# Points M and N lie on the AC side of triangle ABC at distances of 9 and 11 respectively from vertex A. Find the

Points M and N lie on the AC side of triangle ABC at distances of 9 and 11 respectively from vertex A. Find the radius of the circle passing through points M and N and touching the ray AB if cos∠BAC = √11 / 6 In addition, we designate key points and draw segments, as shown in the figure.
By the tangent and secant theorem, we find AD.
AD2 = AM * AN = 9 * 11 = 99
By the cosine theorem, we find DM:
DM2 = AD2 + AM2-2 * AD * AM * cos∠BAC = (√99) ^ 2 + 9 ^ 2-2 * √99 * 9 * √11 / 6 = 99 + 81-18 * 3√11 * √ 11/6 = 180-3 * 3 * 11 = 180-99 = 81
DM = 9
Since DM = AM = 9, it means the triangle ADM is isosceles.
Therefore, by the property of an isosceles triangle ∠BAM = ∠ADM
According to the fourth property of the angles associated with the circle, ∠ADM is equal to half the degree measure of the arc DM.
∠DOM is the central angle, therefore it is equal to the degree measure of the arc DM, i.e. twice as many as ∠ADM.
Consider the DOM triangle.
Since OD = OM = R, this triangle is isosceles.
Draw the height of the OE, as shown in the figure.
By the property of an isosceles triangle: the height of the OE is also a bisector and a median.
Therefore, ∠DOE = ∠DOM / 2 = ∠ADM = ∠BAC
We get that cos∠DOE = cos∠BAC = √11 / 6
sin∠DOE = √1-cos2∠DOE = √1- (√11 / 6) ^ 2 = √1-11 / 36 = √25 / 36 = 5/6 (basic trigonometric formula)
DE = DM / 2 = 9/2 = 4.5 (because OE is the median).
sin∠DOE = DE / DO (by definition).
5/6 = 4,5 / DO
DO = 4.5 / (5/6) = 4.5 * 6/5 = 5.4 = R Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.