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Question Number 210508 by hardmath last updated on 11/Aug/24

Answered by mr W last updated on 11/Aug/24

Commented by mr W last updated on 11/Aug/24

$$\mathrm{\Delta }ABE=\frac{c^2\tan B}{2}$$$$x=\frac{c}{\cos B}-a$$$$y=\frac{x}{\tan B}$$$$\mathrm{\Delta }CDE=\frac{xy}{2}=\frac{1}{2\tan B}{(\frac{c}{\cos B}-a)}^2$$$$\left[ABCD\right]=\mathrm{\Delta }ABC-\mathrm{\Delta }CDE$$$$=\frac{c^2\tan B}{2}-\frac{1}{2\tan B}{(\frac{c}{\cos B}-a)}^2$$$$=\frac{c^2\tan B}{2}-(\frac{1}{2\tan B}+\frac{\tan B}{2})c^2-\frac{a^2}{2\tan B}+\frac{ac}{\sin B}$$$$=\frac{ac}{\sin B}-\frac{a^2+c^2}{2\tan B}✓$$

Commented by hardmath last updated on 11/Aug/24

$$$$Perfect solution as always, thanks dear professor

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