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Question Number 208581 by Tawa11 last updated on 18/Jun/24

Commented by mr W last updated on 18/Jun/24

Commented by Tawa11 last updated on 18/Jun/24

$$\mathrm{Thanks}\mathrm{sir}.$$

Answered by A5T last updated on 18/Jun/24

Commented by A5T last updated on 18/Jun/24

$$2^2-2\times 2\times 5cos\theta =+2\times 3\times 4cos\theta$$$$\Rightarrow cos\theta =\frac{1}{11}$$$$AB=\sqrt{2^2+5^2-2\times 2\times 5\times \frac{1}{11}}=\frac{\sqrt{3289}}{11}$$$$\frac{abc}{4R}=\frac{1}{2}\times 2\times 5\times \frac{\sqrt{120}}{11}\Rightarrow R=\frac{\sqrt{3289}}{2\sqrt{120}}$$

Commented by Tawa11 last updated on 18/Jun/24

$$\mathrm{Thanks}\mathrm{sir}$$

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