 | John Bascombe Lock - 1882 - 378 pages
...multiple of I80jr 180-n-' (7) If a, ft y are in AP, then sin a + sin y = 2 sin /3 . cos (/3 - a). (8) Two parallel chords of a circle, lying on the same...subtend respectively 72° and 144° at the centre. Prove that the distance between the chords is half the radius of the circle. (9) Prove that 4 sin (6... | |
 | Thomas Grenfell Vyvyan - 1882 - 150 pages
...that tan ^45° - ^ + tan ^45° + ^\ = 2 see A. 5. Find the sine of 18°, and deduce the sine of 36°. Two parallel chords of a circle, lying on the same...subtend respectively 72° and 144° at the centre. Shew that the distance between the chords is half the radius of the circle. 6. Find all the angles... | |
 | Woolwich roy. military acad - 1884 - 148 pages
...tan (45° - — J + tan (45° + - J = 2 sec A. 5. Find the sine of 18°, and deduce the sine of 36°. Two parallel chords of a circle, lying on the same...subtend respectively 72° and 144° at the centre. Shew that the distance between the chords is half the radius of the circle. 6. Find all the angles... | |
 | John Bascombe Lock - Trigonometry - 1885 - 368 pages
...multiple of ISOrr 180 -n(7) If a, ft y are in AP, then sin a + sin y = 2 sin /3 . cos (/3 - a). (8) Two parallel chords of a circle, lying on the same...subtend respectively 72° and 144° at the centre. Prove that the distance between the chords is half the radius of the circle. (9) Prove that 4 sin (6... | |
 | E. J. Brooksmith - Mathematics - 1889 - 356 pages
...tan ( 45° -- \ + tan ( 45° + - J = 2 sec A. 5. Find the sine of 18°, and deduce the sine of 36°. Two parallel chords of a circle, lying on the same...subtend respectively 72° and 144° at the centre. Shew that the distance between the chords is half the radius of the circle. 6. Find all the angles... | |
 | Henry Sinclair Hall, Samuel Ratcliffe Knight - Plane trigonometry - 1893 - 434 pages
...(1 - sin a) (1 - sin j8) (1 - sin y), shew that each expression is equal to + cos o cos /3 cos y. 5. Two parallel chords of a circle lying on the same side of the centre subtend 72° and 144° at the centre : prove that the distance between them is one-half of the radius. Also... | |
 | Daniel Alexander Murray - Plane trigonometry - 1899 - 350 pages
...Two parallel chords of a circle lying on the same side of the centre of a circle subtend angles of 72° and 144° at the centre. Show that the distance between the chords is equal to half the radius of the circle, (a) using tables, (6) not using tables. 18. (a) Solve : (i.)... | |
 | Daniel Alexander Murray - Plane trigonometry - 1899 - 226 pages
...18°, without tables' (c) Find the trigonometric functions of 48°. [Hnrr: 48° = 30° + 18°.] 12. Two parallel chords of a circle lying on the same side of the centre of a circle subtend angles of 72° and 144° at the centre. Show that the distance between the chords... | |
 | Charles Hamilton Ashton, Walter Randall Marsh - Trigonometry - 1902 - 186 pages
...angle of elevation 100 yards from the same point. Show that the height of the tower is —¡= ft. V7 25. Two parallel chords of a circle, lying on the...between the chords is half the radius of the circle. X 26. A person standing due south of a lighthouse observes that his shadow, cast by the light at the... | |
 | Daniel Alexander Murray - 1906 - 466 pages
...18°, without tables. (c) Find the trigonometric functions of 48°. [HINT : 48° = 30° + 18°.] 12. Two parallel chords of a circle lying on the same side of the centre of a circle subtend angles of 72° and 144° at the centre. Show that the distance between the chords... | |
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Find the trigonometric functions of 48°. [HINT : 48° = 30° + 18°.] 12. Two parallel chords of a circle lying on the same side of the centre of a circle subtend angles of 72° and 144° at the centre. Show that the distance between the chords is equal... 