$\left.\begin{array}{l}\text { [Inscribed angle, subtending the diameter]} \\ {\left[\begin{array}{l}\text { Auxiliary similar triangles }\end{array}\right]} \\ {[\text { Pythagorean Theorem (direct and inverse). }} \\ {[\text { Trapezoids (other). }}\end{array}\right]$

Given trapezoid $ABCD$, where angle $BAD$ is a right angle. A circle is constructed on side $AB$ as its diameter, intersecting diagonal $BD$ at point $M$. It is known that $AB=3$, $AD=4$, and $BC=1$. Find the angle $CAM$.
Consider the right triangle $A M K$, where $K$ is the intersection point of the diagonals of the trapezoid.

## Solution

Notice that $B D=5, A C^{2}=10$, and $\angle A M B=90^{\circ}$ (it subtends the diameter).

Since $A M \cdot B D=A B \cdot A D$, then $A M=A B \cdot A D / B D=12 / 5$.

Let $K$ be the intersection point of the diagonals of the trapezoid. Then $A K / K C=A D / B D=4$. Therefore, $A K=4 / 5 A C=\frac{4 \sqrt{10}}{5}$.

From the right triangle $A M K$, we find that $\cos \angle K A M=\frac{3}{\sqrt{10}}$. Therefore, $\sin \angle C A M=$

$\sin \angle K A M=\frac{1}{\sqrt{10}}$

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-14.jpg?height=403&width=738&top_left_y=-1&top_left_x=660)

## Answer

$\arcsin \frac{1}{\sqrt{10}}$.