(a) In the diagram (i) The value of x; (ii) (b) If \(2N4_{seven} = 15N_{nine}\), find the value of N.

Answer Details

(a) Finding \(x\).

Reading the diagram. \(Q\), \(R\) and \(S\) lie on a circle with centre \(O\). The straight line \(Q\,O\,S\) passes through the centre, so \(QS\) is a diameter, and it is produced beyond \(S\) to the external point \(T\). The angle at \(Q\) is \(x\), the right-angle mark is at \(R\), and the exterior angle at \(S\) (between \(SR\) and \(ST\)) is \((3x + 15)^\circ\).

Angle in a semicircle. Since \(QS\) is a diameter, the angle it subtends at \(R\) is a right angle:

\[\angle QRS = 90^\circ,\] which agrees with the mark at \(R\).

Exterior-angle theorem in \(\triangle QRS\). The line \(Q\,S\,T\) is straight, so \((3x+15)^\circ\) is the exterior angle at \(S\); it equals the sum of the two remote interior angles \(\angle Q\) and \(\angle R\):

\[3x + 15 = x + 90.\]\[3x - x = 90 - 15 \;\Rightarrow\; 2x = 75 \;\Rightarrow\; x = \mathbf{37.5^\circ}.\]

(The follow-up part (ii) is not legible in the source text, so only \(x\) is evaluated here.)

(b) Number-base equation: \(2N4_{\text{seven}} = 15N_{\text{nine}}\).

Expand each numeral in powers of its base, treating \(N\) as an unknown digit.

Left side (base 7):

\[2N4_{\text{seven}} = 2(7^2) + N(7) + 4 = 98 + 7N + 4 = 102 + 7N.\]

Right side (base 9):

\[15N_{\text{nine}} = 1(9^2) + 5(9) + N = 81 + 45 + N = 126 + N.\]

Equate:

\[102 + 7N = 126 + N \;\Rightarrow\; 6N = 24 \;\Rightarrow\; N = \mathbf{4}.\]

Check: \(N = 4\) is a valid digit in both base 7 and base 9. Left \(= 102 + 28 = 130\); right \(= 126 + 4 = 130\). Both equal \(130_{\text{ten}}\). Correct.