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of the angle included by the two sides, and EZ is drawn parallel to the tangent to the circle at Q. Shew that the remaining side of the spherical triangle is equal to the arc QPZ.

16. If through any point P within a spherical triangle ABC great circles be drawn from the angular points A, B, C to meet the opposite sides at a, b, c respectively, prove that sin Pa cos PA sin Pb cos PB sin Pc cos PC

-1.
sin Aa
sin Bb

sin Cc

+

17. A and B are two places on the Earth's surface on the same side of the equator, A being further from the equator than B. If the bearing of A from B be more nearly due East than it is from any other place in the same latitude as B, find the bearing of B from A.

18. From the result given in example 18 of Chapter v. infer the possibility of a regular dodecahedron.

а

19. Å and are fixed points on the surface of a sphere, and P is any point on the surface. If a and b are given constants, shew that a fixed point S can always be found, in AB or AB produced, such that

a cos AP + b cos BP = 8 cos SP, where s is a constant.

=

20. A, B, C,... are fixed points on the surface of a sphere; a, b, c,... are given constants. If P be a point on the surface of the sphere, such that

a cos AP + b cos BP + c cos CP +

= constant,

shew that the locus of P is a circle.

XVI. NUMERICAL SOLUTION OF SPHERICAL

TRIANGLES.

204. We shall give in this Chapter examples of the numerical solution of Spherical Triangles.

We shall first take right-angled triangles, and then obliqueangled triangles.

Right-Angled Triangles.

205. Given a=37° 48' 12", b= 59° 44' 16", C = 90°. To find c we have

COS C = cos a cos b,
I cos 37° 48' 12'' = 9.8976927
L cos 59° 44' 16' = 9.7023945

L cos C + 10 = 19.6000872

C= 66° 32' 6".

To find A we have

cot A = cot a sin b, L cot 37° 48' 12'' = 10:1102655 L sin 59° 44' 16" = 9.9363770

L cot A + 10 = 20.0466425

A = 41° 55' 45".

To find B we have

cot B=cot 6 sin a,

I cot 59° 44' 16''= 9.7660175
L sin 37° 48' 12" = 9:7874272

L cot B +10= 19.5534447

B=70° 19' 15".

206.

Given A = 55° 32' 45", C = 90°, c= 98° 14' 24".

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Here cosc is negative; and therefore cot B will be negative, and B greater than a right angle. The numerical value of cosc is the same as that of cos 81° 45' 36".

I cos 81° 45' 36" = 9.1563065
L tan 55° 32' 45' = 10.1636102

L cot (180o – B) + 10 = 19.3199167

180° - B= 78° 12' 4"

B= 101° 47' 56".

To find 6 we have

tan b = tan c cos A.

Here tan c is negative ; and therefore tan b will be negative and b greater than a quadrant.

L tan 81° 45' 36" = 10.8391867

L cos 55° 32' 45' = 9.7526221

L tan (180° — 6) + 10 = 20.5918088

180° – b = 75° 38' 32"

b=104° 21' 28".

207. Given A = 46° 15' 25", C = 90°, a= 42° 18' 45".

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To find 6 we have

sin b = tan a cot A,

L tan 42° 18' 45" = 9.9591983

L cot 46° 15' 25" =

9.9809389

I sin 6 + 10 = 19.9401372

b = 60° 36' 10" or 119° 23' 50".

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Oblique-Angled Triangles. 208. Given a=70° 14' 20", b = 49° 24' 10", c= 38° 46' 10". We shall use the formula given in Art. 45,

sin (8-5) sin (8-)) tan : A =

sin s sin (s – a) Here

8=79° 12' 20",

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8° 58',

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I sin 79° 12' 20":

L sin 8° 58'

19.5083472 9.9922465 9.1927342

19.1849807

19.5083472 19.1849807

2) 3233665 L tan1 A-10= "1616832

A = 55° 25'38"
A = 110° 51' 16'.

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