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3 years ago

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Math 1314 lab module 2

1 answer:

denpristay [2]3 years ago

8 0

What, is there a question or no?

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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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Answer:

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Step-by-step explanation:

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