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Question Number 192828 by Ari last updated on 28/May/23

Commented by Ari last updated on 28/May/23

Can we find the side x in the problem given in the figure?

Answered by MM42 last updated on 28/May/23

$$cosE=\frac{25+64-49}{2\times 5\times 8}=\frac{40}{80}=\frac{1}{2}$$$$x^2=64+16-2\times 8\times 4\times \frac{1}{2}=48$$$$\Rightarrow x=4\sqrt{3}✓$$

Answered by HeferH last updated on 29/May/23

Commented by HeferH last updated on 29/May/23

$$SayAC=13k;5k=\frac{35}{13};8k=\frac{56}{13}$$$$BisectorofAEC=\sqrt{40-\frac{35\cdot 56}{169}}=\frac{\sqrt{169\cdot 40-49\cdot 40}}{13}$$$$=\frac{40\sqrt{3}}{13}$$$$AD=\frac{40\sqrt{3}}{13}\cdot \frac{13}{5}=8\sqrt{3}$$$$congruenttriangles\Rightarrow x=\frac{8\sqrt{3}}{2}=4\sqrt{3}$$

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