Answer:
m∠BAD is 20° 20'
m∠ADB is 34° 50'
Step-by-step 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 = [tex]\frac{1}{2}[/tex] × 360° = 180°
To find m∠BAD use the first rule above
∵ m∠BAD = [tex]\frac{1}{2}[/tex] (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 = [tex]\frac{1}{2}[/tex] (110° 20' - 69° 40') = [tex]\frac{1}{2}[/tex] (109° 80' - 69° 40')
∴ m∠BAD = [tex]\frac{1}{2}[/tex] (40° 40')
∴ m∠BAD = 20° 20'
∵ ∠ADB is an inscribed angle subtended by arc BC
- By using the 2nd rule above
∴ m∠ADB = [tex]\frac{1}{2}[/tex] m arc BC
∵ m arc BC = 69° 40'
∴ m∠ADB = [tex]\frac{1}{2}[/tex] (69° 40') = [tex]\frac{1}{2}[/tex] (68° 100')
∴ m∠ADB = 34° 50'
