Elements of Geometry: With Practical Applications to Mensuration |
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Page 311
... solve a geometrical prob- lem by aid of Algebra , draw a figure which shall represent the several parts or conditions of the problem , both known and required . Represent the known parts by the first letters of the alphabet , and the ...
... solve a geometrical prob- lem by aid of Algebra , draw a figure which shall represent the several parts or conditions of the problem , both known and required . Represent the known parts by the first letters of the alphabet , and the ...
Page 41
... solve a plane triangle . The solution of plane triangles depends upon the following FUNDAMENTAL PROPOSITIONS . 109. In a right - angled triangle , the side opposite to an acute angle is equal to the product of the hypothenuse into the ...
... solve a plane triangle . The solution of plane triangles depends upon the following FUNDAMENTAL PROPOSITIONS . 109. In a right - angled triangle , the side opposite to an acute angle is equal to the product of the hypothenuse into the ...
Page 47
... solve the triangle , two elements other than the right angle must be given , one of them being a side . Hence there will be four cases in which there may be given , respectively , I. The hypothenuse and an acute angle . II . A side ...
... solve the triangle , two elements other than the right angle must be given , one of them being a side . Hence there will be four cases in which there may be given , respectively , I. The hypothenuse and an acute angle . II . A side ...
Page 48
... solve the triangle . = Solution . The angle at the perpendicular 90 ° - 59 ° 37′42 ′′ = 30 ° 22′18 ′′ . Let , now , h = 1785.395 feet and A = 59 ° 37′42 ′′ , and we have , by ( 112 ) and ( 113 ) , h1785.395 log 3.251734 A = 59 ° 37 ′ 42 ...
... solve the triangle . = Solution . The angle at the perpendicular 90 ° - 59 ° 37′42 ′′ = 30 ° 22′18 ′′ . Let , now , h = 1785.395 feet and A = 59 ° 37′42 ′′ , and we have , by ( 112 ) and ( 113 ) , h1785.395 log 3.251734 A = 59 ° 37 ′ 42 ...
Page 49
... solve the triangle . To find B. The angle B is the complement of A ; hence , B = 90 ° A. To find p . By ( 87 ) and ( 88 ) , we have or , by logarithms , pb tan A = b cot B ; log p = log blog tan A = log blog cot B. ( 114 ) To find h ...
... solve the triangle . To find B. The angle B is the complement of A ; hence , B = 90 ° A. To find p . By ( 87 ) and ( 88 ) , we have or , by logarithms , pb tan A = b cot B ; log p = log blog tan A = log blog cot B. ( 114 ) To find h ...
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Common terms and phrases
A B C ABCD adjacent angles altitude angle ACB angle equal arc A B base bisect chord circle circumference circumscribed cone convex surface cosec Cosine Cotang cylinder diagonal diameter distance divided drawn equal Prop equilateral triangle equivalent exterior angle feet formed frustum gles greater half the sum hence homologous hypothenuse inches included angle inscribed isosceles less Let ABC line A B logarithmic sine measured by half multiplied number of sides parallel parallelogram parallelopipedon pendicular perimeter perpendicular polyedron prism PROBLEM PROPOSITION pyramid quadrantal radii radius ratio rectangle regular polygon right angles right-angled triangle rods Scholium secant segment side A B similar slant height solve the triangle sphere spherical polygon spherical triangle Tang tangent THEOREM triangle ABC triangle equal trigonometric functions vertex
Popular passages
Page 59 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Page 37 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Page 120 - At a point in a given straight line to make an angle equal to a given angle.
Page 52 - If any number of quantities are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let a : b = c : d = e :f Now ab = ab (1) and by Theorem I.
Page 19 - In an isosceles triangle, the angles opposite the equal sides are equal.
Page 199 - Any two rectangular parallelopipedons are to each other as the products of their bases by their altitudes ; that is to say, as the products of their three dimensions.
Page 121 - Through a given point to draw a straight line parallel to a given straight line, Let A be the given point, and BC the given straight line : it is required to draw through the point A a straight line parallel to BC.
Page 103 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Page 2 - The logarithm of any POWER of a number is equal to the product of the logarithm of the number by the exponent of the power. For let m be any number, and take the equation (Art.
Page 2 - The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power.