Question

Topic :- Applications of Trigonometry An aeroplane is flying horizontally 1km above the ground is observed at an elevation of 60°. If after 10 seconds the elevation is ovserved at 30°. The uniform speed per hour of aeroplane is​

Answer 1

Answer:

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Kkrh??

Step-by-step explanation:

Let CE = x, BC = y, ∠AEB = 30°, ∠DEC = 60°, AB = 1 km and CD = 1 km. Here we have to find the speed of an aeroplane. So we use trigonometric ratios. Hence the speed of aeroplace is 415.68 km/h

Answer 2

✰ Concept :-

In this question, we will use the concept of "trigonometry" to get the required answer. Let's starts :D

✰ The Diagram :-

Let AC be the height of the aeroplane from the ground i.e. 1 km also, let BD be the height of the aeroplane from the group after 10 seconds.

Let E be the observer. So, ∠ AEC = 60° and ∠ BED = 30°.

✰ Solution :-

We know that,

tan60 = √3

Therefore,

$: \implies \sf tan AEC = \frac{AC}{EC} \\ : \implies \sf \: tan30 = \frac{1}{EC} \: \: \: \\ : \implies \sf \: \sqrt{3} = \frac{1}{EC} \: \: \: \: \: \: \: \: \\ : \implies \sf EC = \frac{1}{ \sqrt{3} } \: \: \: \: \: \: \: \: \:$

$: \implies \sf \: tan30 = \frac{BD}{ED} \\ : \implies \sf \: \frac{1}{ \sqrt{3} } = \frac{1}{ED } \: \: \: \\ : \implies \sf \: ED = \sqrt{3} \: \: \: \: \: \:$

Now, we will find the distance between EC and ED i.e. the measure of CD.

$: \implies \sf \: CD = ED - EC \: \: \: \: \: \: \: \: \: \\ : \implies \sf \: CD = \sqrt{3} - \frac{1}{ \sqrt{3} } \: \: \: \: \: \: \: \: \\ : \implies \sf \: CD = \frac{ \sqrt{3} \times \sqrt{3} - 1 }{ \sqrt{3} } \\ : \implies \sf \: CD = \frac{3 - 1}{ \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \sf \: CD = \frac{2}{ \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now, we have already been provided with the time i.e. 10 seconds and we have also found the distance. So we can easily calculate the speed using the formula for speed i.e.,

$\bigstar \: \underline{ \underline{ \boxed{ \sf \: speed = \frac{distance}{time} }}}$

$: \implies \sf \: speed = \frac{ \frac{2}{ \sqrt{3} } }{ \frac{1}{360} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ : \implies \sf \: speed = 240 \sqrt{3} \: km \: hr ^{ - 1}$

$\boxed{ \sf \purple{\therefore \: speed \: of \: the \: aeroplane \: is \: 240√3 km/hr}}$