Determine, with reasons, the size of each of the following angles: a) MOS b) L

The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle. Arc MS subtends $\angle MOS$ at the center O and $\angle STM$ at the circumference.

Reason: Angle at center is twice angle at circumference.

b) $\boldsymbol{\hat{L}}$

Since chords PM and MS are equal, they subtend equal angles at the center.
Therefore, $\angle POM = \angle MOS$.

Reason: Equal chords subtend equal angles at the center.

$\angle PLM$ is the angle subtended by arc PM at the circumference. $\angle POM$ is the angle subtended by arc PM at the center.

Reason: Angle at circumference is half angle at center.

1.1.2 Prove that $SM = 2 \times ON$

A line from the center that bisects a chord is perpendicular to the chord.
Therefore, $\angle ONL = 90^\circ$.

In $\triangle ONL$, we have $\sin(\angle OLP) = \frac{ON}{OL}$.
From part 1.1.1 (b), we know $\angle OLP = \hat{L} = 29^\circ$.
Let $r$ be the radius of the circle, so $OL = r$.

Since OM and OS are both radii, $OM = OS = r$. Thus, $\triangle MOS$ is an isosceles triangle.
From part 1.1.1 (a), we know $\angle MOS = 58^\circ$.
Draw a perpendicular from O to SM, let the intersection point be K. In $\triangle OKM$, $\angle OKM = 90^\circ$.
The line from the center perpendicular to a chord bisects the angle at the center.

In the right-angled triangle $\triangle OKM$:

Since OK is perpendicular to SM, K is the midpoint of SM.

From Equation 1, we have $r \sin(29^\circ) = ON$.
Substitute this into Equation 2:

Thus, it is proven that $SM = 2 \times ON$.