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practical trigonometry

heights and distance
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practical trigonometry heights and distance

From the top of a lighthouse of height 200 feet, the
lighthouse keeper observes a yacht and a barge along the
same line of sight. The angles of depression for the yacht
and the barge are 45 degrees abd 30 degrees respectively.
For safety reasons the two sea vessels should be at least
300 feet apart. If they are less than 300 feet apart, the
keeper has to sound the alarm. Does the keeper have to
sound the alarm

heights and
                  distance

L = top of the lighthouse
B = barge
Y = yacht
LH = 200 feet

from triangle LBH
tan(30 degrees) = LH/HB
[1/sqrt(3)]=200/HB
HB = 200*sqrt(3) feet ----------(1)

from triangle LHY
tan(45 degrees) = LH /HY
1 = 200 / HY
HY = 200 feet -------------(2)

from (1) and (2)
BY = HB - HY = 200*sqrt(3) - 200
       =200[sqrt(3) - 1] = 200*[1.732-1]
       = 200*0.732 =146.4 feet

distance between the barge and yacht is 146.4 feet
which is less than 300 feet

therefore the light house keeper has to sound the alarm



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problem 2
 

In triangle ABC right angled at C find
the values of cos(A+B) and sin (A+B)

C = 90 degree

using angle sum property of a triangle
A+B+C=180 degrees
A+B+90 =180
A+B =180 -90
A+B =90 degrees
therefore
cos(A+B) = cos(90 degrees) =0
sin(A+B)=sin(90 degrees) = 1


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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