Answer the following questions, using trigonometry rules.
Class 10th CBSE, Rd Sharma, some applications of trigonometry
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15. The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is 32°. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]
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Solution:
Let the height of the tower = h m
And the distance BC = x m
Then, from the fig.
In ΔABC
tan 63o = AB/BC
1.9626 = h/x
x = h/ 1.9626
x = 0.5095 h …. (i)
Next, in ΔABD
tan 32o = AB/ BD
0.6248 = h/ (100 + x)
h = 0.6248(100 + x)
h = 62.48 + 0.6248x
h = 62.48 + 0.6248(0.5095 h) ….. [using (i)]
h = 62.48 + 0.3183h
0.6817h = 62.48
h = 62.48/0.6817 = 91.65
Using h in (i), we have
x = 0.5095(91.65) = 46.69
Therefore,
the height of the tower is 91.65 m
Distance of the first position from the tower = 100 + x = 146.69 m