Krishna
0

Step 1:  Examine the given problem and draw a imaginary figure

            


                          Height of the tower PQ = 30 m

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

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

                         Height of the building AB = h say


Step 2:  Find the distance between the building and tower

              NOTE: We know opposite side of the right angle triangle and we have to calculate the adjacent side.

                             So, use \tan \theta = \frac{opposite}{adjacent}


                  From the figure: Right triangle  PQB

                                 \tan 60\degree = \frac{PQ}{BQ}


                                 \tan 60\degree = \frac{30}{BQ}


                                 \sqrt{3} = \frac{30}{BQ}                          \because \tan 60\degree = \sqrt{3}


                                 BQ = \frac{30}{\sqrt{3}}......................(1)


Step 3: Find the height of the building

              From the figure: right triangle ABQ

                                  \tan 30\degree = \frac{AB}{BQ}


                                   \frac{1}{\sqrt{3}} = \frac{h}{BQ}                         \because \tan 60\degree = \frac{1}{\sqrt{3}}


                                   h \sqrt{3} = BQ


                                   h \sqrt{3} = \frac{30}{\sqrt{3}}                      \because Equation (1)


                                   h = \frac{30}{\sqrt{3} * \sqrt{3}}


                                   h = \frac{30}{3}


                                  h = 10 m


                    The height of the building = 10 m