Example 43. Let's find the area of a right-angled triangle if its hypotenuse is equal to $c$, and the sum of the sines of the acute angles is $q$.

Given (Fig. 47): $\triangle A B C, \widehat{C}=90^{\circ},|A B|=c, \sin \widehat{A}+\sin \widehat{B}=q$.

Find: $S_{\triangle A B C}$.
Solution. $\quad S_{\triangle A B C}=\frac{1}{2}|A C| \cdot|B C|$. But $\quad|A C|=c \sin \widehat{B}$, and $|B C|=c \sin \widehat{A}$. Then

$$
|A C|+|B C|=c q
$$

Moreover,

$$
|A C|^{2}+|B C|^{2}=c^{2}
$$

Raising both sides of equation (43.1) to the second power and subtracting equation (43.2) term by term, we find that

$$
2|A C| \cdot|B C|=c^{2}\left(q^{2}-1\right)
$$

from which

$$
S_{\triangle A B C}=\frac{1}{4} c^{2}\left(q^{2}-1\right)
$$

Investigation. Since $q=\sin \widehat{A}+\sin \widehat{B}=\sin \widehat{A}+$ $+\sin \left(90^{\circ}-\widehat{A}\right)=2 \sin 45^{\circ} \cos \left(\widehat{A}-45^{\circ}\right)=\sqrt{2} \cos \left(\widehat{A}-45^{\circ}\right)$, and $0<\cos \left(\widehat{A}-45^{\circ}\right) \leqslant 1$, then

$$
0<q \leqslant \sqrt{2}
$$

Furthermore, since $q=\sin \widehat{A}+\sin \widehat{B}=\frac{|B C|}{|A B|}+\frac{|A C|}{|A B|}=$

$$
=\frac{|B C|+|A C|}{|A B|} \text {, then }
$$

Solving the system composed of (43.3) and (43.4), we conclude that

$$
S_{\triangle A B C}=\frac{1}{4} c^{2}\left(q^{2}-1\right), \text { if } 1<q \leqslant \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-071.jpg?height=320&width=237&top_left_y=248&top_left_x=361)

Fig. 48