Hey <Kuro>

Following are my worked solutions:

Q1. (Refer to attached diagram)

Join the line OA. And triangles OBA, OCA are congruent {(RHS): angle OBA = 90 = angle OCA, OB = OC (radii), OA is common side}.

Angle AOB = angle AOC

Angle COB = 180 - 50 = 130

So angle AOB = 65 = angle AOC

Joining OD (radius), triangle ODB is isosceles {OD = OB (radii) }

Hence angle OBD = 65 ÷ 2 = **37.5** degrees {exterior angle of triangle ODB = sum of interior opposite angles}

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Q2. (Refer to attached diagram)

Triangles PQR, PST are equilateral and SR = SQ {given S is mid-point of QR}

PS is a line segment bisector and angle PSQ = 90 = angle PSR.

angle SPR = 90 - angle PRS = 90 - 60 = 30

Let the two lines PR and ST intersect at a point U.

Angle RSU = 90 - angle PST = 90 - 60 = 30

Angle RUS = angle USP + angle UPS = 60 + 30 = 90

Area RST = (1/2)(ST)(RU)

Area PQR = (1/2)(QR)(PS)

Since ST = PS,

(Area PST) / (Area PQR) = (RU) / (QR)

But QR = PR,

(Area PST) / (Area PQR) = (RU) / (PR)

Refer triangles RUS, PSR (both right-angled triangles).

In triangle RUS,

Sin (30) = (RU / RS) = (1/2)

RU = (1/2)(RS) = (RS)/2

In triangle PSR,

Sin (30) = (RS / PR) = (1/2)

PR = 2(RS)

Hence,

(Area PST) / (Area PQR) = (RU) / (PR)

= (RS)/2 ÷ 2(RS)

= **1/4**

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Trust this helps

Do let me know again if these are different from your <Anskey> or if there's further clarification.

Cheers,

Edward