Circumscribed 6568

In a right triangle ABC with a right angle at the vertex C, it is given: a = 17cm, Vc = 8 cm. Calculate the length of the sides b, c, its area S, the perimeter o, the length of the radii of the circles of the triangle circumscribed by R and inscribed r and the magnitude of the angles alpha and beta.

Correct answer:

b =  9.0667 cm
c =  19.2667 cm
S =  77.0667 cm²
o =  45.3333 cm
A =  61.9275 °
B =  28.0725 °
r =  3.4 cm
R =  9.6333 cm

Step-by-step explanation:

$$\begin{gathered} a=17\text{cm} \\ h=8\text{cm} \\ {c}_1=\sqrt{a^2-h^2}=\sqrt{17^2-8^2}=15\text{cm} \\ {h}^2=c_1\cdot c_2 \\ {c}_2=h^2/c_1=8^2/15=\frac{64}{15}\doteq 4.2667\text{cm} \\ c=c_1+c_2=15+4.2667=\frac{289}{15}\doteq 19.2667 \\ b=\sqrt{c^2-a^2}=\sqrt{19.2667^2-17^2}=9.0667\text{cm} \end{gathered}$$

$c=19.2667=19.2667\text{cm}$

Try calculation via our triangle calculator.

$S=\frac{a\cdot b}{2}=\frac{17\cdot 9.0667}{2}=\frac{1156}{15}{\text{cm}}^2=77.0667{\text{cm}}^2$

$o=a+b+c=17+9.0667+19.2667=45.3333\text{cm}$

$A=\frac{180°}{\pi }\cdot \arcsin (a/c)=\frac{180°}{\pi }\cdot \arcsin (17/19.2667)=61.9275^{\circ }=61°55^{\prime }39"$

$B=\frac{180°}{\pi }\cdot \arcsin (b/c)=\frac{180°}{\pi }\cdot \arcsin (9.0667/19.2667)=28.0725^{\circ }=28°4^{\prime }21"$

$r=\frac{a+b-c}{2}=\frac{17+9.0667-19.2667}{2}=\frac{17}{5}\text{cm}=3.4\text{cm}$

$R=\frac{c}{2}=\frac{19.2667}{2}=\frac{289}{30}\text{cm}=9.6333\text{cm}$

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