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Question Number 146776 by mathdanisur last updated on 15/Jul/21

Commented by mr W last updated on 15/Jul/21

$${AB}^2={20}^2-x^2={(20+5)}^2+{10}^2-2\times 25\times 10\cos C$$$${AD}^2=x^2=5^2+{10}^2-2\times 5\times 10\cos C$$$$\Rightarrow {20}^2-x^2={(20+5)}^2+{10}^2+5(x^2-5^2-{10}^2)$$$$\Rightarrow x^2=50$$$$\Rightarrow x=5\sqrt{2}$$

Commented by mathdanisur last updated on 15/Jul/21

$$thankyouSer$$

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