Question

In the figure above, point O is the center of the circle,
line segment BC is tangent to the circle at point B,
point A lies on the circle, and point O lies on line
segment AC. If the circumference of the circle is 50,
what is the length of minor arc AB ?

Answer 1

Answer:

16.67 or 16 2/3

Step-by-step explanation:

so, as BC is a tangent, it means the angle OBC is 90 degrees.

therefore, the angle BOC is 180 - 90 - 30 = 60 degrees.

that means the angle AOB is 180 - 60 = 120 degrees.

the triangle ABO must be an isosceles triangle, as the sides are both the radius of the circle. so, both remaining angles are the same : half of 180 - 120 = 60 = 30 degrees.

and so, we know that ABC is an isosceles triangle, because both "side angles" are equal = 30 degrees.

so, AB = BC.

and we know the circumference of the circle (50) and can calculate the radius

50 = 2×pi×r

25 = pi×r

r = 25/pi ≈ 7.96

now we have multiple ways to calculate AB.

AB = BC = tan(60) × r

or

AB = sqrt(r² + r² - 2×r×r×cos(120))

or ...

the first one might be easiest.

AB = tan(60) × 7.96 ≈ 13.78

oh, I just recognized, we need the arc AB (not the straight line).

even easier.

50 is the full "arc" of the circle for all 360 degrees.

but we only need the part of the angle AOB = 120 degrees.

so, this is then only the 120/360 part of the full circle circumference.

ABarc = 50 × 120/360 = 50 / 3 ≈ 16.67 or 16 2/3