(2.) The tangent of of the degrees in the greater arc ==; hence Log. tan. arc=10+ Log. 1- Log. 3= 10-477121-9.522879= Log. tan. 18° 26' 6"; therefore the whole arc =73° 44′ 24′′=73·74°.

Also the radius of the circle =

(15)3+53 —25; hence the

2x5

area of the greater segment will be found as follows:

Again, the tangent of of the degrees in the arc of the less segment === tan. 11° 18′ 35′′, and therefore the whole arc -45° 14′ 23′′-45.23972°; hence the area of the segment may be found as follows:

Subtract sin. 45° 14′ 23′′ = 3550295

(2.) 6×6×6=216, surface and solidity.

(3.) (2+1)× 2=7, the perimeter, which being multiplied by the length 12, gives 84, the surface of the sides. To which add 2 × 11 × 2 = 6

2 × 11 × 12

90, surface. = 36, solidity.

(4.) 12×4

=

7,

perimeter.

7x16

(5.) Tabular area

= 112, surface of the sides.

12×12×2 = 61, surface of the ends.

118, whole surface. = 49, the solidity.

433013, page 67.

Multiply by 22—

=

4

42.151686, surface of the ends.

3×5×15 = 268-333333, surface of the sides.

L. 8.79648 L.8,15s.11d. Ans.

(4.) 7854×13×12×51=12.627758, solidity.

3.1416×12x51

surf.

=28-86347 sq. feet, convex.

4 ac. 2 ro. 20 poles =201465

201465

28.86347

volutions.

(1.) 30×4

120 x 35

30 x 30

Sum

sq. feet.

=6979.93, number of re

PROBLEM III.

120, the perimeter.

=1500, slant surface.

900, area of the base. =2400, whole surface.

√(25)—(15)=√400-20, perp. altitude.

... 900 ×

=6000, solidity.