Question

An aircraft is flying at a height of 3400m above the ground. If the angle subtendat ground observation point by the aircraft positions 10 s apart is 30°. then the speed of the aircraft is
a) 19.63 ms'
b) 1963 ms'
c)108 ms'
d) 196.3ms'
e)10.8 ms'​

Answer 1

Answer:

e) 10.8 ms'​

Explanation:

In right angled triangle OAB,  tan30  

o

=  

x

h

​

 

We get  x=  

tan30  

o

 

h

​

 

∴  x=3400  

3

​

=5889 m

Time taken  t=10.0 s

Thus speed of the aircraft  v=  

t

x

​

 

∴  v=  

10.0

5889

​

=588.9  m/s

Answer 2

An aircraft is flying at a height of 3400m above the ground.

Given height of aircraft (h)= 3400. m

Angle subtended after 10 sec = 30°

Using properties of triangle

Height wil be base & distance

covered

$\tan(30) = \frac{perpendicular}{base } = \frac{x}{3400}$

$x = \frac{1}{ \sqrt{3} } \times 3400 = 1962.9$

Now we have x =1962.9

• x is distance covered by aircraft in
• $velocity = \frac{distance}{time}$
• $velocity = \frac{distance}{time}$$velocity = \frac{1962.9}{10} = 196.2$
• $velocity = \frac{distance}{time}$$velocity = \frac{1962.9}{10} = 196.2$Rounding off answer is 196.3m/s