## Task B-3.7.

The lengths of the sides of a cyclic quadrilateral $A B C D$ are $|A B|=3 \mathrm{~cm},|B C|=4 \mathrm{~cm},|C D|=5$ $\mathrm{cm}$ and $|D A|=6 \mathrm{~cm}$. Determine the cosines of the acute angles and the area of the quadrilateral $A B C D$.
## Solution.

Let's consider the sketch.

Let $\varangle B A D=\alpha, \varangle C B A=\beta, \varangle D C B=\gamma, \varangle A D C=\delta$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-20.jpg?height=534&width=551&top_left_y=670&top_left_x=661)

In a cyclic quadrilateral, the sum of opposite angles is $180^{\circ}$, so we have $\beta=180^{\circ}-\delta$ and $\gamma=180^{\circ}-\alpha$.

For their cosines, we have

$$
\begin{aligned}
& \cos \beta=\cos \left(180^{\circ}-\delta\right)=-\cos \delta \\
& \cos \gamma=\cos \left(180^{\circ}-\alpha\right)=-\cos \alpha
\end{aligned}
$$

(*) 1 point

Thus, we are looking for two adjacent angles with a positive cosine.

From triangle $A C D$ using the cosine rule, we get:

$$
\begin{gathered}
e^{2}=|A D|^{2}+|C D|^{2}-2|A D| \cdot|C D| \cdot \cos \delta \\
e^{2}=6^{2}+5^{2}-2 \cdot 6 \cdot 5 \cdot \cos \delta \\
e^{2}=36+25-60 \cdot \cos \delta \\
e^{2}=61-60 \cdot \cos \delta
\end{gathered}
$$

Similarly, from triangle $A B C$ using the cosine rule, we get:

$$
\begin{gathered}
e^{2}=|A B|^{2}+|B C|^{2}-2|A B| \cdot|B C| \cdot \cos \beta \\
e^{2}=3^{2}+4^{2}-2 \cdot 3 \cdot 4 \cdot \cos \beta \\
e^{2}=9+16-24 \cdot \cos \beta \\
e^{2}=25-24 \cdot \cos \beta
\end{gathered}
$$

and due to $(*)$ we have

$$
e^{2}=25-24 \cdot \cos \beta=25-24 \cdot \cos \left(180^{\circ}-\delta\right)=25+24 \cos \delta . \quad(* * *) \quad 1 \text { point }
$$

From the equality $(* *)$ and $(* * *)$ we get

$$
61-60 \cdot \cos \delta=25+24 \cos \delta
$$

from which it follows that $\cos \delta=\frac{3}{7}$.

1 point

Similarly, from triangles $A B D$ and $B C D$ using the cosine rule, we get

$$
6^{2}+3^{2}-2 \cdot 6 \cdot 3 \cos \alpha=4^{2}+5^{2}-2 \cdot 4 \cdot 5 \cos \left(180^{\circ}-\alpha\right)
$$

or $\cos \alpha=\frac{1}{19}$.

Thus, the required cosines are $\cos \alpha=\frac{1}{19}$ and $\cos \delta=\frac{3}{7}$. The remaining two angles have negative cosines due to $(*)$.

From $\cos \delta=\frac{3}{7}$ it follows that

$$
\sin \delta=\sqrt{1-\cos ^{2} \delta}=\sqrt{1-\frac{9}{49}}=\frac{2 \sqrt{10}}{7}
$$

The area of quadrilateral $A B C D$ is equal to the sum of the areas of triangles $A C D$ and $A B C$, so

$$
P_{A B C D}=P_{A C D}+P_{A B C}=\frac{1}{2} \cdot|A D| \cdot|C D| \cdot \sin \delta+\frac{1}{2} \cdot|A B| \cdot|B C| \cdot \sin \beta
$$

From $\sin \beta=\sin \left(180^{\circ}-\delta\right)=\sin \delta$, it follows that

$$
P_{A B C D}=\frac{1}{2} \cdot 6 \cdot 5 \cdot \frac{2 \sqrt{10}}{7}+\frac{1}{2} \cdot 3 \cdot 4 \cdot \frac{2 \sqrt{10}}{7}=\frac{30 \sqrt{10}}{7}+\frac{12 \sqrt{10}}{7}=6 \sqrt{10} \mathrm{~cm}^{2}
$$

1 point

Note: The area can also be calculated in the same way as the sum of the areas of triangles $A B D$ and $B C D$:

$$
\begin{gathered}
\left.\sin \gamma=\sin \left(180^{\circ}-\alpha\right)=\sin \alpha=\sqrt{1-\frac{1}{19^{2}}}\right)=\frac{6 \sqrt{10}}{19} \\
P_{A B C D}=P_{A B D}+P_{B C D}=\frac{1}{2} \cdot|A D| \cdot|A B| \cdot \sin \alpha+\frac{1}{2} \cdot|B C| \cdot|C D| \cdot \sin \gamma= \\
\frac{1}{2} \cdot 6 \cdot 3 \cdot \frac{6 \sqrt{10}}{19}+\frac{1}{2} \cdot 5 \cdot 4 \cdot \frac{6 \sqrt{10}}{19}=6 \sqrt{10} \mathrm{~cm}^{2}
\end{gathered}
$$

Alternatively, the student can apply Heron's formula for the area of a cyclic quadrilateral:

$$
P=\sqrt{(s-a)(s-b)(s-c)(s-d)}=\sqrt{6 \cdot 5 \cdot 4 \cdot 3}=6 \sqrt{10} \mathrm{~cm}^{2}
$$

where $s=9$ is the semiperimeter of quadrilateral $A B C D$.

A correctly calculated area in any way is worth three points.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - B variant
February 5, 2022.

## IF THE STUDENT HAS A DIFFERENT SOLUTION APPROACH, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.