Mathematical Exercises ...: Examples in Pure Mathematics, Statics, Dynamics, and Hydrostatics. With Tables ... and ReferencesLongmans, Green & Company, 1877 - 413 sider |
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Resultat 1-5 av 84
Side 9
... Triangle ABC a sin B a sin c Sin A = = = sin ( B + C ) = b с bc 2 ( 1—0 ) } * ; - { s ( s — a ) ( 8 — b ) - s = a + b + c . 2 ✓ c2 - a2 sin2 C b2 + c2 - a2 Cos A = ± = с 2bc c sin2 B cos Bb2 - c2 sin2 B b S- Sin A = b ) ( s - c ) ; cos ...
... Triangle ABC a sin B a sin c Sin A = = = sin ( B + C ) = b с bc 2 ( 1—0 ) } * ; - { s ( s — a ) ( 8 — b ) - s = a + b + c . 2 ✓ c2 - a2 sin2 C b2 + c2 - a2 Cos A = ± = с 2bc c sin2 B cos Bb2 - c2 sin2 B b S- Sin A = b ) ( s - c ) ; cos ...
Side 10
... triangle are the supple- ments respectively of the angles and sides of the primitive triangle . A + B + C > π < 3 π . In any spherical triangle ; Cos a = cos b cos c + sin b sin c cos A. Sin A sin B sin c :: sin a : sin b : sin c ...
... triangle are the supple- ments respectively of the angles and sides of the primitive triangle . A + B + C > π < 3 π . In any spherical triangle ; Cos a = cos b cos c + sin b sin c cos A. Sin A sin B sin c :: sin a : sin b : sin c ...
Side 11
... triangle A + B + C - T spherical excess 4 right angles = · 2πr2 = 2π × area of hemisphere . FORMULE IN MENSURATION . Area of rectangle = Area of parallelogram ab ; a and b being adjacent sides . ab sin 0 , a and b being adjacent angle 0 ...
... triangle A + B + C - T spherical excess 4 right angles = · 2πr2 = 2π × area of hemisphere . FORMULE IN MENSURATION . Area of rectangle = Area of parallelogram ab ; a and b being adjacent sides . ab sin 0 , a and b being adjacent angle 0 ...
Side 61
... triangle is 6 acres 2 roods and 8 perches , and a perpendicular from one angle on the base measures 524 links ; find the length of the base in chains . Given 1 chain = 100 links = 22 yards . 13. Find the area in acres of a triangle ...
... triangle is 6 acres 2 roods and 8 perches , and a perpendicular from one angle on the base measures 524 links ; find the length of the base in chains . Given 1 chain = 100 links = 22 yards . 13. Find the area in acres of a triangle ...
Side 67
... triangle . 11. A pyramid has a regular hexagon for its base , each side being 20 feet ; find the cubical content of ... triangles and its area found . XLIII . 1. Prove that an + b " is LINE PAPERS . 67.
... triangle . 11. A pyramid has a regular hexagon for its base , each side being 20 feet ; find the cubical content of ... triangles and its area found . XLIII . 1. Prove that an + b " is LINE PAPERS . 67.
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Arithmetic axis ball base bisected body cent centre of gravity circle coefficient of friction compound interest cone cost crown 8vo cube cubic foot curve Define determine diameter Divide dwts ellipse English equal equilibrium expression feet Find the area Find the centre Find the distance Find the equation Find the number Find the sum Find the value fluid forces acting fraction geometrical Grammar horizontal plane hyperbola inches inclined plane inscribed Integrate isosceles latus rectum least common multiple length logarithms miles Multiply parabola parallel particle perpendicular pressure Prove pulleys radius ratio rectangle rectangular Reduce right angles sides simple interest sin² sine spherical triangle square root straight line string subtended Subtract surface tangent theorem tons tower triangle ABC velocity vertical vulgar fraction weight yards
Populære avsnitt
Side 123 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Side 10 - A'B'C', and applying the law of cosines, we have cos a' = cos b' cos c' + sin b' sin c' cos A'. Remembering the relations a' = 180° -A, b' = 180° - B, etc. (this expression becomes cos A = — cos B cos C + sin B sin C cos a.
Side 184 - If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them, is equal to the rectangle contained by the segments of the other.
Side 78 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Side 184 - To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third (20.
Side 184 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.
Side 163 - 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.
Side 184 - In right angled triangles the square on the side subtending the right angle is equal to the (sum of the) squares on the sides containing the right angle.
Side 154 - If two straight lines be cut by parallel planes, they shall be cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D : As AE is to EB, so is CF to FD.