Elementary TrigonometryMacmillan and Company, 1893 - 404 páginas |
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Página 6
... draw BC perpendicular to AQ . Thus a right - angled triangle BAC is formed . With reference to the angle 4 the following definitions are em- ployed . BC opposite side hypotenuse The ratio or AB AC The ratio or AB BC The ratio or AC AC ...
... draw BC perpendicular to AQ . Thus a right - angled triangle BAC is formed . With reference to the angle 4 the following definitions are em- ployed . BC opposite side hypotenuse The ratio or AB AC The ratio or AB BC The ratio or AC AC ...
Página 9
... draw BC and DE perpendicular to OA . Also take any point Fin OP and draw FG at right angles to OP . From the triangle BOC , sin POA = BC OB ' DE from the triangle DOE , sin POA : = OD FG from the triangle FOG , sin POA = OG But the ...
... draw BC and DE perpendicular to OA . Also take any point Fin OP and draw FG at right angles to OP . From the triangle BOC , sin POA = BC OB ' DE from the triangle DOE , sin POA : = OD FG from the triangle FOG , sin POA = OG But the ...
Página 10
... drawn perpendicular to BC and meets CA produced in D : if AB = 12 , AC = 16 , BC = 20 , find BD and CD . From the right - angled tri- angle CBD , BD = tan C ; BC from the right - angled triangle ABC , AB AC = tan C ; .. BD BC = AB AC 20 ...
... drawn perpendicular to BC and meets CA produced in D : if AB = 12 , AC = 16 , BC = 20 , find BD and CD . From the right - angled tri- angle CBD , BD = tan C ; BC from the right - angled triangle ABC , AB AC = tan C ; .. BD BC = AB AC 20 ...
Página 14
... draw BC perpendicular to a b AC , and denote the sides of the right - angled triangle ABC by a , b , c . By definition , and BC α sin A = AB с AC b cos A = AB с a2 ... sin2 A + cos2 A - c2 + | b2 a2 + b2 c2 c2 COR . sin2A = 1 - cos2 A ...
... draw BC perpendicular to a b AC , and denote the sides of the right - angled triangle ABC by a , b , c . By definition , and BC α sin A = AB с AC b cos A = AB с a2 ... sin2 A + cos2 A - c2 + | b2 a2 + b2 c2 c2 COR . sin2A = 1 - cos2 A ...
Página 38
... Draw CD perpendicular to AB produced , and let CD = x . Then CBD = 180 ° - 135 ° = 45 ° ; .. BD = CD = x . Now in ... drawn perpendicular to BC ; solve the triangle , having given AD = 5 , LABD = 60 ° , LACD = 45 ° . In the right ...
... Draw CD perpendicular to AB produced , and let CD = x . Then CBD = 180 ° - 135 ° = 45 ° ; .. BD = CD = x . Now in ... drawn perpendicular to BC ; solve the triangle , having given AD = 5 , LABD = 60 ° , LACD = 45 ° . In the right ...
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Términos y frases comunes
1+cos 1+tan² a+cos A+tan acute angle angle of elevation B+cos centre circle College cos² cos³ cosec cosine cot² cyclic quadrilateral decimal denote diff equal equation ex-central triangle Example expression Fcap feet Find the angle find the distance find the height Find the number Find the value flagstaff following identities formula given log greatest angle Hence horizontal plane hypotenuse inscribed LAOB loga logarithm magnitude miles negative number of radians observer pedal triangle perpendicular polygon positive Prof Prove the following quadrant quadrilateral radian measure radius vector regular polygon right angle right-angled triangle sec² sexagesimal shew sides sin sin sin sin² sin³ sine solution solve the triangle subtends an angle supplementary angles tan² tangent tower triangle ABC trigono trigonometrical functions trigonometrical ratios π π
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