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The problems on Great Circle Sailing and Nautical Astronomy are solved by either right-angled or oblique-angled spherical trigonometry.
If a spherical triangle has one of its angles à right angle, it is called a right-angled triangle ; if one of its sides be a quadrant (90°) it is called a quadrantal triangle ; if two of the sides be equal it is called an isosceles triangle, etc., as in plane trigonometry.
The following properties relate to. spherical triangles
(a) Any side of a spherical triangle is less than a semicircle, and any angle is less than two right angles.
(h) The sum of the three angles is greater than two right angles and less than six right angles.
(c) If the three sides of a spherical triangle be equal, the three angles will also be equal, and vice versâ.
(d) If the sum of any two sides of a spherical triangle be equal to 180°, the sum of their opposite angles will also be equal to 180°, and vice versa.
(e) If the three angles of a spherical triangle be acute, all right, or all obtuse, the three sides will be accordingly all less than 90°, all equal to 90°, or all greater than 90°, and vice versa.
(f) The sum of any two sides is greater than the third side, and their difference is less than the third side.
(g) The sum of any two angles is greater than the supplement of the third angle.
(h) The sum of the three sides is less than the circumference of a great circle.
(i) If any two sides of a triangle be equal to each other, their opposite angles will be equal, and vice versa.
NAPIER'S RULES FOR THE SOLUTION OF RIGHT-ANGLED
In every right-angled spherical triangle there are five “circular parts," exclusive of the right angle, which is not taken into consideration. These five parts consist of the two legs containing the right angle, the complement of the hypotenuse, and the complements of the two angles. They are called “circular parts” because they are each measured by the “arc” of a Great Circle.
Before proceeding to the use of Napier's Rules it is necessary to have a clear idea of what is meant by the “ MIDDLE PART," and the following explanation should make it clear.
In the solution of every problem three parts are concerned, two of which are given and the one required to be found.
If the three parts follow each other in succession, that is, touch each other, that part which is in the middle is the “middle part” and the other two parts are the adjacent parts; if the three parts do not touch each other, two of them must, and that part which is separated from them is the “middle part" and the other two parts are the opposite parts.
It is necessary to point out that the angles separate the hypotenuse from the base and perpendicular, but the perpendicular and base, which contain the right angle, touch each other, that is, the right angle does not separate them.
Another important point to remember is, when a complement comes into an equation the sin., cos., or tan. of the formula are to be changed into cos., sin., or cot., as the case may be, and the following cases will make these points clear.
“ NAPIER'S RULES
The above rules have reference to the natural sines, cosines, and tangents. In practise, however, log. sines, etc., are used in the ordinary way, facilitating the work of multiplication and division by addition and subtraction respectively.
If the student has carefully studied the foregoing the following examples will be clear to him.
In the right-angled spherical triangle A B C right-angled at B, given the side b, and LA, to find side c.
Now it is obvious from what has been said that LA is the middle part and the sides b and c the adjacent parts.
Napier's first rule says the sine of the middle part equals the product of the tangents of the adjacent parts. That is
but as the complements of A and side b are to be taken the formula becomes
Log. tan. C= log. cos. A + 10 log. cot. 6 and by substituting the reciprocal of cot. b we get the following
Log. tan. c = log. cos. A + log. tan. 6 — 10
In the same triangle, given side b and LC, to find side c.
Napier's second rule says the sine of the middle part equals the product of the cosines of opposite parts. That is
Sin. c = cos. b x cos. C
but as the complements of b and C are to be taken the formula becomes
Sin. c =
sin. 6 x sin. C
.: Log. sin. c= log. sin. b + log. sin. C - 10
In the same triangle, given _ C and side a, to find side c.
Sin, a = tan. c x tan. C
but the complement of C is taken and the formula becomes
When an angle and the side opposite are given the case is ambiguous and there will be two answers—the part found and its supplement.
. The rule of signs must be attended to, as like signs give plus and unlike signs give minus when multiplied together or divided by each other.
The sine and cosecant are + up to 180°, that is, in the first and second quadrants.
The cosine, secant, tangent, and co-tangent are plus in the first quadrant, that is, up to 90°, and change to minus in the second quadrant, that is, from 90° to 180°.
The sine and cosecant are minus in the third and fourth quadrants, that is, from 180° to 360°.
The cosine and secant are minus in the third quadrant and plus in the fourth quadrant.
The tangent and co-tangent are plus in the third quadrant and minus in the fourth quadrant.
The student should now be able to solve all the cases in right-angled spherical trigonometry without assistance and to arrive at the required equation at a glance.
Example 1.-In the right-angled spherical triangle A B C (Fig. I) rightangled at C, given the side b 86° 10'00" and the angle A 74° 45' 15", to find the angle B.
< B is the middle part and Z A and
Cos. B = sin. A X COS. Ó.
A 74° 45' 15' sin. 9.984441
In the same triangle, given _ A 74° 45' 15" and side 6 86° 10', to find side a. Side b is the middle part, and _ A and side a adjacent parts.
Sin. 6 tan. a x cot. A
sin. 6 .:. Tan. a=
log. cot. A
a 74° 43' 17" tan. 10°563576 By using the reciprocal of cot. A we get the same result by addition (which is really multiplication).
Log. tan. a log. sin. b + log. tan. A – 10.
b 86° 10' 0" sin. 9.999027
Example 2.-In the right-angled triangle CAB (Fig. 2) right-angled at B, given L C 46° and side c 40°, to find side a.
Side a is the middle part, Z C and side c adjacent parts.
Sin, a = cot. C x tan. c. Log. sin. a = log. cot. C + log. tan. c
cot. 9:984837 C 40
tan. 9.923814 6 = 54° 7' 34" sin. 9.908651
a' 125° 52' 26"
This is the ambiguous case, because the given side is opposite the given angle and there is nothing to indicate which arc is to be taken ; as the angle C is equal to angle C' and c is common to both, therefore, each part found will have two values, that is, the part found and its supplement which is found by subtracting the arc or angle found from 180°.
In the same triangle Fig. 2, given _ C 46° and side c 40°, to find 2 A. < C is the middle part, _ A and side c opposite parts.
.:. Cot. 6
63° 19' 35" cot. 9.701020 180
Fig. 3. b
116 40 25 In this case the cosine is plus and the tangent minus, therefore, the cot. is minus and the arc just found is to be subtracted from 180°, as the minus cot. lies in the second quadrant.
Exercises.-Right-angled Spherical Triangles. 1. In the right-angled spherical triangle A B C, right-angled at C, given A C 65° and B C 51° 36', to find 2 A, B, and the side A B.
Ans. _ A 54° 15' 52" ; < B 70° 5' 36" and A B 74° 53' 52".
2. In the right-angled spherical triangle A B C, right-angled at C, given B C 32° 51' and B 56° 17', to find the other parts.
Ans. L A 45° 40' 14"; A C 39° 6' 20" and A B 49° 18' 54".
3. In the right-angled spherical triangle A B C, right-angled at B, given A B 110° 17' and B C 98° 46', to find the other parts.
Ans. _ C 110° 3' 55"; L A 98° 13' 51" and A C 86° 58' 17".