whence also tan. arc... Log. tan. arc. =10+ Log. 4- Log. 10=10+602060-1-9602060-Log. tan. 21° 48′ 5"; and hence the whole arc =87° 12′ 20′′87-205°; wherefore by Rule II. we have (4.) Since 90° is the fourth part of 360°, the sector of 90° is the fourth part of the whole circle; and the triangle being right-angled, its area will be r2; whence the area of the segment will be = (1.) Since (40-30)2 is less than 4× (35)", the centre is within the zone. OK, or the distance of the centre from the middle of (40+30)(40-30) 700 the zone = (8x35) = =2.5; hence the dis 280 tance of the centre from the less chord +2.5=20. Again the radius =√(20)2+(15)2=25. 35+2·5=17·5 And AD=4(40—30)2+(35)2=35.355337. Wherefore the sine of half the angle at the centre sub35.355337 50 tended by AD= =707107= sin. 45o, and hence the whole angle subtended at the centre by AD=90°. Now the sine of half the angle subtended by AD= =1414213= sin. 8° 7′ 48′′; therefore the whole 5.656854 40 angle 16° 15′ 36′′-16-26°. Next, to find the area of the segment, we have (3.) Since the chords are equal, they must be on opposite sides of the centre; and since the chords AB and DC are equal, it is plain that AD and CB are each equal to QP; but QP is =AB or DC, therefore the figure ABCD is a square, and its area will evidently be 2500. Again, the angle at the centre will evidently be 90°, and the radius is obviously = √(25)2+(25)2= √/1250 =25/2. Hence the area of one of the segments will be (4.) Since the radius is 20 and the less chord 20, the sine of half the angle subtended at the centre by the less chord === sin. 30°, (Trig. 84), and therefore the whole angle 60°; hence the area of the segment cut off by the less chord is And the area of the semicircle ='7854× (40)3 2 (1.) (5+4)(5—4) ×·7854—9 ×·7854=7·0686. Ans. (2.) (16+10)(16-10) ×·7854-26x6x7854= (3.) (24+18) (24-18)x7854-42x6x 7854— 252 x 7854-197-9208. Ans. (4.) (15+12)(15-12) × 7854-27x3x7854= 81 x 7854 63.6174. Ans. PROBLEM XIV. (1.) By Problem XI., the tangent of of the degrees in the arc of the greater segment; Log. tan. arc. 10+ Log. 1— Log. 2=10—301030 = 9·698970— Log. tan. 26° 33′ 54"; therefore the degrees in the arc =106° 15' 36"-106.26°. Also the radius of the circle = (18)2+92 2x9 =22.5; hence the area of the greater segment will be found as fol greater segment. 2236440 226-439550, area of the Again, the tangent of of the degrees in the arc of the less segment; 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 = (18)2+63 2x6 =30; hence the area of the segment will be found as follows: |