More Circle Question's

20....No...the measure of angle ADC  would have to be equal to the measure of angle BDC for the two arcs to  be equal....but....we don't know their measures

24.... Two tangents drawn to a circle  from a point are  equal in length

Therefore....BC  = AB  = 3

And CD  = ED  = 4

So...x  = BC + CD  = 3 + 4  = 7

26

We can apply the secant-tangent theorem, here....since AB  meets radius AP.....then AB  is tangent to the circle at A

And we have this

AB^2  = PB * CB

12^2  = PB  * 8

144  = PB  * 8      divide both sides by  8

18  = PB

So...PB  is the  hypotenuse of  triangle APB.....and by the Pythagorean Theorem, 

PB   = √[ AP^2  + AB^2]       square both sides

PB^2  = AP^2  + AB^2

324 = 144 + AB^2     subtract 144 from both sides

180  = AB^2       take the square root of both sides

√180  =  AB   =  2√45  = AB  =  "x"

Last one

Angle S intercepts  minor arc RT....its measure   = 1/2 of this arc  = 1/2 * 60°  =  30°

Angle R intercepts  minor arc QS ...the measure of minor arc  QS  is twice the measure  of angle R  =2 * 37  = 74°

For the measure of RQ....draw PQ....since PQ  = PR, then in triangle PQR.....the angles opposite these sides are equal

So angle  QRP  = angle RQP  = 37°

So...angle RPQ  = 180  - 37  - 37  = 180 - 74  =   106°

And angle RPQ is a central angle  subtending arc RQ....so....RQ  has the same measure  = 106° 

26.

Since  AB  is tangent to the circle at  A ,  m∠PAB  =  90°

Since  PA  and  PC  are both radii of circle P, they are the same length.

So    the length of PC  =  x    and    the length of PB  =  8 + x

And by the Pythagorean Theorem...

x² + 12²  =  (8 + x)²

x² + 144  =  (8 + x)(8 + x)

x² + 144  =  64 + 16x + x²

144  =  64 + 16x

80  =  16x

5  =  x

(I think that  PB  is not a secant.)