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.