Question

A pole 10m high stands vertically at the centre of gravity of a horizontal equilateral triangle, each side of which sub tends an angle of 60 degrees at the top of the pole. Find the side of the triangle
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pls tell the correct answer ​

Answer 1

Given :- (From image)

• AD = Pole = 10m
• ∠ABC = ∠BAC = ∠BCA = 60° .
• ∠ADB = 90° .

To Find :-

• The side of the equilateral triangle . ?

Solution :-

in right ∆ADB, we have,

→ sin 60° = (perpendicular) / (Hypotenuse)

→ √3/2 = AD / AB

→ √3 / 2 = 10 / AB

→ AB * √3 = 10 * 2

→ √3AB = 20

→ AB = (20/√3)

rationalize the RHS part by multiplying and dividing (√3) ,

→ AB = (20/√3) * (√3/√3)

→ AB = (20√3) / 3 m (Ans.)

Hence, each side of equilateral triangle is (20√3/3)m .

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Answer 2

This is the answer.

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