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: $\frac{tan(28)}{1} =\frac{BC}{85}$

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.

$\frac{tan(31)}{1} =\frac{BD}{85}\\ \\ BD=51.1$

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