Answer:
1. m∠AOB= 75°
105°
285°
2. OB=15cm
3. AB is 9√2 cm
Step-by-step explanation:
1. arc AB= m∠AOB= 75°, the measure of the central angle is equal to the measure of its intercepted arc
arc AD= 180°- arc AB = 180°-75°= 105° , 180° because BD is a straight line and arc BAD is a semicircle
arc ACB = 360°- arc AB= 360°-75°=285°
2. a tangent is perpendicular to the circle at the point of tangency, so ΔBAO is a right triangle and we can use the Phytagorean Theorem here.
[tex](AB)^{2} + (AO)^{2} = (OB)^{2} \\\\(12cm)^{2} + (9cm)=(OB)^{2} \\144cm^{2} +81cm^{2} = (OB)^{2} \\225cm^{2} = (OB)^{2} \\\sqrt{(OB^{2}) } = \sqrt{225cm\\^{2} } \\OB = 15cm[/tex]
3. ΔBOA is an isosceles triangle since two sides are equal since they are both radii of Circle O, if arc AB is 90°, then ∠BOA is also 90°, therefore this is an isosceles right triangle (45-45-90 special right triangle), in a 45-45-90 right triangle, the hypotenuse is √2 times one of the leg, so if one leg measures 9 cm, then AB is 9√2 cm.
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