(a) Two lines AB and CD intersect at x such that \(\stackrel\frown{CAX}\) is equal to...
(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)_LI (1).jpg)
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.
Explanation
(a).jpg)
Note that triangles AXC and BXD are similar.
\(\therefore \frac{6}{|XD|} = \frac{3}{4}\)
\(|XD| = \frac{6 \times 4}{3}\)
= \(8 cm\)
(b)(i)_LI.jpg)
From \(\Delta ZYX\),
\(\stackrel\frown{YZX} = 90° - 22° = 68°\)
\(\stackrel\frown{YXZ} = 312° - 270° = 42°\)
\(\stackrel\frown{ZYX} = 180° - (68° + 42°)\)
= \(180° - 110°\)
= \(70°\)
\(a^{2} = b^{2} + c^{2} - 2bc \cos A\)
\(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \cos 70\)
= \(2500 +1024 - 3200 \times 0.342\)
= \(3524 - 1094.46\)
\(|XZ|^{2} = 2429.54\)
\(|XZ| = \sqrt{2429.54} = 49.29 km\)
\(\approxeq 49.3 km\) (to 1 d.p)
(ii).jpg)
\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)
\(\frac{50}{\sin \theta} = \frac{49.3}{\sin 70}\)
\(\sin \theta = \frac{50 \times \sin 70}{49.3}\)
\(\sin \theta = 0.9530\)
\(\theta = \sin^{-1} (0.9530)\)
= \(72.37°\)
\(\therefore\) The bearing of Z from X = \(312° - 72.37°\)
= \(239.63° \approxeq 239.6°\)