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--------------------------------------------------------------------- practical trigonometry heights and distanceFrom 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 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 LHYtan(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 --------------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---------------------------------------------------------------------------------------------- trigonometric
identity and ratio of certain standard angles* list
of differentiation formula
*integral
calculus integration formula--------------------------------------------------------------------------------------- |

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