268. We have said (Art. 97) that the measurement, with scientific accuracy, of a line of any considerable length involves a long and difficult process.

On the other hand, sometimes it is required to find the direction of a line that it may point to an object which is not visible from the point from which the line is drawn. As for example when a tunnel has to be constructed.


By the aid of the Solution of Triangles we can find the length of the distance between points which are inaccessible; we can calculate the magnitude of angles which cannot be practically observed; we can find the relative heights of distant and inaccessible points.

The method on which the Trigonometrical Survey of a country is conducted affords the following illustration.

269. To find the distance between two distant objects.

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Two convenient positions A and B, on a level plain as far apart as possible, having been selected, the distance between A and B is measured with the greatest possible

This line AB is called the base line. (In the survey of England, the base line is on Salisbury plain, and is about 36,578 feet long).

Next, the two distant objects, P and Q (church spires, for instance) visible from A and B, are chosen.

The angles PAB, PBA are observed. Then by Case II. Chapter XVII, the lengths of the lines PA, PB are calculated.

Again, the angles QAB, QBA are observed ; and by Case II, the lengths of QA and QB are calculated.

Thus the lengths of PA and QA are found.

The angle PAQ is observed ; and then by Case III, the length of PQ is calculated.

270. Thus the distance between two points P and Q has been found. The points P and Q are not necessarily accessible; the only condition being that P and Q must be visible from both A and B.

271, In practice, the points P and will generally be accessible, and then the line PQ, whose length has been calculated, may be used as a new base to find other distances.

272. To find the height of a distant object above the point of observation.

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Let B be the point of observation; P the distant object. From B measure a base line BA of any convenient length, in any convenient direction; observe the angles PAB, PBA, and by Case II. calculate the length of BP. Next observe at B the angle of elevation of P; that is, the angle which the line BP makes with the horizontal line BM, M being the point in which the vertical line through P cuts the horizontal plane through B.

Then PM, which is the vertical height of P above B can be calculated, for PM=BP.sin MBP.

Example 1. The distance between a church spire A and a milestone B is known to be 1764:3 feet ; C is a distant spire. The angle CAB is 94° 54', and the angle CBA is 66° 39'. Find the distance of C from A.

ABC is a triangle and we know one side c and two angles (A and B), and therefore it can be solved by Case II, The angle ACB=180° -- 94° 54' – 66° 39'

=18° 27'. Therefore the triangle is the same as that solved on page 210. Therefore AC=5118•2 feet.

Example 2. If the spire C in the last Example stands on a hill, and the angle of elevation of its highest point is observed at A to be 4° 19'; find how much higher C is than A.

The required height x=AC.sin 4° 19' [Art. 268), and AC is 5118.2 feet,

.: log x=log (4C. sin 4° 19')

=log 5118·2+L sin 4° 19' – 10
=3:7091173+8.8766150 – 10

=2.5857323=log 385.24. Therefore

x=385 ft. 3 in, nearly.


(Exercises xiv. and LXI. consist of easy examples on this subject).

(1) Two straight roads inclined to one another at an angle of 60°, lead from a town A to two villages B and C; B on one road distant 30 miles from A, and C on the other road distant 15 miles from A. Find the distance from B to C. Ans. 25.98 m.

(2) Two ships leave harbour together, one sailing N.E. at the rate of 7} miles an hour and the other sailing North at the rate of 10 miles an hour. Prove that the distance between the ships after an hour and a half is 10.6 miles.

(3) A and B are two consecutive milestones on a straight road and C is a distant spire. The angles ABC and BAC are observed to be 120° and 45° respectively. Show that the distance of the spire from A is 3.346 miles.

(4) If the spire C in the last question stands on a hill, and its angle of elevation at A is 15°, show that it is •866 of a mile higher than A.

(5) If in Question (3) there is another spire D such that the angles DBA and DAB are 45° and 90° respectively and the angle DAC is 45°; prove that the distance from C to D is 24 miles very nearly.

(6) A and B are two consecutive milestones on a straight road, and C is the chimney of a house visible from both A and B. The angles CAB and CBA are observed to be 36° 18' and 120° 27' respectively. Show that C is 2639-5 yards from B,

log 1760 =3.2455127 L sin 36° 18'= 9:7723314
log 2639.5=342152 L cosec 23° 15'=10:4036846.

(7) A and B are two points on opposite sides of a mountain, and C is a place visible from both A and B. It is ascertained that C is distant 1794 feet and 3140 feet from A and B respectively and the angle ACB is 58° 17'. Show that the angle which the line pointing from A to B makes with AC is 86° 55' 49",

log 1346=3•1290451 L cot 29° 8' 30" = 10:2537194

log 4934=3.6931991 L tan 26° 4' 19" = 9•6895654

(8) A and B are two hill-tops 34920 feet apart, and C is the top of a distant hill. The angles CAB and CBA are observed to be 61° 53' and 76° 49' respectively. Prove that the distance from A to C is 51515 feet,

log 34920=4.5430742 L sin 76° 49'= 9.9884008

log 51515=471193 L cosec 41° 18'=101804552. (9) From two stations A and B on shore, 3742 yards apart, a ship C is observed at sea. The angles BAC, ABC are simultaneously observed to be 72° 34' and 81° 41' respectively. Prove that the distance from A to the ship is 8522-7 yards,

log 3742=3.5731038 L sin 81° 41'= 9.9954087 log 8522•7=3.9305774 L cosec 25° 45'=10:3620649.

(10) The distance between two mountain peaks is known to be 4970 yards, and the angle of elevation of one of them when. seen from the other is 9o 14. How much higher is the first than the second ? Sin 9° 14' =.1604555. Ans. 797.5 yards.

(11) Two straight railways intersect at an angle of 60°. From their point of intersection two trains start, one on cach line, one at the rate of 40 miles an hour. Find the rate of the second train that at the end of an hour they may be 35 miles apart. Ans. Either 25 or 15 miles an hour. (Art. 264.)

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