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 =  46.397 °
B =  43.603 °
C =  90 °

Solution:

S=210 o=70  o=a+b+c S=ab/2  o=a+b+a2+b2 2S=ab  b=2S/a  70=a+420/a+a2+(420/a)2 70a=a2+420+a4+4202 70aa2420=a4+4202 a4140 a3+5740 a258800 a+176400=a4+4202 140 a3+5740 a258800 a+176400=4202 140 a35740 a2+58800 a=0  140a25740a+58800=0  140 a25740 a+58800=0 140a25740a+58800=0  p=140;q=5740;r=58800 D=q24pr=57402414058800=19600 D>0  a1,2=q±D2p=5740±19600280 a1,2=5740±140280 a1,2=20.5±0.5 a1=21 a2=20   Factored form of the equation:  140(a21)(a20)=0  a=a1=21S=210 \ \\ o=70 \ \\ \ \\ o=a+b+c \ \\ S=ab/2 \ \\ \ \\ o=a+b+\sqrt{ a^2+b^2 } \ \\ 2S=ab \ \\ \ \\ b=2S/a \ \\ \ \\ 70=a+420/a+\sqrt{ a^2+(420/a)^2 } \ \\ 70a=a^2+420+\sqrt{ a^4+420^2 } \ \\ 70a-a^2-420=\sqrt{ a^4+420^2 } \ \\ a^4 - 140 \ a^3 + 5740 \ a^2 - 58800 \ a + 176400=a^4+420^2 \ \\ - 140 \ a^3 + 5740 \ a^2 - 58800 \ a + 176400=420^2 \ \\ 140 \ a^3 - 5740 \ a^2 + 58800 \ a=0 \ \\ \ \\ 140 a^2 - 5740 a + 58800=0 \ \\ \ \\ 140 \ a^2 - 5740 \ a + 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}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 5740 \pm \sqrt{ 19600 } }{ 280 } \ \\ a_{1,2}=\dfrac{ 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

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b=2 S/a=2 210/21=20 b=a2=20
c=a2+b2=212+202=29
A=180πarcsin(a/c)=180πarcsin(21/29)=46.397=462350"A=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(a/c)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(21/29)=46.397 ^\circ =46^\circ 23'50"
B=180πarcsin(b/c)=180πarcsin(20/29)=43.603=433610"B=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(b/c)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(20/29)=43.603 ^\circ =43^\circ 36'10"
C=90=90



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