Ab is tangent to circle o at a. the diagram is not drawn to scale. if ao=30 and bc=48 what is ab?

1.) The radius which is OB also is 10.0.
   It is a right triangle so c^2 = a^2 + b^2 : In this case, AO^2 = AB^2 + OB^2
   11.4^2 = 6^2 + OB^2
   129.96 = 36 + OB^2
   OB^2 = 93.96
   OB = 9.69

   Note: I think your given is wrong since there's no 9.69 in the choices. But a similar problem is found on the internet and the given are  AB = 6 and AO = 11.7. So If these are the given data, the answer would be 10.0, Letter D.

2.) The perimeter is 64.
   In an incircle, the distances between a vertex and the two nearest tangent points are equal to each other.
   JA = JB
   LA = LC
   KC = KB

   2 × (12 + 15 + 5) = 64

3.) The measure of ZWX is 243º.
   A diameter forms an arc of 180°, so ZRW would equal to 180°. To get ZWX, we'll add ZRW and WX.
   180° + 63° = 243°

4.) The length of AB is 58.5.
   We'll get the length of CB first, by using OB^2 = OC^2 + CB^2, substitute and transpose to get CB.
   CB = sqrt(32^2 - 13^2)
   CB = sqrt(1024 - 169)
   CB = sqrt(855)
   CB = 29.24

   AB is equal to 2CB.
   AB = 2(29.24) = 58.5

1.) The radius which is OB also is 10.0.
   It is a right triangle so c^2 = a^2 + b^2 : In this case, AO^2 = AB^2 + OB^2
   11.4^2 = 6^2 + OB^2
   129.96 = 36 + OB^2
   OB^2 = 93.96
   OB = 9.69

   Note: I think your given is wrong since there's no 9.69 in the choices. But a similar problem is found on the internet and the given are  AB = 6 and AO = 11.7. So If these are the given data, the answer would be 10.0, Letter D.

2.) The perimeter is 64.
   In an incircle, the distances between a vertex and the two nearest tangent points are equal to each other.
   JA = JB
   LA = LC
   KC = KB

   2 × (12 + 15 + 5) = 64

3.) The measure of ZWX is 243º.
   A diameter forms an arc of 180°, so ZRW would equal to 180°. To get ZWX, we'll add ZRW and WX.
   180° + 63° = 243°

4.) The length of AB is 58.5.
   We'll get the length of CB first, by using OB^2 = OC^2 + CB^2, substitute and transpose to get CB.
   CB = sqrt(32^2 - 13^2)
   CB = sqrt(1024 - 169)
   CB = sqrt(855)
   CB = 29.24

   AB is equal to 2CB.
   AB = 2(29.24) = 58.5

1) The correct answer is A) 19.6

See the first picture attached.

A line tangent with a circle (AB) forms a 90° angle with the radius drawn from the tangent point (BO). This means that you can apply the Pythagorean theorem:
BO = √(AO² - AB²)
      = √(21.6² - 9²)
      = √(466.56 - 81)
      = √385.56
      = 19.6

2) The correct answer is B) 64

See the second picture attached.

When you have an incircle (a circle inscribed in a triangle), the distances between a vertex and the two nearest tangent points are equal to each other.

This means that:
JA = JB
LA = LC
KC = KB

Therefore the perimeter of ΔJKL will be: 
2 × (12 + 15 + 5) = 64

3) The question asks for the measure of arc ZWX
The correct answer is D) 243°

See the third picture attached.

We know that a diameter forms an arc of 180°, because it divides the circle into two equal parts.

Therefore, arc ZRW measures 180°.

In order to find arc ZWX you need to add arc ZRW and arc WX:
ZWX = ZRW + WX
         = 180 + 63
         = 243°

1) The correct answer is A) 19.6

See the first picture attached.

A line tangent with a circle (AB) forms a 90° angle with the radius drawn from the tangent point (BO). This means that you can apply the Pythagorean theorem:
BO = √(AO² - AB²)
      = √(21.6² - 9²)
      = √(466.56 - 81)
      = √385.56
      = 19.6

2) The correct answer is B) 64

See the second picture attached.

When you have an incircle (a circle inscribed in a triangle), the distances between a vertex and the two nearest tangent points are equal to each other.

This means that:
JA = JB
LA = LC
KC = KB

Therefore the perimeter of ΔJKL will be: 
2 × (12 + 15 + 5) = 64

3) The question asks for the measure of arc ZWX
The correct answer is D) 243°

See the third picture attached.

We know that a diameter forms an arc of 180°, because it divides the circle into two equal parts.

Therefore, arc ZRW measures 180°.

In order to find arc ZWX you need to add arc ZRW and arc WX:
ZWX = ZRW + WX
         = 180 + 63
         = 243°

the picture in the attached figure

we know that

if AB is tangent to circle O at B

then 

OB is the radius and is perpendicular to AB

therefore

in the right triangle ABO

applying the Pythagoras Theorem

AO²=AB²+OB²-------> solve for OB

OB²=AO²-AB²------> 11.7²-6²------> OB²=100.89

OB=√100.89--------> OB=10.04----------> OB=10.0 units

the answer is the option

C. 10.0

AO = 21
BC = 14
OC = radius of the circle = AO = 21
∴ OB = OC + CB = 21 + 14 = 35
Line AB is tangent to circle O at A
∴ AB is perpendicular to AO
∴ Δ OAB is a right triangle at A
Applying Pythagorean theorem
∴  OB² = AO² + AB²
∴ AB² = OB² - AO² = 35² - 21² = 1225 - 441 = 784
∴ AB = √784 = 28

Recall that a tangent to a circle is perpendicular to the diameter of the circle at the pont of tangency.

Given that AC is tangent to circle O at A, this means that line AB is a diameter of the circle.

Thus, angle BAC is a right angle.

Also, angle BAC = angle BAY + angle YAC.
i.e. 34 + <YAC = 90
<YAC = 90 - 34 = 56 degrees.

AO = 21
BC = 14
OC = radius of the circle = AO = 21
∴ OB = OC + CB = 21 + 14 = 35
Line AB is tangent to circle O at A
∴ AB is perpendicular to AO
∴ Δ OAB is a right triangle at A
Applying Pythagorean theorem
∴  OB² = AO² + AB²
∴ AB² = OB² - AO² = 35² - 21² = 1225 - 441 = 784
∴ AB = √784 = 28