Question

The angle of elevation of a jet plane from A point on the ground is $60^o$. After a flight of 30 seconds, the angle of elevation changes to 30 degrees. if the jet plane is flying at a constant height of $3600\sqrt 3$ meter find the speed of a plane.

Answer 1

AC denotes the jet plane and BD denotes the height of the jet plane. We are interested to determine the speed of the plane that is DE = BC. D and E are the initial and final positions of the plane. The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level that is when we raise our head to look at the object.

Complete step-by-step answer:
Given: height of the jet plane = $3600\sqrt 3$m
Let D and E be the initial and final positions of the plane respectively. A is the point of observation and DB and EC are perpendiculars representing height.

Consider the triangle ABD
tan$60^o$ =$\sqrt 3$ ⇒ tan$60^o$ = $\dfrac{{BD}}{{AB}}$
Now, substitute the value of tan 60 degree
$\sqrt 3$ = $\dfrac{{3600\sqrt 3 }}{{AB}}$ ⇒ AB = 3600
Consider the triangle ACE
tan$30^o$= $\dfrac{{CE}}{{AC}}$
Now, substitute the value of tan 30°, tan $30^o$ = $\dfrac{1}{{\sqrt 3 }}$
$\dfrac{1}{{\sqrt 3 }}$ = $\dfrac{{3600\sqrt 3 }}{ AC \\\ \\\ }$ ⇒ AC = 10800 m
BC = AC – AB = 10800 − 3600
DE = BC = 7200 m
That is, the plane travels a distance of 7200 m in 30 seconds.
Therefore, speed of the plane = $\dfrac{{7200}}{{30}}$

Speed of the plane = 240 m

Note: Look at the figure carefully, there are two triangle angles ACE and angle ABD are alternate angles, and so are equal angle ACE = $30^o$ similarly angle ABD = $60^o$ and must know the value of tan $30^o$ and tan $60^o$.