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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$b=2 \cdot \ S/a=2 \cdot \ 210/21=20 \ \\ b=a_{2}=20$

$c=\sqrt{ a^2+b^2 }=\sqrt{ 21^2+20^2 }=29$

$A={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arcsin (a\mathrm{/}c)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arcsin (21\mathrm{/}29)=46.3972^{\circ }=46^{\circ }23^{\mathrm{\prime }}50\mathrm{"}$

$B={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arcsin (b\mathrm{/}c)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arcsin (20\mathrm{/}29)=43.6028^{\circ }=43^{\circ }36^{\mathrm{\prime }}10\mathrm{"}$

$C=90=90^{\circ }$

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