Solutions to the mathematical examination papers set for admission to the Royal military academy, Woolwich, and for the Royal military college [&c.] by D. Tierney and H. Sharratt1877 |
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... ) Mathematics ( 2 ) 0 33 39 45 , 54 . 55 ས༵ * ཉྙཙ 52 ROYAL CIVIL ENGINEERING COLLEGE , COOPER'S HILL . Arithmetic and Mensuration . Euclid Algebra 66 68 72 8825 Statics and Dynamics Plane Trigonometry Pure Mathematics ( 1 )
... ) Mathematics ( 2 ) 0 33 39 45 , 54 . 55 ས༵ * ཉྙཙ 52 ROYAL CIVIL ENGINEERING COLLEGE , COOPER'S HILL . Arithmetic and Mensuration . Euclid Algebra 66 68 72 8825 Statics and Dynamics Plane Trigonometry Pure Mathematics ( 1 )
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D. Tierney. Statics and Dynamics Plane Trigonometry Pure Mathematics ( 1 ) Pure Mathematics ( 2 ) Mixed Mathematics PAGE 79 85 91 95 100 ROYAL MILITARY ACADEMY , WOOLWICH , Papers in Competitive Examination viii CONTENTS .
D. Tierney. Statics and Dynamics Plane Trigonometry Pure Mathematics ( 1 ) Pure Mathematics ( 2 ) Mixed Mathematics PAGE 79 85 91 95 100 ROYAL MILITARY ACADEMY , WOOLWICH , Papers in Competitive Examination viii CONTENTS .
Page 15
... Define the principal trigonometrical ratios , and trace the changes in sign and magnitude of sin 30 as cos 20 ' e varies from 0 to π . Tod . Trig . , Art . 26 . sin 30 = When 0-0 the value of = 0 TRIGONOMETRY . 15 Trigonometry.
... Define the principal trigonometrical ratios , and trace the changes in sign and magnitude of sin 30 as cos 20 ' e varies from 0 to π . Tod . Trig . , Art . 26 . sin 30 = When 0-0 the value of = 0 TRIGONOMETRY . 15 Trigonometry.
Page 16
... Trig . , Art . 91 . 5. Prove that for all values of A , ( 1 ) 2 cosec 44 + 2 cot 44 = cot A - tan A. cos A + cos 3A ( 2 ) cos 34 + cos 5A - ( 1 ) 2 cosec 14 + 2 cot 4A - 2 × 2 cos2 2A 2 sin 2A cos 2A - 1 2 cos 24 - sec 24 sin 4A · 1 + ...
... Trig . , Art . 91 . 5. Prove that for all values of A , ( 1 ) 2 cosec 44 + 2 cot 44 = cot A - tan A. cos A + cos 3A ( 2 ) cos 34 + cos 5A - ( 1 ) 2 cosec 14 + 2 cot 4A - 2 × 2 cos2 2A 2 sin 2A cos 2A - 1 2 cos 24 - sec 24 sin 4A · 1 + ...
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... Trig . , Art . 107 . and cos B = tan A , prove that cos2A = 2 sin 18 ° . We have cos Atan B = sec2B - 1 = 1 1 ; tan ... Trig . , Art . 215 . D Second part . Let ABC be the triangle , D TRIGONOMETRY . 17.
... Trig . , Art . 107 . and cos B = tan A , prove that cos2A = 2 sin 18 ° . We have cos Atan B = sec2B - 1 = 1 1 ; tan ... Trig . , Art . 215 . D Second part . Let ABC be the triangle , D TRIGONOMETRY . 17.
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Popular passages
Page 55 - If two triangles have two sides of the one equal to two sides of the...
Page 71 - If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Page 11 - ... shall be equal to three given straight lines, but any two whatever of these must be greater than the third.
Page 12 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Page 13 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Page 15 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Page 13 - BAC is cut off from the given circle ABC containing an angle equal to the given angle D : Which was to be done. PROP. XXXV. THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.
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Page 13 - PROP. X. THEOR. IF a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.
Page 70 - And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore the whole angle ABD is equal to the whole angle ACD • (ax.