Question

PLEASE HELP ME ON THIS MATH QUESTION

Answer 1

Problem One

Any angle drawn from the end points of a diameter and whose vertex is on the circumference of a circle is a right angle. <ACB is a right angle. Therefore you can use Pythagoras.

Formula and givens

a = 12

b = 5

a^2 + b^2 = c^2

c^2 = 12^2 + 5^2

c^2 = 144 + 25

c^2 = 169 Take the sqrt of both sides.

sqrt(c^2)=sqrt(169)

c = 13

The radius = d/2 = 13/2 = 6.5 cm

Problem 2

Find <AOC. Begin by drawing OC

The measure of arc AC = 40 degrees, Because AOC is a central angle, it is also 40 degrees. Triangle AOC is isosceles. That comes from OA and OC both being radii. If angle AOC is 40o then <AOC + OAC + OCA = 180

Call <OCA = <OAC = x

40 + x + x = 180

40 + 2x = 180 Subtract 40 from both sides

2x = 180 - 40

2x = 140

x = 70

< OAC = 70