Mathematics

The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.

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From figure,

⇒ tan 52° = $\dfrac{CD}{AC}$

⇒ 1.28 = $\dfrac{100}{AC}$

⇒ AC = $\dfrac{100}{1.28}$ = 78.125 m

⇒ tan 45° = $\dfrac{CD}{BC}$

⇒ 1 = $\dfrac{100}{BC}$

⇒ BC = 100 m

⇒ AB = AC + BC = 78.125 + 100 = 178.125 m

Hence, AB = 178.125 m

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