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


$\red \sf \: explain \: diagram \: clearly \: ..$
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Answer 1

Answer :-

Speed of the aeroplane is 240√3 km/h .

From the diagram :-

Let BX be the horizontal ground and let "B" be the point of observation. Let "A" and "E" be the two positions of the aeroplane.

Let AC ⊥ BX and ED ⊥ BX, then we have :-

∠CBA = 60° ; ∠DBE = 30° ; AC = ED = 1 km

Solution :-

From right Δ BCA, we have :-

⇒ cot 60° = BC/AC

⇒ 1/√3 = BC/1

⇒ BC = 1/√3 km

Now, from right ∆ BDE :-

⇒ tan 30° = ED/BD

⇒ 1/√3 = 1/BD

⇒ BD = √3 km

________________________________

⇒ CD = BD - BC

⇒ CD = √3 - 1/√3

⇒ CD = (3 - 1)/√3

⇒ CD = 2/√3

⇒ CD = 2/√3 × √3/√3

⇒ CD = (2√3)/3 km

So, the aeroplane covers (2√3)/3 km in 10 seconds.

(10 sec = 1/360 hr) . Thus, speed of the aeroplane :-

Speed = Distance/Time

⇒ Speed = (2√3)/3 ÷ (1/360)

⇒ Speed = (2√3)/3 × 360

⇒ Speed = 2√3 × 120

⇒ Speed = 240√3 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,

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

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

Now,

We will find the distance between EC and ED i.e. the measure of CD.

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

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.,

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

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

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

# Happy Learning .