1. An aircraft flies round a triangular course A, B and C. From A to B is 200km on a bearing of 115° and from B to C is 150km on a bearing of 230°. How long is the course from C to A and on what bearing must the aircraft fly?
2. AT is a tangent to the circle ABCD. Angle BAC = 64° and CAT = 72°, calculate angle BCA and CDA
1. D = 200[115o] + 150[230o].
X = 200*Cos115 + 150*Cos230 = -181 km.
Y = 200*sin115 + 150*sin230 = 66.4 km.
D = Sqrt(X^2 + Y^2) = Distance from C to A.
Tan A=Y/X = 66.4/-181 = -0.36685
A = 20.1o N. of W. = 159.9o CCW. = Bearing.
To solve these problems, we can use some basic principles of geometry and trigonometry. Let's break down each problem step by step:
1. To find the length of the course from C to A and the bearing the aircraft must fly, we can use vector addition.
First, let's find the coordinates of each point:
Point A: We're given a distance of 200km on a bearing of 115°. To find the coordinates, we can use trigonometry. The horizontal component is given by 200 * cos(115°), and the vertical component is given by 200 * sin(115°). So the coordinates for A are (200 * cos(115°), 200 * sin(115°)).
Point B: We're given a distance of 150km on a bearing of 230°. Again, we can use trigonometry to find the coordinates. The horizontal component is given by 150 * cos(230°), and the vertical component is given by 150 * sin(230°). So the coordinates for B are (150 * cos(230°), 150 * sin(230°)).
Point C: Since we're looking for the course from C to A, we first need to find the coordinates of point C. We know that it is the starting point for the second leg of the journey, so the coordinates are the same as the coordinates for B.
To find the coordinates for C, we use (150 * cos(230°), 150 * sin(230°)).
Now that we have the coordinates for A and C, we can find the length of the course from C to A using the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Substituting the values, we get:
Distance = √((150 * cos(230°) - 200 * cos(115°))^2 + (150 * sin(230°) - 200 * sin(115°))^2)
Calculating this expression will give you the length of the course from C to A.
To find the bearing the aircraft must fly from C to A, we can use trigonometry again. The bearing is essentially the angle formed between the positive x-axis and the line connecting C to A. We can find this angle using the inverse tangent function:
Bearing = atan2((y2 - y1), (x2 - x1))
Substituting the values, we get:
Bearing = atan2(200 * sin(115°) - 150 * sin(230°), 200 * cos(115°) - 150 * cos(230°))
Calculating this expression will give you the bearing the aircraft must fly.
2. To calculate angle BCA and CDA, we can use some properties of tangents and triangles.
Let's look at triangle BCA first.
We know that the angle BAC is 64°, and CAT is 72°. Since AT is a tangent to the circle, it is perpendicular to the radius drawn at the point of tangency.
Thus, angle ACT is 90°.
Now, we can calculate angle BCA using the following equation:
Angle BCA = (Angle BAC + Angle CAT) - Angle ACT
Substituting the values, we get:
Angle BCA = (64° + 72°) - 90°
Calculating this expression will give you the value of angle BCA.
To find angle CDA, we can use the fact that the opposite angles in a cyclic quadrilateral add up to 180°. Since AT is a tangent, angles BAC and CAT are adjacent angles in the cyclic quadrilateral BCAT, and angles CDA and CAT are opposite angles.
Thus, angle CDA = 180° - Angle CAT
Substituting the value of Angle CAT, we get:
Angle CDA = 180° - 72°
Calculating this expression will give you the value of angle CDA.