An aeroplane flying at a constant speed- parallel to the horizontal ground- -3 km above it- is observed at an elevation of 60- from a point on the ground- If- after five seconds- its elevation from the same point- is 30- then the speed in km-hr of the aeroplane- is -

Let the speed of the plane be s km/hr, We know that, tan 30^∘ = BECE ⇒ CE = 3 km tan 60^∘ = ADCD = nbsp; BECD ⇒ CD = 1 km Now, DE = CE CD = 2 km = AB So, ...

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Byju's Answer

Standard XIII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

An aeroplane ...

Question

An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt{3} k m$ above it, is observed at an elevation of $60^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $30^{\circ}$ , then the speed (in $k m / h r$ ) of the aeroplane, is :

1500

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1440

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750

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720

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Solution

The correct option is B $1440$

Let the speed of the plane be $s k m / h r$ ,
We know that,
$tan 30^{\circ} = \frac{B E}{C E} \Rightarrow C E = 3 k m tan 60^{\circ} = \frac{A D}{C D} = \frac{B E}{C D} \Rightarrow C D = 1 k m$
Now,
$D E = C E - C D = 2 k m = A B$
So,
$s \times \frac{5}{3600} = 2 \Rightarrow s = 1440 k m / h r$

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An aeroplane flying at a constant speed, parallel to the horizontal ground, √3 km above it, is observed at an elevation of 60∘ from a point on the ground. If, after five seconds, its elevation from the same point, is 30∘, then the speed (in km/hr) of the aeroplane, is :

The correct option is B 1440 Let the speed of the plane be s km/hr, We know that, tan30∘=BECE⇒CE=3 kmtan60∘=ADCD=BECD⇒CD=1 km Now, DE=CE−CD=2 km=AB So, s×53600=2⇒s=1440 km/hr