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:

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  a4  140 a3 + 5740 a2  58800 a + 176400 = a4+4202   140 a3 + 5740 a2  58800 a + 176400 = 4202  140 a3  5740 a2 + 58800 a = 0  140a25740a+58800=0  140 a25740 a+58800=0 140a25740a+58800=0 140=2257 5740=225741 58800=2435272 GCD(140,5740,58800)=2257=140  a241a+420=0  p=1;q=41;r=420 D=q24pr=41241420=1 D>0  a1,2=2pq±D=241±1 a1,2=241±1 a1,2=20.5±0.5 a1=21 a2=20  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
c=a2+b2=212+202=29
A=π180°arcsin(a/c)=π180°arcsin(21/29)=46.3972=46°2350"
B=π180°arcsin(b/c)=π180°arcsin(20/29)=43.6028=43°3610"
C=90=90



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