Question

Fine length of BC on the following photo.

Answer 1

`BC=4√(5)\ units`

Explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

`AC^2=AD^2+DC^2`

substitute the given values

`AC^2=16^2+8^2`

`AC^2=320`

`AC=√(320)\ units`

simplify

`AC=8√(5)\ units`

step 2

In the right triangle ACD

Find the cosine of angle CAD

`cos(\angle CAD)=(AD)/(AC)`

substitute the given values

`cos(\angle CAD)=(16)/(8√(5))`

`cos(\angle CAD)=(2)/(√(5))` ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

`cos(\angle BAC)=(AC)/(AB)`

substitute the given values

`cos(\angle BAC)=(8√(5))/(16+x)` ----> equation B

step 4

Find the value of x

In this problem

`\angle CAD=\angle BAC` ----> is the same angle

so

equate equation A and equation B

`(8√(5))/(16+x)=(2)/(√(5))`

solve for x

Multiply in cross

`(8√(5))(√(5))=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units`

`DB=4\ units`

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

`BC^2=DC^2+DB^2`

substitute the given values

`BC^2=8^2+4^2`

`BC^2=80`

`BC=√(80)\ units`

simplify

`BC=4√(5)\ units`