NOTE. Unless the contrary is stated, it will be supposed that the observer's height is disregarded, and that the angles of elevation are measured from the ground. Example I. A person walking along a straight road observes that at two consecutive milestones the angles of elevation of a hill in front of him are 30° and 75°: find the height of the hill. If we take √3=1·732 and reduce to feet, we find that the height is 3606-24 ft. EXAMPLES. XVII. a. 1. From the top of a cliff 200 ft. above the sea-level the angles of depression of two boats in the same vertical plane as the observer are 45° and 30°: find their distance apart. 2. A person observes the elevation of a mountain top to be 15°, and after walking a mile directly towards it on level ground the elevation is 75°: find the height of the mountain in feet. 3. From a ship at sea the angle subtended by two forts A and B is 30°. The ship sails 4 miles towards A and the angle is then 48°: prove that the distance of B at the second observation is 6.472 miles. 4. From the top of a tower h ft. high the angles of depression of two objects on the horizontal plane and in a line passing through the foot of the tower are 45°-A and 45°+A. Shew that the distance between them is 2h tan 24. 5. An observer finds that the angular elevation of a tower is A. On advancing a feet towards the tower the elevation is 45° and on advancing b feet nearer the elevation is 90°-A: find the height of the tower. 6. A person observes that two objects A and B bear due N. and N. 30 W. respectively. On walking a mile in the direction N.W. he finds that the bearings of A and B are N.E. and due E. respectively: find the distance between A and B. 7. A tower stands at the foot of a hill whose inclination to the horizon is 9°; from a point 40 ft. up the hill the tower subtends an angle of 54°: find its height. 8. At a point on a level plane a tower subtends an angle a and a flagstaff c ft. in length at the top of the tower subtends an angle B: shew that the height of the tower is c sin a cosec ẞ cos (a+B). Example II. The upper three-fourths of a ship's mast subtends at a point on the deck an angle whose tangent is 6; find the tangent of the angle subtended by the whole mast at the same point. Let C be the point of observation, and let APB be the mast, AP being the lower fourth of it. Let AB=4a, so that AP-a; also let AC=b, LACB=0, ▲ BCP=ß, NOTE. The student should observe that in examples of this class we make use of right-angled triangles in which the horizontal base line forms one side. Example III. A tower BCD surmounted by a spire DE stands on a horizontal plane. From the extremity d of a horizontal line BA, it is found that BC and DE subtend equal angles. If BC=9 ft., CD=72 ft., and DE=36 ft., find BA. But √39 6.245 nearly; .. x=56.205 nearly. Thus AB=56.2 ft. nearly. 9. A flagstaff 20 ft. long standing on a wall 10 ft. high subtends an angle whose tangent is 5 at a point on the ground: find the tangent of the angle subtended by the wall at this point. 10. A statue standing on the top of a pillar 25 feet high subtends an angle whose tangent is 125 at a point 60 feet from the foot of the pillar: find the height of the statue. 11. A tower BCD surmounted by a spire DE stands on a horizontal plane. From the extremity A of a horizontal line BA it is found that BC and DE subtend equal angles. If BC=9 ft., CD=280 ft., and DE=35 ft., prove that BA=180 ft. nearly. 12. On the bank of a river there is a column 192 ft. high supporting a statue 24 ft. high. At a point on the opposite bank directly facing the column the statue subtends the same angle as a man 6 ft. high standing at the base of the column : find the breadth of the river. 13. A monument ABCDE stands on level ground. At a point P on the ground the portions AB, AC, AD subtend angles a, B, y respectively. Supposing that AB-a, AC=b, AD=c, AP=x, and a+B+y=180°, shew that (a+b+c) x2=abc. Example IV. The altitude of a rock is observed to be 47°; after walking 1000 ft. towards it up a slope inclined at 32° to the horizon the altitude is 77°. Find the vertical height of the rock above the first point of observation, given sin 47° =•731. Let P be the top of the rock, A and B the points of observation; then in the figure PAC=47°, 4 BAC=32°, LPDC= LPBE=77°, AB=1000 ft. x=PA sin PAC-PA sin 47°. We have therefore to find PA in terms of AB. In A PAB, L PAB=47° - 32°=15°; LAPB 77° - 47° = 30°; .. ▲ ABP=135°; E 1000 ft. D C If we take √√2=1·414, we find that the height is 1034 ft. nearly. 14. From a point on the horizontal plane, the elevation of the top of a hill is 45°. After walking 500 yards towards its summit up a slope inclined at an angle of 15° to the horizon the elevation is 75°: find the height of the hill in feet, 15. From a station B at the base of a mountain its summit A is seen at an elevation of 60°; after walking one mile towards the summit up a plane making an angle of 30° with the horizon to another station C, the angle BCA is observed to be 135°: find the height of the mountain in feet. 16. The elevation of the summit of a hill from a station A is a. After walking c feet towards the summit up a slope inclined at an angle ẞ to the horizon the elevation is y: shew that the height of the hill is c sin a sin (y — B) cosec (y — a) feet. 17. From a point A an observer finds that the elevation of Ben Nevis is 60°; he then walks 800 ft. on a level plane towards the summit and then 800 ft. further up a slope of 30° and finds the elevation to be 75°: shew that the height of Ben Nevis above A is 4478 ft. approximately. 200. In many of the problems which follow, the solution depends upon the knowledge of some geometrical proposition. Example I. A tower stands on a horizontal plane. From a mound 14 ft. above the plane and at a horizontal distance of 48 ft. from the tower an observer notices a loophole, and finds that the portions of the tower above and below the loophole subtend equal angles. If the height of the loophole is 30 ft., find the height of the tower. Let AB be the tower, C the point of observation, L the loophole. Draw CD vertical and CE horizontal. Let AB=x. We have CD=14, AD=EC=48, BE=x-14. From AADC, AC2=(14)2+(48)2 = 2500; .. AC=50. From A CEB, CB2= (x − 14)2+(48)2 By squaring, 9(x2-28x+2500) = 25 (x2 - 60x +900). On reduction, we obtain 16x2 - 1248x=0; whence x=' Thus the tower is 78 ft. high. =78. |