This is a geometrical question?

Question

A CHORD OF A CIRCLE IS EQUAL TO THE RADIUS OF THE CIRCLE. FIND THE ANGLE SUBTENDED BY THE CHORD AT A POINT ON THE MINOR ARC AND ALSO AT A POINT ON THE MAJOR ARC.

Answer 1

Let θ be the central angle to the point on the minor arc. Then the chord
x = 2Rsin(θ/2)
and the chord
y = 2Rsin((60 - θ)/2)
Let φ be the angle subtended by the chord of length R intersecting x and y. Then
φ =acos((x^2 + y^2 - R^2)/(2xy)
φ =acos(((2Rsin(θ/2))^2 + (2Rsin((60 - θ)/2))^2 - R^2)/(2*2Rsin(θ/2)2Rsin((60 - θ)/2))
φ =acos((4R^2sin^2(θ/2) + 4R^2sin^2((60 - θ)/2) - R^2)/(8R^2sin(θ/2)sin((60 - θ)/2))
φ =acos((4sin^2(θ/2) + 4sin^2((60 - θ)/2) - 1)/(8sin(θ/2)sin((60 - θ)/2))
φ =acos((sin(θ/2)/2sin((60 - θ)/2) + sin((60 - θ)/2)/8sin(θ/2) - 1/(8sin(θ/2)sin((60 - θ)/2))

This can still be simplified considerably. The angle from a point on the major arc is handled similarly.

Answer 2

If you join the center to the two points where the chord cuts the circle, you would get an equilateral triangle since the chord length is the same as the radius. Thus the angle formed at the center is 60*.
Since the angle at the center is twice the angle at the circumference, the angle subtended by the chord in the major arc would be 30*. As the two opposite angles subtended by a chord add up to 180*, the angle in the minor arc would be 150*

Answer 3

Let x represent the radius of circle O. Draw a radius from the center of circle O to a point A on the circle. From there, draw a chord of length x to a point B on the circle, then a radius from B back to O. Now you have an equilateral triangle, with each side of length x and each angle equal to 60 degrees.

The minor arc thus subtends 60 degrees and the major arc is 300 degrees.

Answer 4

If cord=radius The triangle formed by the extremes of the cord and the center is equilateral.So the center angle is 60 degrees.
From a point of the mayor arc the angle subtended is 60/2=30deg

From a point of the minor arc the angle (360-60)/2 =150 deg

Answer 5

Check it out. If you get it down visually, the problem is cake. The chord is equal to the radius, right? Draw it out. Notice how wherever you put the chord, you can draw 2 radii out to where the chord intersects the circle to form an equilateral triangle.

A property of the equilateral triangle is that each angle is 60 degrees. lucky for you, this is also the angle of the minor arc. subtract that from 360 to find the value of the major arc.

Answer 6

5

Answer 7

45

Answer 8

I don't even know what half of those words mean.