 | George Albert Wentworth - Trigonometry - 1896 - 344 pages
...Show that the height of the tower is V5-1 the tangent of 18° being V(2 + 2 Vo) ' ° ^(10 + 2 Vo) 25. A pole is fixed on the top of a mound, and the angles...length of the pole is twice the height of the mound. 26. At a distance («) from the foot of a tower, the angle of elevation (A) of the top of the tower... | |
 | George Albert Wentworth - Trigonometry - 1896 - 394 pages
...10 + 2V5 and ¡ — 2V5 _(10+2V5)(6+2V5) 62-(2V5)2 = 5 + 2 V5, cot2 30°= 3. Hence 6 = V2 + 2VВ 25. A pole is fixed on the top of a mound, and the angles of elevation of the top and bottom of the pole are 60° and 30°. Prove that the length of the pole is twice the height of the... | |
 | Great Britain. Civil Service Commission - Civil service - 1896 - 112 pages
...angles A and В of the triangle ABC, when a = 176, b = 135, and С = 47° 30'. 11. A pole is fixed at the top of a mound, and the angles of elevation of the bottom and top of the pole, as seen from the foot of the mound, are 30° and 60°. Find the ratio of... | |
 | George Albert Wentworth - Trigonometry - 1897 - 234 pages
...that the height of the tower is a ,,, ,,.,.,,. V5 — 1 ; the tangent of 18° being V(l0-f-2V5) 25. A pole is fixed on the top of a mound, and the angles...length of the pole is twice the height of the mound. 26. At a distance (a) from the foot of a tower, the angle of elevation (A) of the top of the tower... | |
 | George Albert Wentworth - Logarithms - 1897 - 384 pages
...the elevation is 18°. Show that the height of the tower is —32V5) the tfingent of 18° being 25. A pole is fixed on the top of a mound, and the angles...elevation of the top and the bottom of the pole are 6O° and 30° respectively. Prove that the length of the pole is twice the height of the mound. 26.... | |
 | George Albert Wentworth - Trigonometry - 1899 - 392 pages
...b = V2 + 2V5 25. A pole is fixed on the top of a mound, and the angles of elevation of the top and bottom of the pole are 60° and 30°, respectively....length of the pole is twice the height of the mound. Let I = length of pole. h — height of mound. a ~ horizontal distance of observer. Then h - a tan... | |
 | Frank Castle - Mathematics - 1899 - 424 pages
...from a point 30 feet directly above the former point ; find the height and distance of the steeple. 2. A pole is fixed on the top of a mound, and the angles of elevation of the bottom and top of the pole are 30° and 60° respectively ; prove that the height of the pole is twice... | |
 | Pitt Durfee - Plane trigonometry - 1900 - 340 pages
...on top of a knoll. From a point at a distance of 200 ft. from the foot of the knoll, the elevations of the top and the bottom of the pole are 60° and 30°, respectively ; prove that the pole is twice as high as the knoll. 8. A regular hexagon is circumscribed about a circle whose radius... | |
 | George Albert Wentworth - Trigonometry - 1901 - 176 pages
...18°. Show that the height of the tower is VÖ 1 ; the tangent of 18° being — -= , — V(10+2V5) 25. A pole is fixed on the top of a mound, and the angles...length of the pole is twice the height of the mound. 26. At a distance (a) from the foot of a tower, the angle of elevation (A) of the top of the tower... | |
 | George Albert Wentworth - Plane trigonometry - 1902 - 286 pages
...Show that the height of the tower is = ; the tangent of 18° being == • V2 + 2 V5 V 10 + 2 Vo 24. A pole is fixed on the top of a mound, and the angles...length of the pole is twice the height of the mound. 25. At a distance a from the foot of a tower, the angle of elevation A of the top of the tower is the... | |
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A pole is fixed on the top of a mound, and the angles of elevation of the top and the bottom of the pole are 60° and 30° respectively. 
