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:

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=21

Our quadratic equation calculator calculates it.


Try calculation via our triangle calculator.

b=2 S/a=2 210/21=20 b=a2=20
c=a2+b2=212+202=29
A=180πarcsin(a/c)=180πarcsin(21/29)=46.3972=462350"
B=180πarcsin(b/c)=180πarcsin(20/29)=43.6028=433610"
C=90=90



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