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--------------------------------------------------------------------- practical trigonometry heights and distanceA person in an helicopter flying at a height of 500 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30 degrees and 45 degrees. Find the width of the river. H = helicopter A and C the position of the objects on the opposite banks of the river draw HB perpendicular to the AC from triangle AHB tan(30 degrees)=HB/AB 1/[sqrt(3)]=500/AB AB = 500*sqrt(3) m ---------(1) from triangle HBC tan(45 degrees) = HB/BC 1=500/BC BC = 500 m --------(2) from (1) and (2) AC = AB + BC = [500sqrt(3)]+500 = 500{sqrt(3) + 1} m width of the river = 500{1+sqrt(3)}metres
--------------problem 2 An iron right circular cone of diameter 8cm and height 12 cm is melted and recast into spherical lead shots each of radius 4mm. How many lead shots can be made. Cone: R=8/2=4cm H=12cm lead shot(sphere) r = 4mm = ( 4/10 )cm= 0.4cm n = number of lead shots n*volume of one lead shot = volume of cone n*[(4/3)*pi*(r^3)]=(1/3)*pi*[R^2]*H n=[(1/3)*pi*(R^2)*H] / [(4/3)*pi*(r^3)] n=[(R^2)*H] / [4*(r^3)] n = [4*4*12]/[4*0.4*0.4*0.4] n=[3]/[0.004] n=[3*1000]/4 n=750 lead shots other questions and problems: * integral
calculus integration formula---------------------------------------------------------------------------------------------- trigonometric
identity and ratio of certain standard angles* list
of differentiation formula
*integral
calculus integration formula--------------------------------------------------------------------------------------- |

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