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A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of the tower is 60 \degree.

2 viewed last edited 7 months ago
Anonymous
0

From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30 \degree . Find the height of the tower and the width of the road.

Krishna
0

Step 1:  Use the given measurements to make a figure.

            

          FROM THE FIGURE: Height of the tower =  AB  

                                              Width of the road = BD = (BC + CD)

                                              Distance between the two points D and C = 10 cm

                                               Angle of elevation at D  = 30\degree

                                               Angle of elevation at C  = 60\degree


Step 2: Find the common height of the two triangles by using the trigonometric ratios definitions.


EXAMPLE: We have to find the adjacent, and opposite sides so

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

                              From \triangle ABD

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


                                                   \frac{1}{\sqrt{3}} = \frac{AB}{BC + CD} ( \because \tan 30\degree = \frac{1}{\sqrt{3}})


                                                   AB = \frac{BC + CD}{\sqrt{3}}..............................(1)

                              From \triangle ABC

                                                 \tan 60\degree = \frac{AB}{BC}

                                                       AB = BC\sqrt{3} ( \because \tan 60\degree = \sqrt{3} ).............(2)


Step 3: Express BC in terms of CD by solving equation (1) & (2)

                Equations:

                         AB = \frac{BC + CD}{\sqrt{3}}

                         AB = BC\sqrt{3}

             Equate the R.H.S

                       \sqrt{3}BC = \frac{BC + CD}{\sqrt{3}}


                           BC = \frac{BC + CD}{(\sqrt{3})^2}


                         BC + CD = 3BC


                             CD = 2BC


                         BC = \frac{CD}{2} ................................(3)

                      

                        BC  = \frac{10}{2} = 5

Step  4: Calculate the width of the road and height of the tower.

                Width of the road =  BD = BC + CD

                                                BD = 5 + 10 = 15 m


                 Height of the tower AB = BC \sqrt{3}  [ \because equation (2)]

                                                AB = 5 \sqrt{3} m