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Question

An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45. What is the height of the tower? [2 MARKS]


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Solution

Let AB be the tower of height h and CD be the observer of height 1.5 m at a distance of 28.5 m from the tower AB. [12 MARK]



In ΔAED, we have

tan45=h28.5 [1 MARK]

1=h28.5

h=28.5 m

h=AE+BE=AE+DC

=(28.5+1.5)m=30 m

Height of tower = h + 1.5

= 28.5 + 1.5

= 30 m [12 MARK]

Hence, the height of the tower is 30 m.


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