An adult and ayoung boy , standing on the ground are one metre apart The height of adult is - Maths - Some Applications of Trigonometry

Question

An adult and ayoung boy , standing on the ground are one
metre apart The height of adult is 2 times the height of young boy
If at the mid-point of the line joining their feet,the angular
elevation of their tops are complementary,then find the height of
the young boy

Aakash Sharma · Accepted Answer

Let AB be the adult and CD be the young boy.

Then BD = 1m = 100 cm.

Let E be the mid point of line joining there feet.

Then ∠AEB and ∠CED are angles of elevation of their tops

Let ∠AEB = θ

Given: ∠AEB + ∠CED = 90°

⇒ ∠CED = 90° – θ

Let CD = x then AB = 2 CD = 2x

Also $BE = ED = \frac{BD}{2} = \frac{100}{2} cm = 50 cm$

Now in ∆ ABE

$\tan θ = \frac{AB}{BE} = \frac{2 x}{50} = \frac{x}{25} ...... (1)$

In ∆ CDE

$\tan (90 ° - θ) = \frac{CD}{ED} \Rightarrow \cot θ = \frac{x}{50} ...... (2)$

Multiplying (1) and (2), we get

$\tan θ \cot θ = \frac{x}{25} \times \frac{x}{50} \Rightarrow x^{2} = 1250 \Rightarrow x = \sqrt{1250} cm = 25 \sqrt{2} cm$

Hence the height of the young boy is $25 \sqrt{2} cm$

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An adult and ayoung boy , standing on the ground are one metre apart..The height of adult is 2 times the height of young boy.If at the mid-point of the line joining their feet,the angular elevation of their tops are complementary,then find the height of the young boy...