A PERSON OBSERVED THE ANGLE OF ELEVATION OF THE TOP OF A TOWER AS 30°. HE WALKED 50m TOWARDS THE FOOT OF THE TOWER ALONG LEVEL GROUND AND FOUND THE ANGLE OF ELEVATION OF THE TOP OF THE TOWER AS 60°.FIND THE HEIGHT OF THE TOWER.
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the height of the tower is 25 root 3 metres
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We will assume h be the height of the building.
ΔABC tan 60° = AB/BC
√3 = h/x
x = h/√3
And in the Right Angle ABD,
tan 30° = AB/BD
= 1/√3 = h/ x + 50
= h/√3 + 50 = √3h
h = 25√3m
If there is any confusion please leave a comment below.
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