Question-161811

Question Number 161811 by mnjuly1970 last updated on 22/Dec/21

Answered by mnjuly1970 last updated on 22/Dec/21

D E = \sqrt{68}

68 = 64 + 100 - 160 c o s (E \overset{\land}{A D} = α)

c o s (α) = \frac{96}{160} = \frac{3}{5}

c o s (\frac{α}{2}) = \frac{8}{x} \Rightarrow \frac{1 + \frac{3}{5}}{2} = \frac{64}{x^{2}}

\frac{8}{10} = \frac{64}{x^{2}} \Rightarrow x^{2} = 80 \Rightarrow x = 4 \sqrt{5}

Answered by Rasheed.Sindhi last updated on 22/Dec/21

AB = 8

AE = \sqrt{8^{2} + 6^{2}} = 10

∠ BAE = θ

\cos θ = \frac{AB}{AE} = \frac{8}{10} = \frac{4}{5}

θ = \cos^{- 1} (\frac{4}{5})

∠ FAD = \frac{90 - θ}{2}

AD = 8

\cos ∠ FAD = \cos \frac{90 - θ}{2} = \frac{AD}{AF} = \frac{8}{x}

x = \frac{8}{\cos \frac{90 - θ}{2}} = \frac{8}{\cos (\frac{90 - \cos^{- 1} (\frac{4}{5})}{2})}

x = \frac{8}{\cos (\frac{90 - 36.870}{2})} \approx 8.94

Answered by mr W last updated on 22/Dec/21

\cos 2 θ = \frac{6}{10} = \frac{3}{5}

\cos θ = \sqrt{\frac{1 + \cos 2 θ}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \frac{2}{\sqrt{5}}

\Rightarrow x = \frac{8}{\cos θ} = \frac{8}{\frac{2}{\sqrt{5}}} = 4 \sqrt{5}

Commented by mnjuly1970 last updated on 23/Dec/21

v e r y e x c e l l e n t s i r W

Commented by mr W last updated on 23/Dec/21

Commented by Tawa11 last updated on 22/Dec/21

great sir

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