practical trigonometry

heights and distance



practical trigonometry heights and distance

A 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.
heights distance
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
AB = 500*sqrt(3) m ---------(1)

from triangle HBC
tan(45 degrees) = HB/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.
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=[(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=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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