The angle of elevation of a …

Let us suppose that P be the initial position of the plane and let Q be the final position of the plane respectively and suppose A be the point of observation. Let ABC be the horizontal line through A.
According to question, it is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.
{tex} \therefore \quad \angle P A B = 60 ^ { \circ } , \angle Q A B = 30 ^ { \circ }.{/tex}It is also given that PB = 1500{tex} \sqrt { 3 }{/tex} metres
Now, In {tex} \Delta A B P,{/tex} we have

{tex} \tan 60 ^ { \circ } = \frac { B P } { A B }{/tex}
{tex} \Rightarrow \quad \sqrt { 3 } = \frac { 1500 \sqrt { 3 } } { A B }{/tex}
{tex} \Rightarrow{/tex} AB = 1500 m
Again, In {tex} \Delta A C Q,{/tex}we have
{tex} \tan 30 ^ { \circ } = \frac { C Q } { A C }{/tex}
{tex} \Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { 1500 \sqrt { 3 } } { A C }{/tex}
{tex} \Rightarrow \quad A C = 1500 \times 3 = 4500 \mathrm { m }{/tex}
{tex} \therefore{/tex} PQ = BC = AC - AB = 4500 - 1500 = 3000 m
Therefore, the plane travels 3000 m in 30 seconds.
Hence, speed of plane is = {tex} \frac { 3000 } { 30 } = 100 \mathrm { m } / \mathrm { sec } {/tex}