SOLUTIOJt OF TRIANGLES.] 



NAVIGATION. 



1043 



By referring to a table of natural tangents, we find 

 Tan. 50 40'= 1-5204 

 Tan. 34 6'= -6771 



The difference = 8 .4 .3.3) 300 (11 8 '6 = A D. 

 8433 



The division here is by the con- 

 tracted method, recommended 

 before. 



1567 

 843 



724 

 675 



49 

 61 



Hence the height of the tower is 118-6 feet, very 

 nearly ; it is accurately between 118-5 feet and 118-6 

 feet : but nearer the latter than the former. 



To solve such examples as this by logarithms, as is 

 done in the books, the computation of an oblique-angled 

 triangle namely, of the triangle A B C is necessary ; 

 and there will be five references to the logarithmic tables. 

 In the above operation two references to a table suffice ; 

 and the subsequent arithmetical work, whatever number 

 the dividend in the division may be, will always be 

 trifling, if the contracted method of dividing be em- 

 ployed. 



2. From the top of a castle (Fig. 5), 60 feet high, 

 standing upon a hill near the sea-shore, the angle of de- 

 pression H T S of a ship at anchor was observed to be 4 

 62" and the angle of depression O B S, taken from the 

 i of the castle, was found to be 4 2f. What was 

 the distance A S of the ship 1 



Solution wilhyut Logarithms. 



Since T E (Fig. 6) is parallel to A S, the angle AST 

 is equal to the angle H T S 

 (Euc. 29. I.) For a like 

 reason the angle A S B is 

 equal to the angle O B S. 

 Consequently, the angles of 

 elevation AST, A S B are 

 . known namely, A S T . 4 

 5V, and ASB=42'. 

 By right-angled triangles, 



AT = SA tan. AST, AB = SA tan. ASB, 

 .'. AT AB=60=SA(tan. A ST tan. ASB). 

 By the table of natural tangents, we find 

 Tan. A S T=tan. 4 62'= -0851 

 Tan. AS B = tan. 4 2 / = -0705 



Fig. S. 



The difference = 



.01.4,6)60 (4100-SA, 

 684 



16 

 15 



Consequently, the horizontal distance of the ship is 

 4100 feet, or 1367 yards, very nearly. By multiplying 

 S A by tan. A S B, we shall get the height of the hill- 

 namely, 



0705 X 4100= 289= A B. 



3. The angle of elevation of the top of a tower was 

 found to be 46 30 7 , the place of observation from the 

 bottom being 220 feet distant in a horizontal line : re- 

 quired the height of the tower. Ans. 232 feet. 



4. At a horizontal distance of 45 yards from the bottom 

 of a steeple, the angle of elevation of the top was found 

 to be 48 \2f; the height of the observer's eye was 5 feet: 

 required the height of the steeple. Ans. 52 yards. 



5. In order to find the height of a castle surrounded 

 by a moat, the angle of elevation of its top was taken 

 from a convenient station, and found to be 46 10 7 ; then 

 at a station 110 yards further off, but in the same hori- 

 zontal line as the former station and the bottom of the 

 caitle, the angle of elevation was again taken, and found 

 to be 29 66' : required the height of the castle. 



Ans. 141-0 yards. 



6. The height of an inaccessible object being required, 

 the angle of elevation was taken at some distance from 

 it, and found to be 51 30'; and then a further distance of 

 75 feet being measured in the same horizontal line, the 

 angle of elevation was again taken, and found to be 26 

 30'. What was the height of the object, and at what 

 horizontal distance was it from the first place of obser- 

 vation? Ans. Height, 61 "97 feet ; distance, 49 -29 feet. 



7. From the top of a ship's mast, 80 feet above the 

 water, the angle of depression of another ship's hull was 

 taken, and found to be 20. What was the distance be- 

 tween the ships ? Ans. 220 feet. 



8. From the top of a lighthouse, 85 feet high, reckon- 

 ing from the summit of the rock on which it stands, the 

 angle of depression of a ship at anchor was found to be 

 3 38', and the angle at the bottom of the lighthouse, or 

 top of the rock, was found to be 2 43' : required the 

 horizontal distance of the ship and the height of the rock 

 above the level of the sea. 



Ans. Distance, 5296 feet ; height of rock, 251 feet. 



9. From the edge of a ditch, 36 feet wide, surrounding 

 a fort, the angle of elevation of the top of the wall was 

 found to be 62 40' : required the height of the wall, and 

 the length of a scaling-ladder to reach from the edge of 

 the ditch to the top. 



Ans. Wall, 69-6 feet ; ladder, 78'4 feet. 



10. In the year 1784, two observers on Blackheath, at 

 the distance of exactly a mile one behind the other, ob- 

 served the angle of elevation of Lunardi's balloon, at the 

 same instant of time ; these angles were 36 52' and 30 

 58' respectively : required the perpendicular height of 

 the balloon. Ans. 3 miles. 



To CALCULATE THE SIDES AND ANGLES OF OBLIQUE- 

 ANGLED TRIANGLES. In an oblique triangle the given 

 parts must be either 



1 . Two angles and a side ; 



2. Two sides and an angle ; or 



3. The three aides, to find the remaining three parts. 



I. When the given parti are either two angles and an 

 opposite tide, or two side* and an opposite angle. 



Investigation of the Rule. 



Let the triangle be A B C, and let it be either acute- 

 angled or obtuse-angled, as in Figs. 6 and 7 ; and let 

 the perpendicular A D be drawn, meeting the base or 

 the base produced in D. 



Fig. 6. Fig. 7. 



BCD 



As in the former investigations, let the sides of the 

 triangle, or rather their numerical measures, be repre- 

 sented by a, 6, c, these letters applying to the sides oppo- 

 site to the angles A, B, C, respectively. Each triangle 

 presents now two right-angled triangles namely, the 

 triangles A B D, A C t> ; and from each we get a distinct 

 expression for the perpendicular A D common to both ; 

 namely, 



A D = A B sin. B, and A D = A C sin. C ; 

 go that A B sin. B = A C sin. C, that is, 



. c sin. C 



c sin. B = 6 sm. C, . . -r = ~==B 

 b sin. B 



which shows that anyone side of a triangle is to any other, 

 at the sine of the angle opposite to the former, is to the 

 sine of the angle opposite to the latter. The angle C, in 

 the preceding equations, belongs to the right-angled 

 triangle A C D, which, in the second diagram, is not thu 

 angle C of the proposed triangle ABC, but the supple- 

 ment of that angle ; yet, as the sine of an angle is the 

 same as the sine of its supplement, we may regard the 

 C, in sin. C above, as the angle of the proposed triangle 

 in each diagram. 

 The property enounced in italics above, supplier tho 



