This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Another one Intellectual — let's solve it. 1.1.1 Determine, with reasons, the size of each of the following angles: a) MOS Step 1: Identify the relationship between the angle at the center and the angle at the circumference subtended by the same arc. 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 MOS at the center O and STM at the circumference. Step 2: Apply the theorem and substitute the given value. MOS = 2 × STM MOS = 2 × 29^ Step 3: Calculate the angle. MOS = 58^ Reason: Angle at center is twice angle at circumference. b) L Step 1: Use the information that chords PM and MS are equal. Since chords PM and MS are equal, they subtend equal angles at the center. Therefore, POM = MOS. Step 2: Substitute the value of MOS from part (a). POM = 58^ Reason: Equal chords subtend equal angles at the center. Step 3: Identify the relationship between L (which is PLM) and the central angle POM. PLM is the angle subtended by arc PM at the circumference. POM is the angle subtended by arc PM at the center. Step 4: Apply the theorem and calculate L. PLM = (1)/(2) × POM PLM = (1)/(2) × 58^ L = 29^ Reason: Angle at circumference is half angle at center. 1.1.2 Prove that SM = 2 × ON Step 1: Use the information that ON bisects chord LP. A line from the center that bisects a chord is perpendicular to the chord. Therefore, ONL = 90^. Step 2: Consider the right-angled triangle ONL. In ONL, we have ( OLP) = (ON)/(OL). From part 1.1.1 (b), we know OLP = L = 29^. Let r be the radius of the circle, so OL = r. (29^) = (ON)/(r) ON = r (29^) (Equation 1) Step 3: Consider MOS. Since OM and OS are both radii, OM = OS = r. Thus, MOS is an isosceles triangle. From part 1.1.1 (a), we know MOS = 58^. Draw a perpendicular from O to SM, let the intersection point be K. In OKM, OKM = 90^. The line from the center perpendicular to a chord bisects the angle at the center. MOK = (1)/(2) MOS = (1)/(2) × 58^ = 29^ Step 4: Express SM in terms of the radius and angle. In the right-angled triangle OKM: ( MOK) = (MK)/(OM) (29^) = (MK)/(r) MK = r (29^) Since OK is perpendicular to SM, K is the midpoint of SM. SM = 2 × MK SM = 2 × r (29^) (Equation 2) Step 5: Compare Equation 1 and Equation 2. From Equation 1, we have r (29^) = ON. Substitute this into Equation 2: SM = 2 × (r (29^)) SM = 2 × ON Thus, it is proven that SM = 2 × ON. Drop the next question.