Question

answer this question fasttttttt​

Answer 1

Answer:

(Refer to the attachments for reference)

Consider ΔABC (when the sun's angle is 30°)

⇒ Tan 30° = AB/BC

$\implies \dfrac{1}{\sqrt{3}} = \dfrac{45\sqrt{3}}{2 \times x}\\\\\\\text{Cross multiplying we get:}\\\\\implies 2x = 45 \sqrt{3} \times \sqrt{3}\\\\\implies 2x = 45 \times 3\\\\\implies 2x = 135\\\\\implies x = \dfrac{135}{2} = \bf{67.5\;\:m}$

Consider ΔA'B'C' (when the sun's angle is 60°)

⇒ Tan 60° = A'B'/B'C'

$\implies \sqrt{3} = \dfrac{45\sqrt{3}}{2\times y}\\\\\\\text{Cross multiplying we get:}\\\\\implies \sqrt{3} \times (2y) = 45\sqrt{3}\\\\\implies 2y = 45\\\\\implies y = \dfrac{45}{2} = \bf{22.5\;\; m}$

Hence the value of x - y is:

⇒ 67.5 - 22.5 = 45 m

Hence Option (A) is the correct answer.

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that the height of tower be represented as AB.

Let assume that the shadow of tower is represented by BC when sun altitude is 30° and the shadow of tower is represented by BD when sun altitude is 60°.

Now, Given that

$\rm :\longmapsto\:AB = \dfrac{45 \sqrt{3} }{2} \: m$

$\rm :\longmapsto\:BD = y \: m$

$\rm :\longmapsto\:BC = x \: m$

Now,

In right triangle ABD, we have

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{AB}{BD}$

$\rm :\longmapsto\: \sqrt{3} = \dfrac{45 \sqrt{3} }{2y}$

$\rm :\longmapsto\: 1 = \dfrac{45}{2y}$

$\bf\implies \:y = \dfrac{45}{2}$

Now,

In right triangle ABC, we have

$\rm :\longmapsto\:tan30 \degree \: = \: \dfrac{AB}{BC}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} } = \dfrac{45 \sqrt{3} }{2x}$

$\rm :\longmapsto\:2x = 135$

$\bf\implies \:x = \dfrac{135}{2}$

Therefore,

$\bf\implies \:x - y = \dfrac{135}{2} - \dfrac{45}{2} = \dfrac{90}{2} = 45 \: m$

Hence, Option (A) is correct.

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1