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3 years ago

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The number of days in a month is measured in ones

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krok68 [10]3 years ago

4 0

The synodic month, or complete cycle of phases of the Moon as seen from Earth, averages 29.530588 mean solar days in length (i.e., 29 days 12 hours 44 minutes 3 seconds); because of perturbations in the Moon’s orbit, the lengths of all astronomical months vary slightly. The sidereal month is the time needed for the Moon to return to the same place against the background of the stars, 27.321661 days (i.e., 27 days 7 hours 43 minutes 12 seconds)

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What is the length of BC , rounded to the nearest tenth?

Arte-miy333 [17]

Step $1$

In the right triangle ADB

<u>Find the length of the segment AB</u>

Applying the Pythagorean Theorem

$AB^{2} =AD^{2}+BD^{2}$

we have

$AD=5\ units\\BD=12\ units$

substitute the values

$AB^{2}=5^{2}+12^{2}$

$AB^{2}=169$

$AB=13\ units$

Step $2$

In the right triangle ADB

<u>Find the cosine of the angle BAD</u>

we know that

$cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}$

Step $3$

In the right triangle ABC

<u>Find the length of the segment AC</u>

we know that

$cos(BAC)=cos (BAD)=\frac{5}{13}$

$cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{13}{AC}$

solve for AC

$AC=(13*13)/5=33.8\ units$

Step $4$

<u>Find the length of the segment DC</u>

we know that

$DC=AC-AD$

we have

$AC=33.8\ units$

$AD=5\ units$

substitute the values

$DC=33.8\ units-5\ units$

$DC=28.8\ units$

Step $5$

<u>Find the length of the segment BC</u>

In the right triangle BDC

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

we have

$BD=12\ units\\DC=28.8\ units$

substitute the values

$BC^{2}=12^{2}+28.8^{2}$

$BC^{2}=973.44$

$BC=31.2\ units$

therefore

<u>the answer is</u>

$BC=31.2\ units$

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