a girl walks 45m in the direction 50degree from a point Q to the point X , she then walks 24m in the direction 140degree from X to a point Y . How far is she from Q? Solve with a well drawn diagram
I cannot draw a diagram here, but I will describe how to draw one and solve the problem.
1. Draw a point labeled Q.
2. From point Q, draw a line at an angle of 50 degrees from the horizontal, and make it 45m in length. Label the endpoint of this line as X.
3. From point X, draw another line at an angle of 140 degrees from the horizontal, and make it 24m in length. Label the endpoint of this line as Y.
Now we have a triangle QXY, and we want to find the length of the side QY.
4. Find the angle QYX: Since angle QYX and angle QXY are supplementary, we get angle QYX = 180 - 50 = 130 degrees.
5. Find angle YQX: Since the angles of a triangle add up to 180 degrees, we get angle YQX = 180 - 130 - 140 = -90 degrees. Since the angle is negative, it means it's a reflex angle, so the actual angle we want is 360 - 90 = 270 degrees.
Now we can use the law of cosines to find the length of the side QY:
QY^2 = QX^2 + XY^2 - 2 * QX * XY * cos(angle QYX)
QY^2 = (45m)^2 + (24m)^2 - 2 * 45m * 24m * cos(130°)
QY^2 ≈ 2025 + 576 - 2 * 45 * 24 * (-0.6428)
QY^2 ≈ 2601 + 1296.384
QY ≈ √(3897.384)
QY ≈ 62.4 m (approximately)
So, the girl is approximately 62.4 meters away from point Q.