An airplane is flying parallel to the Earth’s surface at a speed of $175 \; m/s$ and at a height of $600 \; m$. The angle of elevation of the airplane from a point on the Earth’s surface is $37^\circ$. After what period of time does the angle of elevation increase to $53^\circ$? Give your answer correct to two decimal places. [Given: $\tan 53^\circ = 1.3270$, $\tan 37^\circ = 0.7536$]
$A_1 =$ Initial position of the airplane from point $O$
$A_2 =$ Final position of the airplane from point $O$
$A_1 P = A_2 Q = 600 \; m =$ Height of the airplane
In $\triangle OA_1P$, $\;$ $\dfrac{A_1P}{OP} = \tan 37^\circ$
$\implies$ $OP = \dfrac{A_1 P}{\tan 37^\circ} = \dfrac{600}{0.7536} = 796.1783 \; m$
In $\triangle OA_2Q$, $\;$ $\dfrac{A_2Q}{OQ} = \tan 53^\circ$
$\implies$ $OQ = \dfrac{A_2 Q}{\tan 53^\circ} = \dfrac{600}{1.3270} = 452.1477 \; m$
Now, $\;$ $A_1 A_2 = OP - OQ = 796.1783 - 452.1477 = 344.0306 \; m$
Given: Speed of plane $= 175 \; m/s$
i.e. $\;$ a distance of $175 \; m$ is traveled in $1 \; s$
$\therefore \;$ a distance of $344.0306 \; m$ is traveled in $\dfrac{344.0306}{175} = 1.9659 \; s$
$\therefore \;$ The time taken for the plane to go from $A_1$ to $A_2$ is $1.97 \; s$ (to 2 decimal places)