A junior high school geometry question - Questions about secant
The radius of a circle is 15, the length of a chord line AB on the circle is 24, extend AB to C, BC=28, the center of the circle is O, find the length of CO.
solution: This question can be solved using three methods.
Method 1: Use Pythagorean theorem , as shown in the figure
Take the midpoint M of AB, then the triangle MOC is a right triangle,
But in the right triangle AMO, one hypotenuse AO=15, one right angled side AM=12, The ratio between the two,
AO/AM=15/12=5:4, according to the Pythagorean theorem, MO=3x3=9,
In the right triangle MOC, MC=12+28=40
Using the Pythagorean theorem again,
, therefore
CO =41
Method 2: Use the secant theorem of the circle
As shown in the figure, extend CO to point D on the circle. CO intersects the circle at point E.
Assume CO=x, then CE=x-15, CD=x+15 According to the secant theorem:
CB·CA=CE·CD
, that is, 28·52=(x-15)(x+15)
. Solve to get First, according to solution 1, since AM=12, AO=15, so MO=9,
, so cos A=AM/AO=12/15=4/5
in triangle ACO, AO=15, AC=52,
According to the Pythagorean theorem
solve CO=41
The sine theorem and the cosine theorem are effective tools for solving triangle problems. They used to be knowledge in junior high schools, but now they are all in high school. It is recommended to master them in advance. After all, they are very simple and easy to understand.
