Question

The horizontal distance between two 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 60 m, find the height of the first tower.

3 points

Your answer

4

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Answer 1

Let AB be the height of second tower and CD be the height of first tower.Given, BD = AE = 140 m& AB = DE = 60 mIn ΔAEC,tan 30°= Perpendicular /Base= CE/AEtan 30 ° = CE/1401/√3= CE /140CE= 140 / √3CE= 140 ×√3/ (√3×√3)[ Rationalising the denominator]CE= 140√3 /3Height of the first tower CD= CE+DE= 140√3/3 + 60=( 140 × 1.73) /3 + 60[ √3=1.73]= 242.2/3 + 60= 80.73 + 60= 140.73 mHeight of the first tower (CD)=140.73 m

Answer 2

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