May be that someone considers my question very trivial but...
... I'd like to know if it is possible to demonstrate that $\displaystyle CAB = \frac{COB}{2}$ without using trigonometry...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
May be that someone considers my question very trivial but...
... I'd like to know if it is possible to demonstrate that $\displaystyle CAB = \frac{COB}{2}$ without using trigonometry...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Do you know that the length of the arc of a circle that an angle subtends is proportional to the measure of the angle times the radius? In fact, radian measure is defined to be that arc length divided by the radius. The arc length is the measure of the angle, in radians, multiplied by the radius. The arclength is the measure of the angle, in degrees, multiplied by the radius also multiplied by [tex]\frac{\pi}{180}[/itex]. But in either case, the arclength is proportional to angle times the radius. And that means that the angle is proportional to arclength divided by the radius. As long as the arclength is constant, the angle is inversely proportional to the radius.
Here, both angles subtend the same arc (from B to C). But the "radius" for one angle is the radius of the circle while for the other it is the diameter.
Hello, chisigma!
I'd like to know if it is possible to demonstrate that: $\displaystyle \angle A \,=\, \tfrac{1}{2}(\angle COB)$ without using trigonometry
$\displaystyle OA = OC$ . Both are radii.
$\displaystyle \Delta AOC$ is isosceles: .$\displaystyle \angle A = \angle C$
$\displaystyle \angle COB$ is an exterior angle of $\displaystyle \Delta AOC $.
Hence, $\displaystyle \angle COB \:=\:\angle A + \angle C \:=\:2\angle A$ .**
Therefore: .$\displaystyle \angle A \:=\:\tfrac{1}{2}(\angle COB)$
**
An exterior angle of a triangle is equal to the sum of the two nonadjacent angles.
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