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The Horizontal Distance Between Towers is 140 M. the Angle of Elevation of the Top of the First Tower When Seen from the Top of the Second Tower is 30°. If the Height of the Second Tower is - Mathematics

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प्रश्न

The horizontal distance between towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60m, find the height of the first tower. 

योग

उत्तर

Let AB and PQ be the two towers.

BQ = AR = 140 m

AB = RQ = 60 m

Let PR = h

We need to find the  height of the first tower, i.e. , PQ.

In ΔPRA,

`"PR"/"AR" = tan30^circ`

`"h"/140 = 1/sqrt(3)`

`"h" = 140/sqrt(3) = 80.83` 

∴ PQ = PR + RQ = 80.83 m + 60 m = 140.83 m

Thus , the height of the first tower is 140.83 m.

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Heights and Distances - Solving 2-D Problems Involving Angles of Elevation and Depression Using Trigonometric Tables
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 22: Heights and Distances - Exercise

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फ्रैंक Mathematics - Part 2 [English] Class 10 ICSE
अध्याय 22 Heights and Distances
Exercise | Q 39

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