Question
The angle of elevation of the top of a 70 feet tall
tower from the initial position of a person on the ground was 30°. She walked towards the tower in a manner that the foot of the tower, her initial position and the final position were all in the same straight line. If she walked 140(√3/3) feet from her initial position, what was the angle of elevation of the top of the tower from her final position?Solution
tan B = 70/AB tan 30˚ = 70/AB => 1/√3 = 70/AB AB = 70√3 Now, => AD = AB - BD => AD = 70√3 - 140√3/3 => AD = 70√3/3 Now, => tan θ = 70/(70√3/3) = 3/√3 => tan θ = √3/1 = tan 60˚
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