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Question Number 189270 by Rupesh123 last updated on 14/Mar/23

Commented by maths_plus last updated on 14/Mar/23

$$\mathrm{good}$$

Answered by HeferH last updated on 14/Mar/23

$${\mathrm{x}}^2-{\mathrm{a}}^2={\mathrm{b}}^2-{(\mathrm{a}-\mathrm{b})}^2$$$${\mathrm{x}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-({\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab})$$$${\mathrm{x}}^2=2\mathrm{ab}$$

Commented by manxsol last updated on 14/Mar/23

$$excelent$$

Answered by manxsol last updated on 14/Mar/23

$$x^2=b^2+b^2-2b.bcos(180-\theta )$$$$x^2=2b^2+2b^{2}cos\theta$$$$cos\theta =\frac{a-b}{b}$$$$x^2=2b^2(1+\frac{a-b}{b})$$$$x^2=2b^2\frac{a}{b}$$$$x^2=2ab$$

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