Question

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

Answer 1

Step $1$

In the right triangle ADB

Find the length of the segment AB

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

Find the cosine of the angle BAD

we know that

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

Step $3$

In the right triangle ABC

Find the length of the segment AC

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 $4$

Find the length of the segment DC

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 $5$

Find the length of the segment BC

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

the answer is

$BC=31.2\ units$

Answer 2

The answer is 31.2 units