Page 43 - CITS - WCS - Mechanical
P. 43

WORKSHOP CALCULATION - CITS




           Solution:
           Given,
           cos A = ⁴/5
           As we know, from trigonometry identities,
           1 + tan A = sec A
                 2
                        2
           sec A – 1 = tan A
                        2
              2
           (1/cos A) -1 = tan A
                2
                           2
           Putting the value of cos A = ⁴/5.
           (5/4)2 – 1 = tan  A
                         2
           (25 – 16)/16 = tan A
                           2
           tan A =    9/16
              2
           tan A = 3/4.
           2. If x sin θ + y cos θ = sin θ cos θ and x sin θ = y cos θ , then prove that x  + y  = 1.
                            3
                   3
                                                                             2
                                                                                 2
                                            3
                                    3
                   3
                xsin
                    θ

             Solution:  ycos 3 θ   sinθ   xsincosθ   θ    ycos  θ    sinθ   cosθ
                                   Solution
                                    3
                           3 3
                                            3
                   3 3

                                             θ
                                   cosθycosθ
                            θ   xsin  sinθθ
                       ycos
               xsin  xsin xsin 3 θ    ycos 3 θ    sinθ   cosθ opp  sinθ   cosθ
                    θ   ycosθ
                                     cosθ
                               sinθ
                              3
                                   tanθ
                                            2
                               θ
                                     3
                  xsin
                          ycos
                   3

                                         sinθ
                                   cosθ cos-(1 θ  θs
                       ycos

                                             θ)


               yco  xsins θ   -(1 θ   3 θ   cos 32 θ   ycos  ycoθ)   sinθ sinθ   cosθ adj cosθ ycos 3 θ   sinθ   cosθ
                                   3
                                    3

                  ycos

                           2 3
                                               cosθ θycos
               ycosθ  ycos    θ    (1yco   -   θ    (1 - cos 2 θ)    ycos 3 θ     3 sinθ 600  3     sinθ   cosθ
                        cos θs
                                     θ   ycos  sinθ
                            θ)   ycosθ ycos
                                        sinθ cosθ
                                            cosθ
                                             θ
                                ycosθ
                                         o
                                   tan37
                                   3

                                              cosθ
                         ycosy

               sinθy   ycosθ    ycos 3 θ   3 θ   sinθ   ycos 3 θ    sinθ x    y
                  ycosθ
                                    θ

                                           cosθ
                                       sinθ
                               ycos
                                             600
                                   0.7536
             y   cosθx   y   sinθ  x   cosθ  x    y
                sinθ
                    2
                         2
                                     2 2
                                         2
                                            cos y)
             LHS
                                   x   θn
             x   x   cosθ y    co LHS  siθs 2     (0.7536)(x  2 θ   600  sin 2 θ   1  RHS
                                        y   RHS1
                cosθ x
                                             600
                           2
                                 2
                       2
                                        2
                                     sin
                                     2
                LHS
                                               RHS
                              cos
                              2
                      x
                          y
                    2
                                            1
                                         θ
                                  θ
                        2
                                         o
                                            RHS
                           cos
                                      θ
                       y
                   x
                               θ
                                  sin 53 tan
             LHS
                                         1
           3.  An aeroplane is flying parallel to the Earth’s surface at a speed of 175 m/sec and at a height of 600 m. The
                                              x
              angle of elevation of the aeroplane from a point on the Earth’s surface is 37° at a given point. After what
                                         600
                                   x
              period of time does the angle of elevation increase to 53°? (tan 53° = 1.3270, tan 37° = 0. 7536).
                                       1.3270
                                   x    452.15
            Solution:              0.7536x    7536y    600
             Solution
                    opp            0.7536    452.15    0.7536y    600
             tanθ
                    adj
                                   0.75.36y    600    340.74
             tan37 o     600       0.7356y    259.26
                      x    y
                                       256.26
                       600         y
             0.7536                    0.7356
                      x    y
                                   y    344.03
             (0.7536)(x    y)    600
                                   Distance    344.03
             tan   53 o     600    Time    speed    344.03
                        x
                  600                      344.03    344.03
             x                     Time                       1.97sec
                1.3270                     speed      175
             x    452.15           Time    1.97sec
             0.7536x    7536y    600
             0.7536    452.15    0.7536y    600
             0.75.36y    600    340.74
             0.7356y    259.26
                 256.26
             y
                 0.7356
             y   344.03
             Distance   344.03                             30
             Time    speed    344.03       CITS : WCS - Mechanical - Exercise 6
                    344.03    344.03
             Time                       1.97sec
                     speed      175
             Time    1.97sec
   38   39   40   41   42   43   44   45   46   47   48