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.
Isosceles triangle: ∠OAD=∠ODAExterior Angle Theorem: ∠OAP=∠OAD+∠ODA∠OAP=∠OAD+∠OAD∠OAP=2⋅∠OAD|:2∠OAP2=∠OAD41∘2=∠OAD20.5∘=∠OAD
see: https://www.mathsisfun.com/geometry/triangle-exterior-angle-theorem.html
27.
Secant-Tangent Rule: x2=3⋅(3+9)x2=3⋅12x2=36x=6
see: http://www.regentsprep.org/regents/math/geometry/gp14/circlesegments.htm
can someone help me solve numbers 25 and 27? Thanks you :)
without mistake:
25.
Isosceles triangle: ∠OAD=∠ODAExterior Angle Theorem: ∠AOP=∠OAD+∠ODA∠AOP=∠OAD+∠OAD∠AOP=2⋅∠OAD|:2∠AOP2=∠OAD41∘2=∠OAD20.5∘=∠OAD
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 △ABDand△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
∴△ABD∼△CBA
Now I am going to use the ratios of similar triangles:
ABCB=BDBAx12=3xx2=36x=6units