(a) Triangle PQR is right-angled at Q. PQ = 3a cm... | WAEC SSCE - General Mathematics - 1989 (Essay)

Question 1 Report

(a) Triangle PQR is right-angled at Q. PQ = 3a cm and QR = 4a cm. Determine PR in terms of a.

(b) Ayo travels a distance of 24km from X on a bearing of 060° to Y. He then travels a distance of 18km to a point Z and Z is 30km from X.

(i) Draw the diagram to show the positions of X, Y and Z ; (ii) What is the bearing of Z from Y ; (iii) Calculate the bearing of X from Z.

Answer

(a)

Since triangle \(PQR\) is right-angled at \(Q\), by Pythagoras' theorem,

\[PR^2=PQ^2+QR^2=(3a)^2+(4a)^2=9a^2+16a^2=25a^2.\]

\[\therefore\ PR=5a\text{ cm}.\]

(b)(i) Diagram

The diagram below is drawn to scale in the correct relative positions. \(XY=24\text{ km}\) is on a bearing of \(060^\circ\), \(YZ=18\text{ km}\), and \(XZ=30\text{ km}\).

Scale diagram of the journeys from X to Y and from Y to Z.

(ii) Bearing of \(Z\) from \(Y\)

\[24^2+18^2=576+324=900=30^2.\]

Therefore, \(\angle XYZ=90^\circ\). The bearing of \(X\) from \(Y\) is \(240^\circ\). Hence,

\[\text{bearing of }Z\text{ from }Y=240^\circ-90^\circ=150^\circ.\]

Bearing of \(Z\) from \(Y\) = \(150^\circ\).

(iii) Bearing of \(X\) from \(Z\)

Using the cosine rule at \(Z\),

\[\cos \angle XZY=\frac{30^2+18^2-24^2}{2(30)(18)}=\frac{648}{1080}=0.6.\]

\[\angle XZY=\cos^{-1}(0.6)=53.13^\circ.\]

Since the bearing of \(Y\) from \(Z\) is \(150^\circ+180^\circ=330^\circ\),

\[\text{bearing of }X\text{ from }Z=330^\circ-53.13^\circ=276.87^\circ\approx277^\circ.\]

Bearing of \(X\) from \(Z\) = \(277^\circ\).

Read lesson note on Circles (WAEC)

Read lesson note on Triangles And Polygons (WAEC)

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