In the given figure, $ABCD$ is a square with a side length of one unit, and the triangle $\triangle BCE$ is equilateral. The point $M$ is the midpoint of the segment $CE$, the segment $DN$ is perpendicular to $BM$, and the segment $BM$ is perpendicular to $CE$.

(a) Calculate the lengths of the sides of the triangle $\triangle DBN$.

(b) Use part (a) to calculate the cosine, sine, and tangent of the angles $15^{\circ}$ and $75^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-032.jpg?height=320&width=508&top_left_y=434&top_left_x=1322)
$ Since $D B$ is the diagonal of a square with a side length of $1 \mathrm{~cm}$, the Pythagorean Theorem ensures that $D B^{2}=1^{2}+1^{2}=2$, or in other words, $D B=\sqrt{2}$. Let's recall that

$$
\begin{array}{ll}
\cos 60^{\circ}=\sin 30^{\circ}=\frac{1}{2} ; & \tan 60^{\circ}=\frac{\sin 60^{\circ}}{\cos 60^{\circ}}=\sqrt{3} \\
\sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} ; & \tan 30^{\circ}=\frac{\sin 30^{\circ}}{\cos 30^{\circ}}=\frac{\sqrt{3}}{3}
\end{array}
$$

(a) The triangle $\triangle B C E$ is equilateral, so its internal angles measure $60^{\circ}$. From this information, we obtain the angles marked in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-108.jpg?height=437&width=716&top_left_y=1261&top_left_x=750)

In the triangle $\triangle C D F$ we have $\sin 60^{\circ}=\frac{C D}{D F}=\frac{1}{D F}$. Since $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$, it follows that $\frac{1}{D F}=\frac{\sqrt{3}}{2}$ and, therefore, $D F=\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$. Still in the triangle $\triangle C D F$ we have $\cos 60^{\circ}=\frac{C F}{D F}=\frac{C F}{2 \sqrt{3} / 3}$. But $\cos 60^{\circ}=\frac{1}{2}$, from which we conclude that $\frac{1}{2}=\frac{C F}{2 \sqrt{3} / 3}$, or in other words, $C F=\frac{\sqrt{3}}{3}$. It follows that $B F=1-C F=1-\frac{\sqrt{3}}{3}$. We have, now,

$$
\frac{1}{2}=\sin 30^{\circ}=\frac{F N}{B F}=\frac{F N}{1-\sqrt{3} / 3}, \quad \text { so that } \quad F N=\frac{3-\sqrt{3}}{6}
$$

and

$$
\frac{\sqrt{3}}{2}=\cos 30^{\circ}=\frac{B N}{B F}=\frac{B N}{1-\frac{\sqrt{3}}{3}}, \quad \text { so that } \quad B N=\frac{\sqrt{3}-1}{2}
$$

Thus, we calculate the three sides of the triangle $\triangle D B N$, as follows.

- $D B=\sqrt{2}$;
- $D N=D F+F N=\frac{2 \sqrt{3}}{3}+\frac{3-\sqrt{3}}{6}=\frac{1+\sqrt{3}}{2}$;
- $B N=\frac{\sqrt{3}-1}{2}$.
(b) In the triangle $\triangle D B N$ we have $D \widehat{B} N=45^{\circ}+30^{\circ}=75^{\circ}$, from which we conclude that $B \widehat{D} N=15^{\circ}$. Thus, we have

$$
\cos 75^{\circ}=\frac{B N}{D B}=\frac{\frac{\sqrt{3}-1}{2}}{\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}
$$

and

$$
\cos 15^{\circ}=\frac{D N}{D B}=\frac{\frac{1+\sqrt{3}}{2}}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}
$$

It remains to observe that $\sin 75^{\circ}=\cos 15^{\circ}$, $\sin 15^{\circ}=\cos 75^{\circ}$, and that

$\tan 75^{\circ}=\frac{\sin 75^{\circ}}{\cos 75^{\circ}}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}$ and $\tan 15^{\circ}=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-109.jpg?height=526&width=406&top_left_y=445&top_left_x=1476)