(a) Two lines AB and CD intersect at x such that (stackrelfrown{CAX}) is equal to (stackrelfrown{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.
Explanation
(a)
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)
From (Delta ZYX),
(stackrelfrown{YZX} = 90° – 22° = 68°)
(stackrelfrown{YXZ} = 312° – 270° = 42°)
(stackrelfrown{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)
(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°)