Angle LEBF

In the figure, we have a circle with points A, B, E, and F. AB is the diameter of the circle, and ABCD and CEF are straight lines. We are given that the angle LBE (LBEC) is 40° and the angle ZFC (ZFCD) is 150°.
To find the angle LEBF, we need to look at the angles formed by the intersecting lines. The angle LEBF is formed by the lines EB and FB.
First, we can use the fact that the angle in a semicircle is always a right angle. Since AB is the diameter, the angle AEB is a right angle (90°). Therefore, the angle LEB is complementary to 90° and can be calculated as follows:
LEB = 90° – LBEC LEB = 90° – 40° LEB = 50°
Next, we can use the property of angles on a straight line. The sum of the angles on a straight line is always 180°. Therefore, the angle CEB can be calculated as follows:
CEB = 180° – LBEC CEB = 180° – 40° CEB = 140°
Now, to find the angle LEBF, we need to consider the angle ZFC (ZFCD) as well. The angle ZFC is an external angle to the triangle CEB, and it is equal to the sum of the two remote interior angles, LEB and CEB. Therefore, we can calculate ZFC as follows:
ZFC = LEB CEB ZFC = 50° 140° ZFC = 190°
Finally, to find the angle LEBF, we can use the property of angles around a point. The sum of all angles around a point is always 360°. Therefore, we can calculate LEBF as follows:
LEBF = 360° – ZFC LEBF = 360° – 190° LEBF = 170°
So, the angle LEBF is 170°.