Question-164712

Question Number 164712 by cortano1 last updated on 21/Jan/22

Answered by mr W last updated on 21/Jan/22

Commented by mr W last updated on 21/Jan/22

R^{2} = {(b + c)}^{2} + b^{2}

R^{2} = {(b + 2 c)}^{2} + c^{2}

{(b + 2 c)}^{2} + c^{2} = {(b + c)}^{2} + b^{2}

4 c^{2} + 2 b c - b^{2} = 0

\Rightarrow c = \frac{(\sqrt{5} - 1) b}{4}

R = \sqrt{{(b + \frac{(\sqrt{5} - 1) b}{4})}^{2} + b^{2}} = \frac{b \sqrt{6 \sqrt{5} + 30}}{4}

a = R - b

\frac{a}{b} = \frac{R}{b} - 1 = \frac{\sqrt{6 \sqrt{5} + 30}}{4} - 1 ✓

Commented by Tawa11 last updated on 21/Jan/22

Great sir

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