ADB =
`(1)/(2)` (69° 40') =
`(1)/(2)` (68° 100')

∴ m∠ADB = 34° 50'