The answer
(i) shown
(ii) shown
O-Level A-Math 2022 Paper 1 Question 7 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 7 of the O-Level A-Math 2022 Paper 1. It tests equal tangents from an external point, in the Proofs in plane geometry area. It is worth 6 marks: 3 + 3. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
(i) \(S\) is the point where the tangent at \(R\) meets the tangent at \(P\), so \(SP\) and \(SR\) are the two tangents drawn from the external point \(S\); hence \(SP = SR\) and triangle \(SPR\) is isosceles. Its base angles are therefore equal: \(\angle RPS = \angle SRP = x\). The line \(ST\) (the tangent at \(P\) extended to \(T\)) makes \(\angle RST\) an exterior angle of triangle \(SPR\) at \(S\)... more directly, \(\angle RST\) is the exterior angle of triangle \(RPS\) at \(S\) on the straight line \(PT\), equal to the sum of the two interior opposite angles: \(\angle RST = \angle RPS + \angle PRS = x + x = 2x\). (shown)
(ii) Since \(PQ\) is a diameter, \(\angle QRP = 90^{\circ}\) (angle in a semicircle). Since \(PT\) is the tangent at \(P\) and \(OP\) is a radius, \(\angle QPT = 90^{\circ}\) (tangent perpendicular to radius). The angle between the tangent \(SR\) and the chord \(RP\) equals the angle in the alternate segment, so \(\angle SRP = x\) corresponds to \(\angle PQR = x\). In triangle \(PQR\)... working with the straight line \(PRT\) along the chord produced: because \(\angle QRP = 90^{\circ}\), the angle \(\angle PRT = 180^{\circ} - 90^{\circ} = 90^{\circ}\) (angles on the straight line through \(R\)). Then \(\angle SRT = \angle PRT - \angle SRP = 90^{\circ} - x\). In triangle \(PRT\), \(\angle RTS = 180^{\circ} - \angle QPT - \angle PQR = 180^{\circ} - 90^{\circ} - x = 90^{\circ} - x\). Thus \(\angle SRT = \angle RTS = 90^{\circ} - x\), so triangle \(RST\) has two equal base angles and is isosceles. (shown)
Answer: (i) shown
(ii) shown
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level A-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a equal tangents from an external point question from Proofs in plane geometry, worth 6 marks: 3 + 3.
Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.
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