#1**+3 **

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

CPhill May 8, 2018

#2**+2 **

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"

CPhill May 8, 2018

#3**+3 **

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°

CPhill May 8, 2018

#6**+2 **

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...

x^{2} + 12^{2} = (8 + x)^{2}

x^{2} + 144 = (8 + x)(8 + x)

x^{2} + 144 = 64 + 16x + x^{2}

144 = 64 + 16x

80 = 16x

5 = x

(I think that PB is not a secant.)

hectictar May 8, 2018