What is it. Please do you carry the negative or the square first.
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In the parallelogram ABCD, join BD.
Consider the triangle Δ ABD.
It is given that AB > AD.
Since, in a triangle, angle opposite to longer side is larger, we have,
∠ ADB > ∠ ABD. --- (1)
Also, AB || DC and BD is a transversal.
Therefore,
∠ ABD = ∠ BDC
Substitute in (1), we get,
∠ ADB > ∠ BDC.
We have to use the rule of cosx° to solve this problem. Attached is a diagram of the navigator's course for the plane. It is similar to the shape of a triangle. We know the plane is 300 miles from its destination, so that will be one of the sides. On the current course, it is 325 miles from its destination, so that will be another one of the sides. The last side is 125 because that is the distance between the destination and the anticipated arrival. Cosx° is what we are looking for.
To find how many degrees off course the plane is, we must use the rules of Cosx°, which is shown in the attached image.
The plane is approximately 23° off course.
To find how many degrees off course the plane is, we must use the rules of Cosx°, which is shown in the attached image.
The plane is approximately 23° off course.