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Question Number 164973 by mnjuly1970 last updated on 24/Jan/22

Answered by mr W last updated on 24/Jan/22

$${AC}^2={\left(2\sqrt{7}\right)}^2+{\left(3\sqrt{7}\right)}^2-2\left(2\sqrt{7}\right)\left(3\sqrt{7}\right)\cos 120°$$$${AC}^2=28+63+42=133$$$$AC=\sqrt{133}$$$$\cos \alpha =\frac{{\left(2\sqrt{7}\right)}^2+{\left(\sqrt{133}\right)}^2-{\left(3\sqrt{7}\right)}^2}{2\left(2\sqrt{7}\right)\left(\sqrt{133}\right)}=\frac{7\sqrt{19}}{38}$$$$\sin \alpha =\frac{3\sqrt{57}}{38}$$$$\cos \beta =\frac{{\left(4\right)}^2+{\left(\sqrt{133}\right)}^2-{\left(9\right)}^2}{2\left(4\right)\left(\sqrt{133}\right)}=\frac{17\sqrt{133}}{266}$$$$\sin \beta =\frac{9\sqrt{399}}{266}$$$${BD}^2={\left(2\sqrt{7}\right)}^2+{\left(4\right)}^2-2\left(2\sqrt{7}\right)\left(4\right)(\frac{7\sqrt{19}}{38}\times \frac{17\sqrt{133}}{266}-\frac{3\sqrt{57}}{38}\times \frac{9\sqrt{399}}{266})$$$${BD}^2=28+16-16\times \sqrt{7}\times \frac{\sqrt{7}}{14}=36$$$$\Rightarrow ?=BD=6$$

Commented by mr W last updated on 24/Jan/22

Commented by mnjuly1970 last updated on 24/Jan/22

$$thanksalotsir$$

Commented by Tawa11 last updated on 24/Jan/22

$$\mathrm{Weldone}\mathrm{sir}$$

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