But HCB, having its vertex at the center, is measured by the intercepted arc DH; and DAF, being an inscribed angle, is measured by one-half the arc DH; hence, DAF = \ HCB = } (A — B).

In the right triangles ADE and ADF we have

From the similar triangles, ABE and FBD, we have

BE: BD :: AE : DF.

Since CECA, BE=BC+CA=a+b.

Since CD CA, BD= BC-CA-a-b. Substituting the values of BE, BD, AE, and DF in the above proportion, and omitting the common factor AD in the second couplet, we have

a+bab: tan (A+B): tan (A- B).

Hence, In any plane triangle, the sum of the sides including an angle is to their difference as the tangent of half the sum of the other two angles is to the tangent of half their difference.

We find from the proportion, the equation

tan (AB)=

(ab) tan (A+B)

a + b

... log tan (A—B) = log (a—b) + log tan † (A+B)

.. log clog a + log sin C+ a. c. log sin A-10.

S. N. 6.

a+ba-b: tan (A+B): tan (A-B).

.. tan (A— B) :

(A-B)

(a - b) tan (A+B)

a + b

... log tan (A—B)=log (a—b) + log tan †(A+B)

log tan (A+B) (55° 47′ 30′′) = 10.16761

a. c. log (a+b) (61.31)

log tan (A-B)

8.21247

=

A = 1⁄2 (A + B) + 1⁄2 (A — B) — 74° 7′ 25′′.

B = {(A + B) — ¦ (A — B) — 37° 27′ 35′′.

=

log clog a + log sin C+ a. c. log sin A-10.

Given the three sides of a triangle, required the angles.

Let ABC be a triangle, take the longest side for the base, and draw the perpendicular p from the vertex B to the base.

B

A

Denote the segments of the base by s and s' respect

ively.

Then, (1) c2-s'′2=p2,

and (2) a2-s2= p2.

... (3) c2—8'2=a2—82, ... (4) 82—8′2—a2—c2.

.'. (5) (8+8′) (8 — s'′) = (a + c) (a—c).

.'. (6) s + s′ : a + c :: a-c: s—s'.

Hence, The sum of the segments of the base is to the sum of the other sides as the difference of those sides is to the difference of the segments.

.. (8) log (8-8)=log (a+c) + log (a−c)

a. c. log (s+8') 10.

In case the sides of the triangle are small, find s-s' from (7); otherwise, it will be more convenient to employ (8).

Having ss and s-s', we find s and s' thus,

(9) 8 = 1⁄2 (8 + 8′ )+1(s—s′), (10) s′ = ↓ (s+s′ ) — 4(8—8′).

(14) log cos C 10+ log slog a.

S

s′ = 1 (8 + 8′) — 1 (8-8)=75-18.75=56.25.

1. A horizontal plane is a plane parallel to the horizon.

2. A vertical plane is a plane perpendicular to a horizontal plane.

3. A horizontal line is a line parallel to a horizontal plane.

4. A vertical line is a line perpendicular to a horizontal plane.

5. A horizontal angle is an angle whose plane is horizontal.