Question

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.

Answer 1

See the attached file for solution. Thanks

Answer 2

Answer:

The distance between the two planes is  6250 m

To find:

The distance between the two planes vertically one above the other

Solution:

Let P is the observation point, which makes an elevation to the planes. Let x cm is the distance between the two planes and d be the distance between the observed point and from the ground level of the plane.

Let us consider flight B,

$\begin{array} { c } { \tan 30 ^ { \circ } = \frac { 3125 } { d } } \\\\ { d = \frac { 3125 } { \tan 30 ^ { \circ } } } \\\\ { d = \frac { 3125 } { \sqrt {\frac{1}{3} }} } \\\\ { d = 3125 \times \sqrt { 3 } } \end{array}$

Let us consider flight A,

$\begin{array} { c } { \tan 60 ^ { \circ } = \frac { 3125 + x } { d } } \\\\ { \tan 60 ^ { \circ } = \frac { 3125 + x } { 3125 \times \sqrt { 3 } } }\\ \\ { \sqrt { 3 } = \frac { 3125 + x } { 3125 \times \sqrt { 3 } } } \\\\ { 3 \times 3125 = 3125 + x } \\\\ { x = 9375 - 3125 = 6250 m } \end{array}$

Therefore, the distance between the flights A and B be 6250 m.