An aeroplane is flying at 3400m height above the ground. If the time taken when the angle subtended between the two positions of aircraft is 30° is 10s . Calculate the speed of the aircraft. (Observation point is on the vertical line, which is mid between positions of aircraft, if tan15° =0.2679 ]

Question

An aeroplane is flying at 3400m height above the ground. If the time taken when the angle subtended between the two positions of aircraft is 30° is 10s . Calculate the speed of the aircraft. (Observation point is on the vertical line, which is mid between positions of aircraft, if tan15° =0.2679 ] (A) 262.4ms- 1 (B) 131.2ms- 1 (C) 91.1ms- 1 (D) 182.2ms- 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-jhntyqitrtqzwacu.jpg" /> tan15° =(x/3400) ⇒ x=3400× 0.2679=910.86m Speed of aircraft =(2 x/10) =(2 × 910 . 86/10)=182.2ms- 1

$262.4 m s^{- 1}$

$131.2 m s^{- 1}$

$91.1 m s^{- 1}$

$182.2 m s^{- 1}$

$t an 1 5^{\circ} = \frac{x}{3400}$
$\Rightarrow x = 3400 \times 0.2679 = 910.86 m$
Speed of aircraft $= \frac{2 x}{10}$
$= \frac{2 \times 910.86}{10} = 182.2 m s^{- 1}$

262.4ms−1

131.2ms−1

91.1ms−1

182.2ms−1

tan15∘=3400x​ ⇒x=3400×0.2679=910.86m Speed of aircraft =102x​ =102×910.86​=182.2ms−1

$262.4 m s^{- 1}$

$131.2 m s^{- 1}$

$91.1 m s^{- 1}$

$182.2 m s^{- 1}$

$t an 1 5^{\circ} = \frac{x}{3400}$
$\Rightarrow x = 3400 \times 0.2679 = 910.86 m$
Speed of aircraft $= \frac{2 x}{10}$
$= \frac{2 \times 910.86}{10} = 182.2 m s^{- 1}$