(a) Two lines AB and CD intersect at x such that \(\stackrel\frown{CAX}\) is equal to \(\stackrel\frown{BDX}\). If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm,...
Question 1 Report
(a) Two lines AB and CD intersect at x such that \(\stackrel\frown{CAX}\) is equal to \(\stackrel\frown{BDX}\). If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm, find |XD|.
(b) 
The diagram shows the positions of three points X, Y and Z on a horizontal plane. The bearing of Y from X is 312° and that of Y from Z is 022°. If |XY| = 32 km and |ZY| = 50 km, calculate, correct to one decimal place : (i) |XZ| ; (ii) the bearing of Z from X.
Answer Details
(a) Finding |XD|. Since \(\angle CAX=\angle BDX\) (given) and \(\angle AXC=\angle DXB\) (vertically opposite), triangles \(AXC\) and \(DXB\) are similar (AA). Matching the equal angles gives
\[\frac{AX}{DX}=\frac{CX}{BX}\Rightarrow \frac{6}{XD}=\frac{3}{4}\]
\[XD=\frac{6\times4}{3}=8\text{ cm}.\]
(b) The angle at Y. The bearing of Y from X is \(312°\), so the bearing of X from Y is \(312°-180°=132°\). The bearing of Y from Z is \(022°\), so the bearing of Z from Y is \(022°+180°=202°\). Hence
\[\angle XYZ=202°-132°=70°.\]
(i) |XZ|. By the cosine rule,
\[XZ^2=XY^2+ZY^2-2\,XY\cdot ZY\cos70°=32^2+50^2-2(32)(50)\cos70°\]
\[XZ^2=3524-3200(0.3420)=2429.5\Rightarrow XZ=\sqrt{2429.5}=49.3\text{ km}.\]
(ii) Bearing of Z from X. By the sine rule,
\[\frac{\sin\angle YXZ}{ZY}=\frac{\sin70°}{XZ}\Rightarrow \sin\angle YXZ=\frac{50\sin70°}{49.3}=0.9532\]
\[\angle YXZ=72.4°.\]
From the diagram Z lies to the west of Y as seen from X, so the bearing of Z from X is measured anticlockwise from that of Y:
\[312°-72.4°=239.6°.\]
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