Question

D. 5.9
Math Question No Guessing Please help

Answer 1

We will make use of our trigonometric functions in this problem.

`\tan(31^(\circ)) = \frac{\overline{DB}}{\overline{AB}}`

`\tan{31^(\circ)} = \frac{\overline{DB}}{85}`

`85 \cdot \tan({31^(\circ)}) = \overline{DB}`

Now that we know
`\overline{DB}` we need to find
`\overline{CB}`, since
`\overline{CD} = \overline{DB} - \overline{CB}`.

`\tan(28^(\circ)) = \frac{\overline{CB}}{\overline{AB}}`

`\tan(28^(\circ)) = \frac{\overline{CB}}{85}`

`85 \cdot \tan(28^(\circ)) = \overline{CB}`

Therefore, we can come to our final answer:

`\overline{CD} = \boxed{85\tan(31^(\circ)) - 85\tan(28^(\circ))} \approx \boxed{5.88}`

Answer 2

So for this, we will be using tangent (tan = opposite/adjacent) to solve for the sides BD and BC, then subtract the two sides to get CD.

Triangle ABC has ∠A = 28 degrees. Since we have the adjacent side (AB = 85) and we need to solve for the opposite side, we are using tangent. Our equation will look as such:
`(tan(28))/(1) =(BC)/(85)`

For this, just cross-multiply and your answer will be (Rounded to the tenths):
`BC=45.2`

Next, Its the same process with triangle ABD as ABC except that the angle is 31 degrees instead of 28.

`(tan(31))/(1) =(BD)/(85)\\ \\ BD=51.1`

Next, subtract the two sides together and your answer will be 5.9, or D.