Question-164973

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

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

{A C}^{2} = {(2 \sqrt{7})}^{2} + {(3 \sqrt{7})}^{2} - 2 (2 \sqrt{7}) (3 \sqrt{7}) \cos 120 °

{A C}^{2} = 28 + 63 + 42 = 133

A C = \sqrt{133}

\cos α = \frac{{(2 \sqrt{7})}^{2} + {(\sqrt{133})}^{2} - {(3 \sqrt{7})}^{2}}{2 (2 \sqrt{7}) (\sqrt{133})} = \frac{7 \sqrt{19}}{38}

\sin α = \frac{3 \sqrt{57}}{38}

\cos β = \frac{{(4)}^{2} + {(\sqrt{133})}^{2} - {(9)}^{2}}{2 (4) (\sqrt{133})} = \frac{17 \sqrt{133}}{266}

\sin β = \frac{9 \sqrt{399}}{266}

{B D}^{2} = {(2 \sqrt{7})}^{2} + {(4)}^{2} - 2 (2 \sqrt{7}) (4) (\frac{7 \sqrt{19}}{38} \times \frac{17 \sqrt{133}}{266} - \frac{3 \sqrt{57}}{38} \times \frac{9 \sqrt{399}}{266})

{B D}^{2} = 28 + 16 - 16 \times \sqrt{7} \times \frac{\sqrt{7}}{14} = 36

\Rightarrow ? = B D = 6

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

Commented by mnjuly1970 last updated on 24/Jan/22

t h a n k s a l o t s i r

Commented by Tawa11 last updated on 24/Jan/22

Weldone sir

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