Question #e4968

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Question #e4968

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Answer 1 · 2016-04-03T19:13:15

The line AH is tangential to the earth's surface, hence angle AHO will be $90^{0}$ $90^0$

Angle HAO will be $(90^{0} - {2.23}^{0})$ $(90^0-2.23^0)$ , so angle HOA will be ${2.23}^{0}$ $2.23^0$

We know that:
- line OH has length $r$ $r$ , and
- line OA has length $r + 4.83 k m$ $r+4.83km$

We can now use the cosine rule in relation to angle HOA
${cos 2.23}^{0} = \frac{r}{r + 4.83} = \frac{adjacent}{hypotenuse}$ $cos2.23^0=r/(r+4.83) ="adjacent"/"hypotenuse"$

rearranging to find $r$ $r$ :

$(r + 4.83) \cdot {cos 2.23}^{0} = r$ $(r+4.83)*cos2.23^0=r$

$r \cdot {cos 2.23}^{0} + 4.83 \cdot {cos 2.23}^{0} = r$ $r*cos2.23^0+4.83*cos2.23^0=r$

$4.83 \cdot {cos 2.23}^{0} = r - r \cdot {cos 2.23}^{0}$ $4.83*cos2.23^0=r -r*cos2.23^0$

$4.83 \cdot {cos 2.23}^{0} = r (1 - {cos 2.23}^{0})$ $4.83*cos2.23^0=r(1 -cos2.23^0)$

$\frac{4.83 \cdot {cos 2.23}^{0}}{1 - {cos 2.23}^{0}} = r$ $(4.83*cos2.23^0)/(1 -cos2.23^0)=r$

Using the above in a calculator gives r=6,373km

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Question #e4968

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