Three towns \(P\), \(Q\) and \(R\) are joined by straight roads. \(PQ = 63\) km, \(QR = 47\) km and \(PR = 88\) km. (a) Calculate angle \(PQR\). [3] (b) Cal...

Assessment: Mathematics 0580 | Paper 4 Mock 01 | Calculator (Extended) Subject: Mathematics - 0580

Question 1 Report

Three towns \(P\), \(Q\) and \(R\) are joined by straight roads. \(PQ = 63\) km, \(QR = 47\) km and \(PR = 88\) km.

(a) Calculate angle \(PQR\). [3]

(b) Calculate the area of triangle \(PQR\). [2]

(c) Calculate the shortest distance from \(Q\) to the road \(PR\). [2]

(d) Calculate angle \(QPR\). [3]

Give each answer correct to 3 significant figures.

Answer Details

All three sides of triangle \(PQR\) are given and no angles, so every angle must come from the cosine rule in its rearranged form. Parts (b) and (c) then reuse the results rather than starting again.

  1. (a) Angle \(PQR\) is at \(Q\), enclosed by \(PQ\) and \(QR\), with \(PR\) opposite it: \[\cos PQR=\frac{PQ^{2}+QR^{2}-PR^{2}}{2\times PQ\times QR}=\frac{63^{2}+47^{2}-88^{2}}{2\times 63\times 47}\] [M1] The numerator is \(3969+2209-7744=-1566\) and the denominator is \(5922\), so \[\cos PQR=-0.2644\ldots\] [M1] \[PQR=105^\circ\] [A1] (accept \(105.3\)) The negative cosine tells you at once that the angle is obtuse, which fits \(PR=88\) km being much the longest road.
  2. (b) With the included angle now known, the area rule uses the two sides that enclose it: \[\text{Area}=\frac{1}{2}\times 63\times 47\times\sin 105.33^\circ=1480.5\times 0.96441\ldots\] [M1] \[=1427.8\ldots=1430\text{ km}^{2}\] [A1] (accept \(1428\))
  3. (c) The shortest distance from \(Q\) to the road \(PR\) is the perpendicular distance, which is the height of the triangle when \(PR\) is taken as the base. Since the area is already known, \(\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}\) rearranges to \[h=\frac{2\times 1427.8}{88}\] [M1] \[=32.4\text{ km}\] [A1]
  4. (d) Angle \(QPR\) is at \(P\), enclosed by \(PQ\) and \(PR\), with \(QR\) opposite it: \[\cos QPR=\frac{63^{2}+88^{2}-47^{2}}{2\times 63\times 88}=\frac{3969+7744-2209}{11088}\] [M1] \[\cos QPR=\frac{9504}{11088}=0.857142\ldots\] [M1] \[QPR=31.0^\circ\] [A1]

Part (c) rewards recognising that "shortest distance to a line" always means the perpendicular distance, and that the area already found is the quickest route to it. As a check on parts (a) and (d), the third angle would be \(180-105.3-31.0=43.7^\circ\), and the sides 88, 63, 47 km rank in the same order as the angles opposite them, \(105.3^\circ\), \(43.7^\circ\) and \(31.0^\circ\).

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