# Right triangle eq2

Find the lengths of the sides and the angles in the right triangle. Given area S = 210 and perimeter o = 70.

Correct result:

a =  21
b =  20
c =  29
A =  46.3972 °
B =  43.6028 °
C =  90 °

#### Solution:

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$b=2 \cdot \ S/a=2 \cdot \ 210/21=20 \ \\ b=a_{2}=20$
$c=\sqrt{ a^2+b^2 }=\sqrt{ 21^2+20^2 }=29$
$A=\frac{18{0}^{\circ }}{\pi }\cdot \mathrm{arcsin}\left(a\mathrm{/}c\right)=\frac{18{0}^{\circ }}{\pi }\cdot \mathrm{arcsin}\left(21\mathrm{/}29\right)=46.397{2}^{\circ }=4{6}^{\circ }2{3}^{\mathrm{\prime }}50\mathrm{"}$
$B=\frac{18{0}^{\circ }}{\pi }\cdot \mathrm{arcsin}\left(b\mathrm{/}c\right)=\frac{18{0}^{\circ }}{\pi }\cdot \mathrm{arcsin}\left(20\mathrm{/}29\right)=43.602{8}^{\circ }=4{3}^{\circ }3{6}^{\mathrm{\prime }}10\mathrm{"}$
$C=90=9{0}^{\circ }$

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