Question-169088

Question Number 169088 by cortano1 last updated on 24/Apr/22

Answered by mr W last updated on 24/Apr/22

a = 18

b = 15

c = 12

s = (18 + 15 + 12) / 2 = 22.5

Δ = \sqrt{22.5 \times 4.5 \times 7.5 \times 10.5} = \frac{135 \sqrt{7}}{4}

\sin B = \frac{2 Δ}{a c} = \frac{2 \times 135 \sqrt{7}}{4 \times 18 \times 12} = \frac{5 \sqrt{7}}{16} \Rightarrow \cos B = \frac{9}{16}

\sin C = \frac{2 Δ}{a b} = \frac{2 \times 135 \sqrt{7}}{4 \times 18 \times 15} = \frac{\sqrt{7}}{4} \Rightarrow \cos C = \frac{3}{4}

\tan \frac{B}{2} = \frac{\frac{5 \sqrt{7}}{16}}{1 + \frac{9}{16}} = \frac{\sqrt{7}}{5}

\tan \frac{C}{2} = \frac{\frac{\sqrt{7}}{4}}{1 + \frac{3}{4}} = \frac{\sqrt{7}}{7}

{(18 - \frac{3}{\tan \frac{B}{2}} - \frac{R}{\tan \frac{C}{2}})}^{2} = {(3 + R)}^{2} - {(3 - R)}^{2}

{(18 - \frac{15 \sqrt{7}}{7} - \sqrt{7} R)}^{2} = 12 R

{(18 - \frac{15 \sqrt{7}}{7})}^{2} - 18 (2 \sqrt{7} - 1) R + 7 R^{2} = 0

R = \frac{1}{7} [9 (2 \sqrt{7} - 1) - \sqrt{81 {(2 \sqrt{7} - 1)}^{2} - 7 {(18 - \frac{15 \sqrt{7}}{7})}^{2}}]

= \frac{9 (2 \sqrt{7} - 1) - 6 \sqrt{2 (3 \sqrt{7} - 2)}}{7} \approx 2.564

Commented by mr W last updated on 24/Apr/22

Commented by Tawa11 last updated on 24/Apr/22

Great sir

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