An aircraft which is maintaining a constant altitude of 1000 meters is travelling directly away from an observer. At a particular instant the sngle of elevation is 52 degrees and 12 seconds later the angle of elevation is 16 degrees. Find the distance travelled in this time.
(Use sine rule)
HELP!!!!!!!!
Hmmm. I don't see an obvious application of the law of sines here. However, if we consider the distance directly to the plane, and call the first one x and the second one y, then
1000/x = sin52°
1000/y = sin16°
Then we want the distance z given by the law of cosines,
z^2 = x^2+y^2-2xy cos(52-16)°
Or,
z/sin36° = x/sin16° = y/sin128°
To find the distance traveled by the aircraft, we can use the sine rule. The sine rule states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant.
Let's consider the triangle formed by the observer, the aircraft, and the point where the aircraft was 12 seconds ago. Label the angle of elevation at the initial point as A, the distance traveled by the aircraft as x, and the distance between the observer and the initial position of the aircraft as y.
According to the problem, the angle of elevation at the initial point is 52 degrees, and after 12 seconds, the angle of elevation is 16 degrees.
Using the sine rule, we can write:
sin(A) / x = sin(16°) / y
And,
sin(180° - A) / (x + y) = sin(52°) / y
Now, we need to solve these equations to find the values of x and y.
To make the calculations easier, you can convert degrees to radians. To convert from degrees to radians, use the formula: radians = (degrees * π) / 180
Substituting the known values and converting degrees to radians:
sin(16°) / y = sin((52° * π) / 180) / y
sin(180° - A) / (x + y) = sin((52° * π) / 180) / y
Now, we can solve these equations to find the values of x and y.