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The angle of elevation of the top of a tower at a point on the line through the foot of the tower is `45^@`. After walking a distance towards the foot of the tower along the same horizontal line elevation of the top of the tower changes to `60^@`. Find the height of tower. |
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Answer» With the given details, we can create a diagram. Please refer to video to see the diagram. From the digram, In `Delta OBC`, `h/x = tan60^@ = sqrt3` `=>x = h/sqrt3->(1)` In `Delta OAC`, `h/(80+x) = tan45^@ = 1` `=> h = 80+x` From (1), `=>h = 80+h/sqrt3` `=>sqrt3h-h = 80sqrt3` `=>h = (80sqrt3)/(sqrt3-1)**(sqrt3+1)/(sqrt3+1)` `=>h = 40sqrt3(sqr3+1)m` So,the height of the tower is `40sqrt3(sqrt3+1)m`. |
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