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practical trigonometry

heights and distance
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  heights and distance

A jet fighter at a height of 3000m from the ground
passes directly over another jet fighter . At that instant
the angles of elevation of the two fighters from a
point on the ground are 60 degrees and 45 degrees,
Find the vertical distance between the two jets.

heights and distance
O = observer
A= first jet fighter
B=second jet
AC= 3000 m given

in triangle OAC
tan(60 degrees) = 3000 / OC
sqrt(3) = 3000 / OC
OC =3000/ sqrt(3) ----------(1)

from triangle OBC
tan(45 degrees ) = BC / OC
1 = BC / OC
BC = OC
from eqn(1)
BC =3000/sqrt(3)
BC=3000*[sqrt(3)]/3 introducing conjugate
BC = 1000 sqrt(3) m
BC = 1000*1.732=1732 m-----(2)

using the picture
AB = AC - BC
from (2)
AB = 3000-1732 = 1268 m

distance between the jets is 1268 m.

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problem 2
 
If tanA +sinA = m , tanA - sinA = n
show that (m^2)-(n^2)=4sqrt(mn)
use the identities for (a+b)^2 and (a-b)^2
for the LHS
and (a+b)(a-b) for the RHS

L.H.S =(m)^2-(n)^2
=[tanA+sinA]^2 + [tanA-sinA ]^2

=[{tanA}^2+2tanAsinA+{sinA}^2 ]-
[{tanA}^2-2tanAsinA+{sinA}^2

=4tanAsinA--------------(1)

R.H.S = 4sqrt(mn)
=4sqrt[(tanA+sinA)(tanA-sinA)]
=4sqrt[ (tanA)^2 - (sinA)^2 ]
=4sqrt[{sinA/cosA}^2 - (sinA)^2 ]
=4sqrt[ (sinA)^2  [{1/(cosA)^2}-1 ]  ]
=4sqrt[ (sinA)^2 [(secA)^2 - 1 ] ]
=4sqrt[ (sinA)^2 [ (tanA)^2 ] ]
=4sinAtanA -----------(2)

from (1) and (2)
L.H.S = R.H.S



other questions and problems:


*integral calculus integration formula





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trigonometric identity and ratio of  certain standard angles

*list of differentiation formula     *integral calculus integration formula
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