Question-76087

Question Number 76087 by aliesam last updated on 23/Dec/19

Answered by mr W last updated on 23/Dec/19

Commented by mr W last updated on 23/Dec/19

A B = \sqrt{2^{2} + 6^{2}} = 2 \sqrt{10}

A C = \sqrt{10}

A O = R = \sqrt{50}

\cos ∠ O A C = \frac{\sqrt{10}}{\sqrt{50}} = \frac{1}{\sqrt{5}}

\cos ∠ D A C = \frac{6}{2 \sqrt{10}} = \frac{3}{\sqrt{10}}

\cos α = \cos (∠ O A C - ∠ D A C)

= \frac{1}{\sqrt{5}} \times \frac{3}{\sqrt{10}} + \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{10}} = \frac{\sqrt{2}}{2}

x^{2} = {O D}^{2} = {(\sqrt{50})}^{2} + 6^{2} - 2 \sqrt{50} \times 6 \times \cos α

= 50 + 36 - 2 \sqrt{50} \times 6 \times \frac{\sqrt{2}}{2} = 26

\Rightarrow x = \sqrt{26}

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