Question

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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?



Sure dear :)
@kumarkapil143825​

Answer 1

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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)

$\binom{ \sqrt[10]{3} }{l} \times \binom{ \sqrt[10]{3} }{l + 20} = 1$

⇒ 10 (10 + 20) = l (l + 20)

⇒ l = 10 m.

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

Thank you!

@itzshivani

Answer 2

Answer:

See the diagram.

Angles of depression are x and 90-x.

Tan x = 10√3 / PR    =>  PR = 10√3 /Tan x

Tan (90-x) = Cot x = 10√3 /QR    =>  QR = 10√3 tan x

PR - QR = 20 m = 10√3 (1/tanx  - tan x)

=> Tan² x + (2/√3) tan x - 1 = 0

=>  Tan x = 1/√3   by using the solution of quadratic equation.

=>  x = 30°    and hence,  90 - x = 60°

Now QR = 10√3 Tan x = 10 m

=> PR = 20 +10 = 30 meters