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:

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

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