Hypotenuse and center

Point S is the center of the hypotenuse AB of the right triangle ABC. Calculate the area of triangle ABC if the line on the hypotenuse is 0.2 dm long and if angle ∢ACS is 30°.

Correct answer:

S =  3.4641 cm²

Step-by-step explanation:

$$\begin{gathered} t=0.2\text{dm}\to \text{cm}=0.2\cdot 10\text{cm}=2\text{cm} \\ \angle ACS=30{}^{\circ } \\ \angle SCB=90-\angle ACS=90-30=60{}^{\circ } \\ \frac{\sin ACS}{\sin \alpha }=\frac{c/2}{t} \\ \frac{\sin SCB}{\sin \beta }=\frac{c/2}{t} \\ \beta =90°-\alpha \\ \frac{\sin ACS}{\sin \alpha }=\frac{\sin SCB}{\sin (90°-\alpha )} \\ \sin ACS\cdot \sin (90°-\alpha )=\sin SCB\cdot \sin \alpha \\ \sin (\alpha -\beta )=\sin \alpha .\cos \beta -\cos \alpha .\sin \beta \\ \sin ACS\cdot (\sin 90°\cos \alpha -\cos 90°\sin \alpha )=\sin SCB\cdot \sin \alpha \\ \sin ACS\cdot \cos \alpha =\sin SCB\cdot \sin \alpha \\ \sin ACS/\sin SCB=\sin \alpha /\cos \alpha \\ \tan \alpha =\sin ACS/\sin SCB \\ \alpha =\frac{180°}{\pi }\cdot \arctan (\sin ACS/\sin SCB)=\frac{180°}{\pi }\cdot \arctan (\sin 30°/\sin 60°)=\frac{180°}{\pi }\cdot \arctan (0.5/0.866025)=30 \\ \beta =90-\alpha =90-30=60{}^{\circ } \\ c=2\cdot t\cdot \frac{\sin ACS}{\sin \alpha }=2\cdot t\cdot \frac{\sin 30°}{\sin 30°}=2\cdot 2\cdot \frac{\sin 30°}{\sin 30°}=2\cdot 2\cdot \frac{0.5}{0.5}=4\text{cm} \\ a=c\cdot \sin \alpha =c\cdot \sin 30°=4\cdot \sin 30°=4\cdot 0.5=2\text{cm} \\ b=c\cdot \sin \beta =c\cdot \sin 60°=4\cdot \sin 60°=4\cdot 0.866025=3.4641\text{cm} \\ S=\frac{a\cdot b}{2}=\frac{2\cdot 3.4641}{2}=2\sqrt{3}=3.4641{\text{cm}}^2 \end{gathered}$$

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