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 °

S = 210 o = 70 o = a + b + c S = a b / 2 o = a + b + a^{2} + b^{2} 2 S = a b b = 2 S / a 70 = a + 420 / a + a^{2} + (420 / a)^{2} 70 a = a^{2} + 420 + a^{4} + 42 0^{2} 70 a - a^{2} - 420 = a^{4} + 42 0^{2} a^{4} - 140 a^{3} + 5740 a^{2} - 58800 a + 176400 = a^{4} + 42 0^{2} - 140 a^{3} + 5740 a^{2} - 58800 a + 176400 = 42 0^{2} 140 a^{3} - 5740 a^{2} + 58800 a = 0 140 a^{2} - 5740 a + 58800 = 0 140 a^{2} - 5740 a + 58800 = 0 140 a^{2} - 5740 a + 58800 = 0 140 = 2^{2} \cdot 5 \cdot 7 5740 = 2^{2} \cdot 5 \cdot 7 \cdot 41 58800 = 2^{4} \cdot 3 \cdot 5^{2} \cdot 7^{2} GCD (140, 5740, 58800) = 2^{2} \cdot 5 \cdot 7 = 140 a^{2} - 41 a + 420 = 0 p = 1; q = - 41; r = 420 D = q^{2} - 4 p r = 4 1^{2} - 4 \cdot 1 \cdot 420 = 1 D > 0 a_{1, 2} = \frac{- q \pm D}{2 p} = \frac{4 1 \pm 1}{2} a_{1, 2} = \frac{4 1 \pm 1}{2} a_{1, 2} = 20.5 \pm 0.5 a_{1} = 21 a_{2} = 20 a = a_{1} = 21

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

c = a^{2} + b^{2} = 2 1^{2} + 2 0^{2} = 29

A = \frac{1 8 0 °}{π} \cdot arcsin (a / c) = \frac{1 8 0 °}{π} \cdot arcsin (21 / 29) = 46.397 2^{\circ} = 46 ° 2 3^{'} 50 "

B = \frac{1 8 0 °}{π} \cdot arcsin (b / c) = \frac{1 8 0 °}{π} \cdot arcsin (20 / 29) = 43.602 8^{\circ} = 43 ° 3 6^{'} 10 "

C = 90 = 9 0^{\circ}

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