Question

A plane flying at a distance of 3000m above the ground. Firstly a person observe angle of elevation to be 60 degree than when it move forward it angle of elevation became 30 degree after 6 sec. Find the time take nby plane to travel that distance

Answer 1

Answer: The speed of the plane is 577.36 m/s.

Explanation:

Given that,

Height  h = 3000 m

Angle of elevation $\theta = 60^{0}$

Angle of elevation after some time $\theta = 30^{0}$

Time t = 6 sec

According to figure,

In Δ ABE, ∠ AEB = 90°

$\tan\theta = \dfrac{BE}{AE}$

$\tan 60^{0} = \dfrac{3000}{AE}$

$AE=\dfrac{3000}{\sqrt{3}}$

In Δ ACD, ∠ ADE = 90°

$\tan\theta = \dfrac{CD}{AD}$

$\tan 30^{0} = \dfrac{3000}{AD}$

$AD= 3000\sqrt{3}$

Now, BC= ED

ED = AD-AE

$ED= 3000\sqrt{3}-\dfrac{3000}{\sqrt{3}}$

$ED = \dfrac{6000}{\sqrt{3}}m$

So, plane travels $\dfrac{6000}{\sqrt{3}}m$ in a 6 sec

We know that,

$v = \dfrac{d}{t}$

$v = \dfrac{\dfrac{6000}{\sqrt{3}}}{6}$

$v = \dfrac{1000}{\sqrt{3}}$

$v = 577.36 m/s$

Hence, The speed of the plane is 577.36 m/s.