HELP

can someone help me solve numbers 25 and 27? Thanks you :)

without mistake:

25.
\(\begin{array}{rcll} \text{Isosceles triangle: } & \angle{OAD} &=& \angle{ODA} \\\\ \text{Exterior Angle Theorem: } & \angle{AOP} &=& \angle{OAD} + \angle{ODA} \\ & \angle{AOP} &=& \angle{OAD} + \angle{OAD} \\ & \angle{AOP} &=& 2\cdot \angle{OAD} \qquad | \qquad : 2\\ & \frac{\angle{AOP} } {2} &=& \angle{OAD} \\ & \frac{ 41^{\circ} } {2} &=& \angle{OAD} \\ & 20.5^{\circ} &=& \angle{OAD} \\ \end{array}\)

25.

<POD is 180 degrees,  

<POA=41

so <AOD=139

triangle AOD is isoceles so 

<OAD= 0.5*(180-139)=0.5*41=   20 degrees and 30 minutes

27.   Mmm

Well it has to be more than 3 because the two tangents to a circle subtended from a point will be equal.

(I not sure how I should word that) 

I can't be 9 either.  For the same reason.

So it has to be between 3 and 9.

What about 8 ......

No, it'll have to be less than that.

OK I don't think it can be any of those but I would like someone to give a more mathematical precise......

can someone help me solve numbers 25 and 27? Thanks you :)

25.

\(\begin{array}{rcll} \text{Isosceles triangle: } & \angle{OAD} &=& \angle{ODA} \\\\ \text{Exterior Angle Theorem: } & \angle{OAP} &=& \angle{OAD} + \angle{ODA} \\ & \angle{OAP} &=& \angle{OAD} + \angle{OAD} \\ & \angle{OAP} &=& 2\cdot \angle{OAD} \qquad | \qquad : 2\\ & \frac{\angle{OAP} } {2} &=& \angle{OAD} \\ & \frac{ 41^{\circ} } {2} &=& \angle{OAD} \\ & 20.5^{\circ} &=& \angle{OAD} \\ \end{array} \)

see: https://www.mathsisfun.com/geometry/triangle-exterior-angle-theorem.html

27.

\(\begin{array}{rcll} \text{Secant-Tangent Rule: } & x^2 &=& 3\cdot (3+9) \\ & x^2 &=& 3\cdot 12 \\ & x^2 &=& 36 \\ & \mathbf{x} & \mathbf{=} & \mathbf{6} \end{array}\)

see: http://www.regentsprep.org/regents/math/geometry/gp14/circlesegments.htm

I did it as follows:

.

Question 27.

I am looking at an alternative to Heureka's and Alan's answers.  

Heureka And Alan did both inspire this answer.    Thanks  guys 

I cannot remember all theorems relating to circes so I am going to use this more common theorem.

The angle between a tangent and a chord is equal to the angle subtended by the chord in the alternate segment.

Proof is here:

https://www.youtube.com/watch?v=LCfgfg8Jv8g

Now to find x  

Consider   \(\triangle ABD \;\;and\;\; \triangle CBA\)

<BAD = <BCA   The angle bteween a chord and a tangent is equal to the anle subtended by the chord in the alternate segment    

<ABD = <CBA   Common angle

\(\therefore \triangle ABD \sim \triangle CBA\)

Now I am going to use the ratios of similar triangles:

\(\frac{AB}{CB}=\frac{BD}{BA}\\ \frac{x}{12}=\frac{3}{x}\\ x^2=36\\ x=6\;units\)