We have \(\widehat{POS}=180^o -2\widehat{SPO}\)

Also \(\widehat{POS}=\widehat{POR}+\widehat{ROS}\)

We also have \(\widehat{POR} = 180^o -2\widehat{RPO}\)

Hence \(\widehat{POS}=\widehat{POR}+\widehat{ROS} \\
\implies 180^o -2\widehat{SPO} = 180^o -2\widehat{RPO} +\widehat{ROS} \\
\implies \widehat{ROS} = 2\widehat{RPO} - 2\widehat{SPO} \\
\implies \widehat{ROS} = 2\widehat{RPS}\)

(1) If \(\widehat{ROS} = 40^o \implies \widehat{RPS} = 20^o\)

Since \(\widehat{PST} = 90^o \implies \widehat{PTS} = 90^o - \widehat{RPS} = 70^o\). Sufficient.

(2) We can't know the value of \(\widehat{PQT}\) so we can't know the value of \(\widehat{PTS}\). Insufficient.

The answer is A.
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Statement (1) can be solved exploring two facts only:

(i) Angle PSQ (hence angle PST) is (are) right (because PQ is a diameter)

(ii) Angle RPS is inscribed with the corresponding 40-degrees central angle ROS

Statement (2) is not sufficient, as proven with the geometric bifurcation.




This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Hi, jorgetomas9 !

Thank you for your interest in my solution.

Every inscribed angle (in this case RPS) is half the value of the central angle (in this case ROS) corresponding to the same arc (in this case arc RS).

Regards,
Fabio.
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Hi, castiel.

Thank you for the nice compliment (and for the kudos)!

You are absolutely right. Bifurcations are extremely useful and, to be honest, they change intuition (I feel...) into certainty (I am sure...)!

This "shield" is also possible in algebraic and geometric-algebraic bifurcations, by the way. (You can find them in my previous posts... and in my course.)

Regards and success in your studies,
Fabio.
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Hi Bunuel could you please provide an explanation for this problem ?