b^2\left[\frac{\pi }{4}-\:arcos\left(\frac{15\sqrt{2}}{2b}\right)\right]+\left(\frac{15\sqrt{2}}{2}\right)\sqrt{b^2-\frac{225}{2}}=101.25

asked by guest
on Mar 16, 2025 at 5:44 am



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Investigate the equation: b2(π4−arcos⁡(1522b))+1522b2−2252=101.25{b}^{2} \left(\frac{\pi}{4} - a r \cos\left( \frac{15 \sqrt{2}}{2 b} \right)\right) + \frac{15 \sqrt{2}}{2} \sqrt{{b}^{2} - \frac{225}{2}} = 101.25.

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