Question

${\rm{\blue{\underline{\underline{\huge{Question}}}}}} \huge{\pink{࿐}}$

p and q are two point observed from the top of a building 10√3 m hight . if the angle of depression of the point are complementary and pq = 20m , then the distance of p from the building is




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Answer 1

Let OR = 10√3 m.

The angles of depression of the points are complementary and PQ = 20 m.

⇒ ∠OQR + ∠OPR = 90°

Let ∠OQR = x and ∠OPR = y

⇒ x + y = 90°

And let RQ = l m.

⇒ tan x = 10√3/l …….. eq(1)

⇒ tan y = 10√3/(l + 20)

⇒ tan (90 – x) = 10√3/(l + 20)

⇒ cot x = 10√3/(l + 20)

We know that tan θ . cot θ = 1.

⇒ 100 × 3 = l (l + 20)

⇒ 10 (10 + 20) = l (l + 20)

⇒ l = 10 m.

∴ The distance of P from the building = 10 + 20 = 30 m.

SIS please give me multiple thanks on my all ANSWER please sis

Answer 2

Answer:

Let OR = 10√3 m.

The angles of depression of the points are complementary and PQ = 20 m.

⇒ ∠OQR + ∠OPR = 90°

Let ∠OQR = x and ∠OPR = y

⇒ x + y = 90°

And let RQ = l m.

⇒ tan x = 10√3/l …….. eq(1)

⇒ tan y = 10√3/(l + 20)

⇒ tan (90 – x) = 10√3/(l + 20)

⇒ cot x = 10√3/(l + 20)

We know that tan θ . cot θ = 1.

⇒ 100 × 3 = l (l + 20)

⇒ 10 (10 + 20) = l (l + 20)

⇒ l = 10 m.

∴ The distance of P from the building = 10 + 20 = 30 m.

SIS please give me multiple thanks on my all ANSWER please sis