Question

A man, 1.7 meters tall, standing at the foot of the tower sees the top of the building 50 meters away at an elevation of 60 degrees. On climbing the top of the tower he sees it at an elevation of 50 degrees. Draw a rough figure according to the given data. Compute the height of the tower and the building. ($\sin 50^0$ = 0.77, $\cos 50^0$ = 0.64, $\tan 50^0$ = 1.19)

Answer 1

Hint - In order to solve this problem, draw a diagram and analyze it. And use the values of $\sin 50$ = 0.77, $ \cos 50$ = 0.64,$\tan 50$ = 1.19. Using this will solve your problem and give you the right answer.

Complete Step-by-Step solution:
The diagram for this equation can be drawn as:

DF is the height of man it is 1.7m
CF is the height of the tower and AG is the height of the building.
On considering the diagram we can say CF = BG and DF = EG = 1.7m and DE = FG = CB = 50m(Given).
In triangle ADE,
$
  {\text{tan 60 = }}\dfrac{{{\text{AE}}}}{{{\text{DE}}}}{\text{ = }}\dfrac{{{\text{AE}}}}{{{\text{50}}}} \\
  {\text{AE = 50 }\times }\sqrt {\text{3}} \\
  {\text{AE = 50 }\times 1}{\text{.73}} \\
  {\text{AE = 86}}{\text{.5m}} \\
$
So the height of the building will be AG = AE + EG = 86.5 + 1.7 = 88.2 m
Then in triangle ABC we get,
It is given that tan50 = 1.19
\[
  {\text{tan50 = }}\dfrac{{{\text{AB}}}}{{{\text{BC}}}}{\text{ = }}\dfrac{{{\text{AB}}}}{{{\text{50}}}} \\
  {\text{AB = 50} \times 1}{\text{.19 = 59}}{\text{.5m}} \\
\]
We know from above data height of the tower BE = CD = AE – AB = 86.5 – 59.5 = 27m
Hence the height of the building and tower is 88.2m and 27m.

Note – Whenever you struck with this types of problems where you have to find the lengths of sides of triangle where angles and any of the lengths is given then use the concept of trigonometry in triangles and solve using the values of angle use of tan will solve your most of the problems and drawing diagram will also solve your problem.