Circle & Law of Sines

[mathxyz]

I tried solving the following question using my little knowledge of law of sines.

HERE IT IS:

A) Find AB

B) Find the measure of OBTUSE angle BAC correct to the nearest tenth of a degree.

HERE IS WHAT I KNOW ABOUT CIRCLE O:

CDB is a secant on circle O.
AB is a tangent on circle O.
BD = 12, CD = 15 and angle AD = 60 degrees

I will use d to represent degrees.

MY WORK:

Angle C = 1/2(angle AD or 60d)

Angle C = 30d

NEXT:

I found the measure of arcs CD and AC by doing this:

arc CD + arc AC + 60d = 360d

Thus, arc CD = 150d = arc AC

Am I right thus far?

I then needed to find angle B, which lies OUTSIDE circle O.

I found angle B to be 45d.

I then used the law of sines to find length AB.

AB/sin30d = 15/sin45d OR sin 135d, right?

I found AB = 10.61

Book's answer is: AB = 18

Knowing that I also need OBTUSE angle BAC, this is what I did:

Angle BAC = 180d - 45d - 30d

Angle BAC = 105d

Book's answer is: 48.6d

Now, 48.6 degrees is NOT obtuse, right?

What did I do wrong?

 

[Gene]

Your terminology needs work. A point doesn't have degrees and can only be considered an angle if EXACTLY two lines meet at it. CD is a line. To refer to segment CD you have to say segment.
Now:
I got a diagram from your work

    .  C.
 .       \ . 
.         \ .
.     O    \.D
 .         .\
   .     .   \
      A_______\B

<OCD = 30° Good. (most computers will take Alt-0 as °. < can be used for angle.)

arc CD + arc AC + 60d = 360d

Also good.
arc CD + arc AC = 300°
You go astray when you assume they are equal.
Triangle ODA is equilateral.
<ODA = 60°
Triangle ODC is isosceles so <OCD = <ODC = 30°
<ODC + <ODA = 90°
<ADB = 90°
<DBA = 60°

Why don't you try it from there?

 

[mathxyz]

yes

Thank you Gene. I can take it from here.