log sin p" = log p-log r = 6.685566 + log P.

1...4. Solve the right triangles (preferably by aid of natural functions), and check the work, given:

5, 6. Solve the right triangles by aid of the table for small angles, and check the work, given:

7...9. Solve the isosceles triangles, and check the work, given:

7. base = 26.13, side=127.8.

231.1, base angle = 27° 19'.

9. side = 49.25, vertical angle = 57° 33'.

10...13. Solve the oblique triangles, and check the work,

14. Find the ratio of the areas of two regular decagons, the one inscribed in, and the other circumscribed about, the same circle.

So, of two regular heptagons.

So, of two regular n-gons.

[.9045

So, of a regular n-gon inscribed and a regular m-gon circumscribed.

15. In a circle of unit radius a regular pentagon is inscribed; find its side, its apothem, and its area.

So for a regular n-gon inscribed in a circle of radius r.

16. Find the angle at which the lateral face of a pyramid is

inclined to the base, the faces being equilateral triangles
and the base a square: thence find the diedral angle of
a regular octaedron.
[54° 44' 8", 109° 28' 16"

17. In a regular tetraedron whose edge is unity, find the diedral
angle of an edge, the perpendicular from the vertex to
the base, and the distance apart of two opposite edges.
[70° 31' 44", .8165, .7071

So in a regular pentagonal pyramid whose lateral edge is three times an edge of the base.

18. If two circles, of radii r, r', touch externally, and if ◊ be the angle between their common tangents, find the value of sin in terms of r, r'.

For, drop a perpendicular p from c to the base c at D, dividing

b2= p2+q2, a2 = p2 + (c− q)2 = b2 + c2 — 2 cq;

q = b cos A,

then and

A= √[(s — b) (s — c) : bc]........ [s=(a+b+c)

COR. 1. sin

For 2 sin2

A = 1- cos A

sin

and so for sin

[II th. 13 cr.

Q. E. D.

COR. 3. tan

=1+[(b2+c2 − a2): 2bc]

= [(b+c)2 - a2]: 2 bc

=(b+c+a) (b+c-a): 2 bc

=4s (sa): 2bc,

A =√[8 (s − a) : bc] ;

B, COS C.

[th. 13 cr.

Q. E. D.

▲ = √/[(s — b) (s — c): s ( s − a)] ....... [crs. 1, 2 NOTE. In the limited triangle no angle is negative or greater than two right angles, no half-angle is negative or greater than a right angle; and the radicals of crs. 1–3 are all positive.

COR. 1. (a+b): c = cos (A — B): sin C,

(ab): c=sin(A — B): cos c.

For

·.· a:b:c=sin a : sin B : sin c,

[th. 2

... (a+b):c= (sin A + sin B): sin c

= 2 sin (A+B) COS (AB): 2 sin c cos c

So

= cos(A-B): sin c. Q. E.D. [sin(A+B) = cos c (a - b): csin (AB): cos c.

*GEOMETRIC PROOF. On BC take D, E such that CD, CE = CA;

then

BD = α-b, BE = a+b.

E

A

F B

The reader may prove that AEC= c, that
BAD (AB); and may then apply the

that

to the triangles Bad and BAE.

COR. 2. tan (AB) = (a - b): (a+b). cot c.

DAE is right,

law of sines

[cr. 1

*GEOMETRIC PROOF. In the figure of cr. 1 draw DF parallel to

EA and therefore perpendicular to AD;

then tan (A — B) =

DF DF EA BD
= cotc =
AD EA AD BE

a-b a+b

cotc.

EXAMPLES.

1...4. In any plane triangle ABC, show that:

1. s:c=cos A COSB: sin c.

2. (sa): c = cosa sin B: COS C. 3. (sb): csin A COSB: cos C. 4. (sc): c= sina sin B: sin c.

[s = (a+b+c)