Theory question 113
Triangles and Quadrilaterals(a) Two lines AB and CD intersect at X such that CÂX is equal to BD̂X. 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.
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(a) ∠CAX = ∠BDX and ∠AXC = ∠DXB (vertically opposite), so ΔAXC ~ ΔDXB with A↔D, C↔B.
AX/DX = CX/BX: 6/|XD| = 3/4, so |XD| = 8 cm.
(b)(i) At Y: bearing of X from Y = 132°, bearing of Z from Y = 202°, so ∠XYZ = 70°.
|XZ|2 = 322 + 502 − 2(32)(50) cos 70° = 3524 − 1094.5 = 2429.5, so |XZ| ≈ 49.3 km.
(ii) sin ∠YXZ = 50 sin 70°/49.29 ≈ 0.9532, ∠YXZ ≈ 72.4°.
Bearing of Z from X = 312° − 72.4° = 239.6°.
Back bearings differ by 180°: use them to find the angle at Y, then the cosine and sine rules.