10.425 A chord $AB$ is drawn through a point $M$ located on the diameter of a circle with a radius of 4 cm, forming an angle of $30^{\circ}$ with the diameter. A chord $BC$ is drawn through point $B$, perpendicular to the given diameter. Find the area of triangle $ABC$, if $AM: MB=2: 3$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-256.jpg?height=307&width=350&top_left_y=164&top_left_x=101)

Let $M D \perp B C$. According to the problem, segment $M D$ lies on the diameter of the given circle. Therefore, $M D$ bisects chord $B C$ $\Rightarrow$ $\Rightarrow M B=M C \Rightarrow \angle B M C=60^{\circ}$ (by the condition, $\left.\angle B M D=30^{\circ}\right) \Rightarrow \triangle M B C$ is equilateral.

Let $M B=3 x$, then $B C=3 x, A M=2 x$ (by the condition). By the Law of Sines, $A C=2 \cdot 4 \sin \angle B=4 \sqrt{3}$.

By the Law of Cosines, $A C^{2}=48=(5 x)^{2}+(3 x)^{2}-30 x^{2} \cos 60^{\circ} \Rightarrow x^{2}=\frac{48}{19}$, $S_{A B C}=\frac{1}{2} A B \cdot B C \cdot \sin 60^{\circ}=\frac{15 x^{2} \sqrt{3}}{4}=\frac{180 \sqrt{3}}{19}$.

Answer: $\frac{180 \sqrt{3}}{19} \mathrm{~cm}^{2}$.

## PROBLEMS IN STEREOMETRY