Krishna
0

Step 1: According to the given information construct an imaginary figure.

              

                      

                  Given height of tower CD = 60 m

                  Let the height of the building, AB = h

                    In right angled triangle BDC,

                  The angle of elevation of the top of the building from the foot of the tower \angle ADB = 30\degree

                  The angle of elevation of the top of the tower from the foot of the building \angle CBD = 60\degree


Step 2:  Find the height of the building use the suitable trigonometric ratios

            From right triangle BCD

                   \tan \theta = \frac{opposite}{adjacent}


                   \tan 60\degree = \frac{CD}{BD}


                   \tan 60\degree = \frac{60}{BD}


                        \sqrt{3} = \frac{60}{BD}


                         BD = \frac{60}{\sqrt{3}}


                         BD = \frac{60}{\sqrt{3}} *\frac{\sqrt{3}}{\sqrt{3}}


                         BD = \frac{60\sqrt{3}}{3}


                          BD = 20\sqrt{3}


          From right triangle ABD


                        \tan \theta = \frac{AB}{BD}


                         \tan 30\degree = \frac{h}{20\sqrt{3}}


                                     \frac{1}{\sqrt{3}} = \frac{h}{20\sqrt{3}}

                                             h = \frac{20\sqrt{3}}{\sqrt{3}}

        

                                                 h = 20 m

    

              Thus height of the building is 20 m.