CPhill

First  one  :

A radius meeting a tangent does so at right angles.....so

angle ACB  and angle ABC  are complementary angles  in right triangle  ABC

So  

24  +  angle ABC  =  90   subtract 24 from  both sides

 angle ABC  =  66° 

PS  - if this is from VixenGrace....I answered your other multi-part question from  yesterday.......

Inscribed Angle Theorem

Correct

Correct  !!!!!

Third  one :

S  = arc length....so...

S   =2 * pi  * radius  *  (central  angle measure / 360)    ....so....

S   =  2 * pi  * (12.6)  *  ( 120 / 360)  =

2 * pi * 12.6  *  (1/3)  =

2 * 12.6 / 3  * pi   =

8.4 pi  cm      CORRECT -A - MUNDO  !!!!

Second one :

If  XYZ  = 76°.....then minor arc  XZ  = twice this   =  152°

So....major arc XYZ  = 360  - 152  =   208°

First one....

Connect  BC......

This sets up triangle BPC   where angle PCB  = angle PBC....so  BP  = CP....so BP  = 8

Next....connect  AD

This sets up triangle  APD  with  angle DAP  = angle ADP.......so  AP  = PD  = 6

Then  BD  = BP  +  PD    =   8  +  6   =  14     and you are correct   !!!!!

4)Let BC and DE be chords of a circle, which intersect at A, as shown. If AB = 3, BC = 15, and DE = 3, then find AE.

Draw BE  and  CD

We have two similar triangles

Angle BCD  = angle BED   and angle A is common to both

Therefore  

ΔCAD  is similar to Δ EAB      ⇒    CA / AD  =  EA / AB

Let   AD  = x   and we have

18 / x  =  ( x + 3 ) / 3        cross-multiply

18 * 3  =  x ( x + 3)

54  = x^2  + 3x        rearrange

x^2  + 3x  - 54  = 0       factor

(x + 9)  ( x - 6 )  =  0

Setting the second factor  = 0  and solving for x   produces a positive result for x  ⇒    6   =  AD

3)Chords PQ and RS of a circle meet at X inside the circle. If RS = 38, PX = 6, and QX = 12, then what is the smallest possible value of RX?

RX  * XS  = PX  * QX

RX * XS  = 6 * 12

RX * XS  = 72

And RX + XS    = 38    ⇒   XS = 38 - RX.....so....

RX  * XS    = 72

RX *  (38 - RX )  =  72   rearrange

RX^2 - 38RX +  72  = 0    Factor

(RX - 36) (RX - 2)  = 0

And its clear that the smallest  value for RX  = 2

OK....try this

(3x - 1) ( x - 4) ( x - 3)  = 

3 x^3 - 22 x^2 + 43 x - 12

2)Chord BA bisects chord OP at S. If AS = 3 and BS = 4, find OP

          O

    B    S    A

          P

Intersecting Chord Theorem

BS * AS  =   OS  * PS           but   OS  = PS....so

BS * AS  =  OS * OS

3 * 4  =  OS^2

12  = OS^2

√12  = OS   =  2√3

But  OP  = OS + PS = OS + OS  =    2OS    =   2 *  2√3     = 4√3  

1.  Intersecting  Chord Theorem :

WV  * VY   = XV * VZ

3 * VY  = 4 * 9

VY  =  4 * 9 /3     =  12

Since it appears the  VY  and VZ  are perpendicular....then....by the Pythagorean Theorem

VY^2  + VZ^2  =  YZ^2

12^2  +  9^2  = YZ^2

225  =  YZ^2   take the square root of both sides

15  = YZ