Question

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

13.0 units
28.8 units
31.2 units
33.8 units

Answer 1

Option C on edge 2020 :)

Answer 2

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)=(adjacent\ side )/(hypotenuse)=(AD)/(AB)=(5)/(13)`

Step
`3`

In the right triangle ABC

Find the length of the segment AC

we know that

`cos(BAC)=cos (BAD)=(5)/(13)`

`cos(BAC)=(adjacent\ side )/(hypotenuse)=(AB)/(AC)`

`(5)/(13)=(AB)/(AC)`

`(5)/(13)=(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`