Topical Revision Paper 12 Unit 12 Bearings Sine Cosine Rule Area of Triangle

Topical Revision Paper 12 Term 3 Unit 12: Bearings, Sine Rule, Cosine Rule and Area of Triangle Question 1: The diagram below shows the position of three locations A, B and C with respect to a reference point O. The bearings of A, B and C from O are 050 , 135 and 240 respectively. It is also given that 120OA m and 150OB m. (i) Find the bearing of B from A. [2] (ii) If the bearing of B from C is 100 , calculate the length of OC. [2] (iii) A mast stands vertically at B such that the angle of elevation of the top of the mast from O is 62 . Find the height of the mast. [2] Solution: (i) 135 50 85BOA   22150 120 2(150)(120)cos85 183.7AB    sin sin85 54.4150 OAB OABAB   Hence bearing of B from A 360 (180 50 ) 54.4 175.6     (ii) 240 135 105COB   and 100 (180 120) 40OCB    B A C O N 50 120 150

Hence by sine rule : 150 sin 35 sin 40 133.8 m OC OC   (iii) heighttan 62 150 height 282.1 m Question 2: In the diagram A, B and S represent three points in the sea. B is 3 km from A on a bearing of 060 . The bearings of S from A and B are 110 and 155 respectively. (i) Calculate the distance AS. [2] (ii) A ship at S sails directly to A at a steady speed of 2 1kmh . After 40 minutes, the ship is at T, a point on AS. (a) Calculate the distance BT. [3] (b) Find the bearing of B from T. [2] (iii) A hot air balloon, H, is lowering at a point 600 m vertically above B. Calculate the greatest angle of elevation of H from an observer on the ship as the ship sails from S to A. [2] Solution: (i) 000 5060110 BAS 0000 85)155180(60 ABS 0000 455085180 ASB

00 0 0 3 sin 85 sin 45 3 85 4.226 4.23sin 45 AS siinAS km     (iv) (a) Distance = Speed x Time 3 4 60 402  km AT = 4.226 – 1.333 km = 2.893 km Using cosine rule, 2 2 2 03 2.893 2(3)(2.893)cos50 2.492 2.49 BT BT BT km      (b) 0 0 0 sin sin 50 2.892 2.492 2.892sin 50 62.742.492 ABT ABT      000 74.26074.62 y The bearing of B from T is 0 0 0 0360 2.74 357.26 357.3   (iii) Shortest distance from observer to B = 03sin 50 1 0 0 0.6tan 3sin 50 14.6      Question 3: The diagram shows three points, A, B and C on a piece of horizontal land. It is given that AB = 400 m, BC = 330 m, AC = 710 m and B is due North of A.

(i) Calculate (a) ABC , [2] (b) the area of triangle ABC, [2] (c) the bearing of C from A. [2] (ii) BZ is a tower crane standing vertically at B and the angle of elevation of Z from A is 18 . Calculate the height of the tower crane. [2] (iii) A car moves along the path AC until it reaches a point P which is nearest to the tower crane BZ. Calculate the largest angle of depression of the point P from Z. [2] Solution: (i) (a) 222400 330 710cos 2 400 330 153.0 (1 d.p) ABC ABC      (b) 2 1Area of 400 330 sin(152.99)2 = 30000 m (3 s.f) ABC     (c) sin sin(152.988) = 330 710 BAC 12.2BAC   North B C A 710 m 330 m 400 m

Bearing is 012.2 (ii) Let BZ be the height of the tower crane tan18 400 130 m (3s.f) BZ BZ   (iii) BP is the shortest distance from B to path AC 1 710 29975.692 84.44 m BP BP     Let largest angle of depression be  , 129.97tan 84.44 57.0 (1 d.p)     Question 4: A man, at point M, is 80 metres at a bearing of 200 from an observation deck, D, which stands 25 metres tall. He begins to walk to a point P which is due East of D at a bearing of 050 . Ignoring the man’s height, calculate (i) the angle of elevation of the top of the deck from the man at M, [2] M North East D P 200 North 50

(ii) the distance he walks when the angle of elevation of the top of the deck from his position is the greatest, and [2] (iii) the distance DP. [2] Solution: (i) Angle of elevation 1 25tan 80  17.4 (1 d.p.) (ii) By Sine Rule, 80 m sin 30 sin(180 30 110 ) DP     80 m2sin 40DP  62.2 (3 s.f.) mDP (iii) 50 (200 180 ) DMP     30 Distance he walks 80cos30 m 40 3 or 69.3 (3 s.f.) m