Answer:
m∠CAB = 90°
BD = 9
m∠ABC = 90°
arc AEC = 180°
AD = 14.5
Step-by-step explanation:
Question 1
The angle between the radius and a tangent to the circle is always 90°.
Therefore, as AC is the radius and AB is a tangent segment, m∠CAB=90°
As m∠CAB = 90°, we can use Pythagoras' Theorem to find BC (hypotenuse):
AC² + AB² = BC²
⇒ 8² + 15² = BC²
⇒ BC² = 289
⇒ BC = 17
As CD is the radius, CD = 8
Therefore, BD = BC - CD = 17 - 8 = 9
Question 2
If D is the center, then AC is the diameter. Angles at the circumference in a semicircle are always 90°. Therefore, m∠ABC = 90°
As AC is the diameter, arc AEC is half of the circle.
Therefore, arc AEC = 360° ÷ 2 = 180°
As m∠ABC = 90°, we can use Pythagoras' Theorem to find AC (hypotenuse):
BC² + AB² = AC²
⇒ 20² + 21² = AC²
⇒ AC² = 841
⇒ AC = 29
AD is the radius. A radius is half the diameter.
Therefore, AD = AC ÷ 2 = 29 ÷ 2 = 14.5
