Question

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Answer 1

Question:

The angle of elevation of the top of the tower from two points at a distance of a metres and b metres from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is √(a•b) .

Note:

• In a right angled triangle, two of its sides are mutually perpendicular (orthogonal).

• is right angled triangle, one of its angle is right angle ,ie; 90° and the other two are acute angles.

• In a right angled triangle, the side lying opposite to the right angle is considered to be hypotenuse.

• Hypotenuse is the longest side of a right angled triangle.

• In a right angled triangle, the side lying opposite to an angle (any acute angle of that right angled triangle) is considered to be perpendicular and the another orthogonal side is considered to be perpendicular for that angle.

• tan@ = perpendicular/base

• cot@ = base/perpendicular

• cot@ = 1/tan@

• tan@•cot@ = 1

• tan(90°-@) = cot@

• cot(90°-@) = tan@

• Two angles are said to be complementary if their sum is 90°.

• If we considered and angle @, then its complementary angle will be given by (90°-@).

Proof:

First of all , let's plot a rough diagram to describe the given situation.

Let OC be the tower (with O as its foot) of height h metres.

Let the point A is a metres away from the foot of the tower O and let the angle of elevation be @ from the point A.

Let the point B is b metres away from the foot of the tower O and let the angle of elevation be (90°-@) from the point B.

Now,

In ∆OAC,

=> tan@ = OC/OA

=> tan@ = h/a ------------(1)

Also,

In ∆OBC,

=> tan(90°-@) = OC/OB

=> tan(90°-@) = h/b

=> cot@ = h/b ---------------(2)

Now,

Multiplying eq-(1) and (2) , we get;

=> tan@•cot@ = (h/a)•(h/b)

=> 1 = (h^2)/a•b

=> a•b = h^2

=> h = √(a•b)

Hence, proved.

Answer 2

☆Refer to the attachment☆