QQuestion
The angles of elevation of the top of a 100 m high tree from two points A and B on opposite sides of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.
ICSEClass 10MathematicsHeights and Distances06 Mark2024 Moderate
✓Answer
Let C be the foot and D the top of the tree.
CD = 100 m
In right-angled △ACD:
tan 52° = CD/AC
⇒ 1.28 = 100/AC
⇒ AC = 100/1.28
⇒ AC = 78.125 m
In right-angled △BCD:
tan 45° = CD/BC
⇒ 1 = 100/BC
⇒ BC = 100 m
Since A and B lie on opposite sides of the tree:
AB = AC + BC
⇒ AB = 78.125 + 100
⇒ AB = 178.125 m
To the nearest metre, AB = 178 m.
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