Question

pls solve

ligit ans only​

Answer 1

$\large \sf \underline\purple{ Answer: - }$

• 200 m/s

$\large \sf \underline\purple{ solution: - }$

Let , D and E be the initial and final position of the plain respectively,

Consider the ∆ABC,

$\: \: \: \: \: \: \: \: \: \: \rm \: tan \: 60 \degree = \frac{BD}{AB}$

$\implies \rm{ \sqrt{3} = \frac{1500 \sqrt{3} }{AB} }$

$\implies \boxed{\sf AB = 1500}$

Consider the ∆ACE,

$\rm \: tan \: 30 \degree = \frac{CE}{AC }$

$\implies \rm{ \frac{1}{ \sqrt{3} } = \frac{1500 \sqrt{3} }{AC} }$

$\implies \boxed{\sf AC = 4500}$

Here,

BC = AC - AB

= 4500 - 1500

= 3000 m

i.e. train travels a distance of 3000 m in 15 seconds,

$\sf\therefore \boxed{ \sf \: speed \: of \: plane = \frac{3000}{15} = 200 meter \: per \: second}$

Answer 2

Answer:

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Step-by-step explanation:

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