Centre of the hypotenuse

For the interior angles of the triangle ABC, alpha beta and gamma are in a ratio of 1: 2: 3. The longest side of the AB triangle is 30 cm long. Calculate the perimeter of the triangle CBS if S is the center of the side AB.

Correct answer:

x =  45 cm

Step-by-step explanation:

$$\begin{gathered} \alpha :\beta :\gamma =1:2:3 \\ d=180/(1+2+3)=30{}^{\circ } \\ \alpha =1\cdot d=1\cdot 30=30{}^{\circ } \\ \beta =2\cdot d=2\cdot 30=60{}^{\circ } \\ \gamma =3\cdot d=3\cdot 30=90{}^{\circ } \\ AB=30\text{cm} \\ \sin \alpha =\frac{CB}{AB} \\ \sin \beta =\frac{AC}{AB} \\ CB=AB\cdot \sin \alpha =AB\cdot \sin 30°=30\cdot \sin 30°=30\cdot 0.5=15\text{cm} \\ AC=AB\cdot \sin \beta =AB\cdot \sin 60°=30\cdot \sin 60°=30\cdot 0.866025=25.98076\text{cm} \\ \angle CBS:60°-60°-60° \\ CS=CB=15\text{cm} \\ x=CB+AB/2+CS=15+30/2+15=45\text{cm} \end{gathered}$$

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