 | John Bascombe Lock - 1892 - 354 σελίδες
...isin 44° = 9-8417713, L sin 70° =9-9729858, L sin 66° = 9-9607302, log 1006-62 = 3-0028656, 89. Find the length of the arc of a circle whose radius is 8 feet which subtends at the centre an angle of 50°, having given 7r=3-1416. 90. Prove that sin A... | |
 | John Bascombe Lock - 1896 - 242 σελίδες
...sin 66° = 9.9607302, log 1006.62 = 3.0028656, log 7654321 = 6.8839067, log 103543 = 5.0151212. 89. Find the length of the arc of a circle whose radius is 8 feet which subtends at the centre an angle of 50°, having given TT =3.1416. 90. Prove that sin A... | |
 | 1915 - 278 σελίδες
...— Find the length of the arc of a circle when R = 9 inches and N = 74 degrees. Thus, L = .0174RN = .0174 X 9 X 74 = 11.58 inches. Example. — Find the...15 minutes is equal to .25 of one degree. Thus, L = .0171RN = .0174 X 7.5 X 67.25 = 8.77 inches. 332. Inscribed Angle. An inscribed angle is one whose... | |
 | 1915 - 278 σελίδες
...incheB め ndN ヰ 74de ち r 飴・ Thus, 乙ヰ .0171RN = .0174 X 9 X 74 = 11.58 inches Example.@Find the length of the arc of a circle whose radius is...number of degrees subtended by the arc is 67 15'. NOTE.@15 minutes is equal to .25 of one degree Thus. 乙ヰ .0174fl# = .0174 X 7.5 X 67.25 = 8.77 inches... | |
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Example. — -Find the length of the arc of a circle when R = 9 inches and N — 74 degrees. Thus, L = .0174RN = .0174 X 9 X 74 = 11.58 inches. Example. — Find the length of the arc of a circle whose radius is 7.5 inches and the number of degrees subtended... 