Question

he ground, the
at the angle of
1. The angles of elevation of the top of a tower from two points at a distance of 4 m and
9 m from the base of the tower and in the same straight line with it are complementary,
Prove that the height of the tower is 6 m.
s 30° and the
P. If the to

Answer 1

Answer:

$\bigstar{\bold{Height\:of\:tower=6\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\rm{Given:}}}}$

• Angle of elevation from two pints 4 m and 9 m away from the tower are complementary

$\Large{\underline{\underline{\rm{To\:Prove:}}}}$

• Height of the tower is 6 m

$\Large{\underline{\underline{\rm{Identities\:used:}}}}$

➙ tan (90 - θ) = cot θ

➙ cot θ = 1/tanθ

$\Large{\underline{\underline{\rm{Proof:}}}}$

➙ Let the height of the tower be AB

➙ Let ∠C be θ

➙ By given,

    ∠C and ∠D are supplementary.

➙ Hence,

    ∠D = 180 - θ

➙ Here BC = 4 m and BD = 9 m

➙ Consider Δ ABC

    tan θ = AB/BC

    tan θ = AB/4---------(1)

➙ Consider Δ ABD

    tan (90 - θ) = AB/BD

    tan (90 - θ) = AB/9

    cot θ = AB/9

     1/tan θ = AB/9

     tan θ = 9/AB------(2)

➙ From equation 1 and 2, the LHS are equal, hence RHS are also equal

    AB/4 = 9/AB

➙ Cross multiplying,

    AB × AB = 9 × 4

    AB² = 36

    AB = √36

    AB = 6

➙ Hence the height of the tower is 6 m

➙ Hence proved.

$\Large{\underline{\underline{\rm{More\:Identities:}}}}$

➙ cosec θ = 1/sinθ

➙ sec θ = 1/cosθ

➙ cos (90 - θ) = sin θ

➙ cosec (90 - θ) = sec θ

Answer 2

Answer :-

★Height of the tower = 6 m

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★ Concept :-

• Here the concept of trigonometric ratios and their relation with acute angles is used. We already know, that Trigonometry can be applied in right triangles only where one angle = 90° and others are acute.

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★Solution :-

Let us use the given information and the diagram to get raw info.

We get,

• Height of the tower = h = AB

• Distance between the foot of tower to the first point of observation = BC = 4 m

• Distance between the foot of the tower to the second point of observation = BD = 9 m

• Distance between the two points of observations = BD - BC = 9 - 4 = 5 m

________________________________

★ To prove : Height of the tower, h = 6 m

________________________________

★ Proof :-

• Let the angle of elevation to top of tower from first point of observation = <C = Ø

Since, its given that <C and <D are complementary. Then,

•Let the angle of elevation to the top of tower from the second point of observation = <D

= (90° - Ø)

» Now from right ∆ ACB, we get,

• Tan Ø = Perpendicular/Base = AB/BC

So,

▶Tan Ø = AB / 4 ...(i)

» Now from right ∆ ADB, we get,

• Tan(90° - Ø) = Perpendicular/Base

= AB/BD

So,

▶ Tan(90° - Ø) = AB / 9

We know that,

✏ cot Ø = tan(90° - Ø)

▶ cot Ø = AB / 9

Also, cot Ø = 1 / tan Ø

So,

▶ Tan Ø = 9 / AB ...(ii)

From equation (i) and equation (ii), we get,

✒ AB / 4 = 9 / AB

✒ AB × AB = 9 × 4

✒ (AB)² = 36

✒ AB = ± √36 = +6 , -6

Here, since the length cannot be negative,we are neglecting the negative value.

So, height of the tower = AB = 6 m

★ Hence proved.

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★More to know :-

• Trigonometry is the branch of mathematics that deals with the calculation of sides and angles of a triangle especially right angled triangle.

• Trigonometric differences for acute angles :-

▶ cos Ø = (90° - sin Ø)

▶ sec Ø = (90° - cosec Ø)

▶ cot Ø = (90° - tan Ø)

• Trigonometric ratios :-

▶ sin Ø = Perpendicular / Hypotenuse

▶cos Ø = Base / Hypotenuse

▶ tan Ø = Perpendicular / Hypotenuse

▶ cosec Ø = 1 / (sin Ø)

▶ sec Ø = 1 / (cos Ø)

▶ cot Ø = 1 / (tan Ø)

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