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 answer:

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

Step-by-step explanation:

$$\begin{gathered} S=210 \\ o=70 \\ o=a+b+c \\ S=ab\mathrm{/}2 \\ o=a+b+\sqrt{a^2+b^2} \\ 2S=ab \\ b=2S\mathrm{/}a \\ 70=a+420\mathrm{/}a+\sqrt{a^2+(420\mathrm{/}a{)}^2} \\ 70a=a^2+420+\sqrt{a^4+420^2} \\ 70a-a^2-420=\sqrt{a^4+420^2} \\ {a}^4-140a^3+5740a^2-58800a+176400=a^4+420^2 \\ -140a^3+5740a^2-58800a+176400=420^2 \\ 140a^3-5740a^2+58800a=0 \\ 140a^2-5740a+58800=0 \\ 140a^2-5740a+58800=0 \\ 140a^2-5740a+58800=0 \\ p=140;q=-5740;r=58800 \\ D=q^2-4pr=5740^2-4\cdot 140\cdot 58800=19600 \\ D>0 \\ {a}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{5740\pm \sqrt{19600}}{280}} \\ {a}_{1,2}={\displaystyle \frac{5740\pm 140}{280}} \\ {a}_{1,2}=20.5\pm 0.5 \\ {a}_1=21 \\ {a}_2=20 \\ \text{ Factored form of the equation: } \\ 140(a-21)(a-20)=0 \\ a=a_1=21 \end{gathered}$$

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$$\begin{gathered} b=2\cdot S\mathrm{/}a=2\cdot 210\mathrm{/}21=20 \\ b=a_2=20 \end{gathered}$$

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

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

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

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

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