Question-164973

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

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

AC^2 =(2(√7))^2 +(3(√7))^2 −2(2(√7))(3(√7)) cos 120° AC^2 =28+63+42=133 AC=(√(133)) cos α=(((2(√7))^2 +((√(133)))^2 −(3(√7))^2 )/(2(2(√7))((√(133)))))=((7(√(19)))/(38)) sin α=((3(√(57)))/(38)) cos β=(((4)^2 +((√(133)))^2 −(9)^2 )/(2(4)((√(133)))))=((17(√(133)))/(266)) sin β=((9(√(399)))/(266)) BD^2 =(2(√7))^2 +(4)^2 −2(2(√7))(4)(((7(√(19)))/(38))×((17(√(133)))/(266))−((3(√(57)))/(38))×((9(√(399)))/(266))) BD^2 =28+16−16×(√7)×((√7)/(14))=36 ⇒?=BD=6

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

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