An aeroplane at an altitude of 200m observes the angles of depression of opposite points on the two banks of a river to be 45 degree and 60 degree. Find in metres, the width of river.
Let H be the altitude (which you already know), and D be the horizontal distance from the plane to the nearest bank of the river. Let w be the width of the river.
H/D = tan 60
H/(D + w) = tan 45
Not that the second equation implies that
H = D + w
and the first equation implies that
H = sqrt3 * D
Solve for w in terms of H, using substitution to eliminate D.
H = H/sqrt3 + w
H(1 - 1/sqrt3) = w
To find the width of the river, we can use trigonometry.
Let's label the points as follows:
A - position of the airplane
P - point on the first bank of the river
Q - point on the second bank of the river
R - point directly below the airplane on the river
Let's denote the width of the river as x.
From point A, the angle of depression to point P is 45 degrees, and the angle of depression to point Q is 60 degrees. This means that angles APQ and ARQ are right angles, as shown below:
P
/|
/ |
x / | 200m
/ |
/ |
/ R
/ |
/_______|
A x Q
In right triangle APQ, we can use the tangent function to relate the angle of depression to the width of the river:
tan(45°) = x / 200
1 = x / 200
x = 200 meters
Therefore, the width of the river is 200 meters.